EvaluatingLimits(Limit Laws)

ThE LIMIt OF A FUNCTIONEVALUATING LIMITS – THE LIMIT LAWS

Let's quickly reflect on what we covered in the previous section.

We tried to evaluate limits using numerical and graphical methods, but we saw that these methods could only be used to

make approximations. We also found that there are a number of pitfalls to guessing the values of limits, which usually

leads to making wrong guesses.

In this section, we will use foolproof methods to compute the EXACT values of limits. We will also examine

➔ how to combine the various laws to evaluate seemingly complicated limits, and

➔ Theorems that ease simplification of limits

We start with the limit laws. The textbook I'm using highlights eleven limit laws. The first five deal with basic limit

operations: addition, subtraction, multiplication, division and multiplication by a constant. The first five laws are stated

below.

Note that all eleven laws are valid if and only if the limits

exist, and also if c is a constant:

f(x) limx→a

and g(x) limx→a

[f(x) + g(x)]limx→a

= f(x) limx→a

+ g(x) limx→a

Law 1 – Addition Law:

Law 2 – Difference Law:

[f(x) – g(x)]limx→a

= f(x) limx→a

– g(x) limx→a

Law 3 – Constant Multiple Law:

f(x) limx→a

c=[c f(x)] limx→a

Law 4 – Product Law:

[f(x) g(x)] limx→a

= f(x) limx→a × g(x) lim

x→a

Law 5 – Quotient/Division Law:

f(x)

g(x)limx→a

=f(x) lim

x→a

g(x) limx→a

[On the condition that the denominator IS NOT ZERO]

The sixth law is the power law, which is derived from the product law, by putting g(x) = f(x) and using the product law

repeatedly:

Laws 7 and 8 are critically important, if we are to apply any of the limit laws:

Law 9 is obtained from Law 6 by putting f(x) = x and applying Law 8.

From Law 6,

If we put f(x) = x, we have

Applying Law 8 gives

The next law is the Root Law:

Finally, we have the eleventh law, which is basically a slight modification of law 10. Specifically, this law is obtained from

law 10 by putting f(x) = x and applying law 8.

From Law 10, we know that

Putting f(x) = x gives

Law 6 – Power Law:

=[f(x)]n limx→a f(x) lim

x→a

n

c = climx→a

Law 7

x = alimx→a

Law 8

=[f(x)]n limx→a f(x) lim

x→a

n

=[x]n limx→a

x limx→a

n

xn = anlimx→a

Law 9

√f(x) limx→a

n= f(x) lim

x→a√n

Law 10

Where n is a positive integer. We use this law with the assumption that the limit is greater than zero, if n is aneven number.

√f(x) limx→a

n= f(x) lim

x→a√n

From Law 8,

Therefore,

For the rest of this section, we'll use these 11 laws in computing the exact values of limits. In applying any of these 11

limit laws, there's one general rule that applies to all.

If f is a polynomial/rational function, and a is in the domain of f, then,

Functions that satisfy Law 12 are said to be continuous. Thus, their limits can be evaluated by using direct substitution.

Bear in mind that there are exceptions, as we'll see later in this section.

Example 1Estimate the value of the limit

(a) by evaluating g(x) = 5x2 – 2x + 3 for values of x that approach 4.

(b) from a graph of g.

(c) by using the appropriate limit laws. Justify each step.

Solutionx g(x) x g(x)

3.5 57.25 4.5 95.25

3.7 64.05 4.3 86.85

3.9 71.25 4.1 78.85

3.99 74.62 4.01 75.38

3.995 74.81 4.001 75.04

3.999 74.96 4.0005 75.01

3.9995 74.98 4.0001 75.00

3.9999 74.99 4.00001 75.00

√x limx→a

n= x lim

x→a√n

x = alimx→a

limx→a √x

n= √a

n

Law 11

f(x) = f(a)limx→a

Law 12

(5x2 – 2x + 3)limx→4

The question asks that we use numerical evidence to guess the valueof the limit.

As usual, we draw up a table of values for g(x) using values of x thatapproach 4 from both sides.

The table clearly shows that g(x) approaches 75 as x gets closer andcloser to 4 from either side. Thus, using this table, we assume that

Therefore, numerical evidence shows that the value of the limit is 75.

(5x2 – 2x + 3) = 75limx→4

Next, we'll use graphical evidence to guess the value of the limit.

If we graph the function g(x) = 5x2 – 2x + 3 in the viewing rectangle [–6, 6] by [–4, 85], we get this:

From the graph, it is clear that g(x) → 75 as x → 4 from the left. Thus,

Likewise, g(x) → 75 as x → 4 from the right. So,

Since the left and right hand limits exist and are equal, it is safe to guess that

OK, Enough with the guessing. Let's do some real math. This time, we'll apply the limit laws; no graphs, no tables. Here

goes:

Using the addition and subtraction laws (laws 1 and 2),

You can see that this step is pretty much like expanding an algebraic expression. Next, we deal with the three limits on

the RHS of (1) above. Now study the following steps carefully, as we combine the limit laws:

g(x) → 75 as x → 4 from both sides

(5x2 – 2x + 3) = 75limX→4–

(5x2 – 2x + 3) = 75limX→4+

(5x2 – 2x + 3) = 75limx→4

(5x2 – 2x + 3)limx→4

= 5x2limx→4

– 2xlimx→4

+ 3limx→4 (1)

5x2limx→4

– 2xlimx→4

+ 3limx→4

= Law 3x2limx→4

– xlimx→4

+ 3limx→4

5 2

Now that the limits have been reduced to their simplest possible forms, the next step is to evaluate them via direct

substitution:

At this point, we can categorically say that

Example 2Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).

SolutionFrom the question, we see that

f(x) = (x3 + 2)(x2 – 5x)

There are two ways of evaluating this limit:

● Since the function is a product, we can start by breaking the limit into a product of two limits. From the product

law,

On the other hand,

● We can simply expand f, and evaluate:

Either way, we get the same answer. Let's use the first method:

Try using the second method. Here's a hint: use laws 1, 2, 3, 8, and 9.

Bear in mind, that your answer MUST BE –174.

x2limx→4

– xlimx→4

+ 3limx→4

5 2 = 5(42) – 2(4) + 3

= 80 – 8 + 3

= 75

(5x2 – 2x + 3) = 75limx→4

(x3 + 2)(x2 – 5x)limx→3

(x3 + 2)limx→3

(x2 – 5x)limx→3

= × (x3 + 2)(x2 – 5x)limx→3

(x3 + 2)(x2 – 5x)limx→3

= (x5 – 5x4 + 2x2 – 10x)limx→3

(x3 + 2)limx→3

(x2 – 5x)limx→3

= × (x3 + 2)(x2 – 5x)limx→3

= x3limx→3

+ 2limx→3

× x2limx→3

– xlimx→3

5

(Law 4)

(Laws 1, 2, 3)

= [(3)3 + 2] × [(3)2 – 5(3)]

= [27 + 2] × [9 – 15] = – 174

(Laws 7, 8, 9)

Note:

When you're dealing with polynomials, you can always verify your answer by evaluating f at a; that is, by computing

f(a). This is called direct substitution. It also works for rational functions on ONE CONDITION: THAT THE

DENOMINATYOR DOES NOT EVALUATE TO ZERO).


SolutionGenerally, understanding the kind of function you're dealing with helps you decide which limit law to use first. In this

case, it is obvious that we're dealing with a quotient function. Hence, we apply the quotient law first:

Then, we use the power law:

Using Laws 1, 2 and 3, we have

Using direct substitution, and laws 7, 8, and 9, we have

Therefore,

=

(x4 + x2 – 6)

(x4 + 2x + 3)limx→1

2

(x4 + x2 – 6)

(x4 + 2x + 3)limx→1

2 (x4 + x2 – 6)limx→1

(x4 + 2x + 3)limx→1

2

(x4 + x2 – 6)limx→1

(x4 + 2x + 3)limx→1

2

= (x4 + x2 – 6)lim

x→1

(x4 + 2x + 3)limx→1

2

2

(x4 + x2 – 6)limx→1

(x4 + 2x + 3)limx→1

2

2=

x4limx→1 + x2lim

x→1– 6lim

x→1

2

x4limx→1

+ xlimx→1

2 + 3limx→1

2

x4limx→1 + x2lim

x→1 – 6limx→1

2

x4limx→1

+ xlimx→1

2 + 3limx→1

2

=((1)4 + (1)2 – 6)2

((1)4 + 2(1) + 3)2

=(–4)2

(6)2

=4

9

(x4 + x2 – 6)

(x4 + 2x + 3)limx→1

2

=4

9


SolutionHere, the function is f(t) = (t + 1)9 (t2 – 1).

One way of solving this limit is by expanding f, but the drawback is that f will be too long when expanded, and let's not

forget that that there is a slight chance of making a mistake during the expansion,

An alternative method is to split the limit into two, since f is a product (like example 2):

Then,

Using Laws 7, 8 and 9, we have

Therefore,


SolutionSince f(u) = u4 + 3u + 6, then Law 10 (the Root Law) gives

We then use the addition law, together with the constant multiple law to simplify f:

(t + 1)9 (t2 – 1)limt→ –2

(t + 1)9 (t2 – 1)limt→ –2

= (t + 1)9limt→ –2 × (t2 – 1)lim

t→ –2(Product Law)

(t + 1)9limt→ –2 × (t2 – 1)lim

t→ –2= (t + 1)lim

t→ –2 × (t2 – 1)limt→ –2

9

(Power Law)

= tlimt→ –2

+ 1limt→ –2

× t2limt→ –2

– 1lim t→ –2

9

(Addition & Subtraction Laws)

tlimt→ –2

+ 1limt→ –2

× t2limt→ –2

– 1lim t→ –2

9

= [(–2) + 1]9 × [(–2)2 – 1]

= – 3

(t + 1)9 (t2 – 1)limt→ –2

= – 3

u4 + 3u + 6limu→ –2 √

u4 + 3u + 6limu→ –2 √ = [u4 + 3u + 6]lim

u→ –2 √

[u4 + 3u + 6]limu→ –2 √ = (u4)lim

u→ –2 √ + (3u)limu→ –2 + (6)lim

u→ –2

Evaluating the limits gives

Hence,

Example 6 What is wrong with the equation?

In view of part (a), explain why the equation

is correct.

SolutionClearly, there is nothing wrong with the equation

because, if we evaluate the expression on the left, we have

Hence, the equation is correct. For this very reason, the equation

is also correct. Bear in mind that,although (ii) is correct, the limit on the left, that is

DOES NOT EXIST. This is because the denominator would evaluate to zero using either the limit laws or direct

substitution. But if the function is simplified, we get the limit on the right side of (ii) as shown from (i).

We can prove that (ii) is correct in two ways:

➔ By simplifying the function itself:

(u4)limu→ –2 √ + (3u)lim

u→ –2 + (6)limu→ –2 = √(–2)4 + 3(–2) + 6

= √16 – 6 + 6 = 4

u4 + 3u + 6limu→ –2 √ = 4

x2 + x – 6

x – 2 = x + 3

limx→2

x2 + x – 6

x – 2 = lim

x→2(x + 3)

(a)

(b)

x2 + x – 6

x – 2 = x + 3

x2 + x – 6

x – 2 = x + 3

(x + 3)(x – 2)

x – 2 = (i)

limx→2

x2 + x – 6

x – 2 = lim

x→2(x + 3) (ii)

limx→2

x2 + x – 6

x – 2

=limx→2

x2 + x – 6

x – 2

(x + 3)(x – 2)

(x – 2)limx→2

= limx→2

(x + 3)

➔ OR by using the limit laws to simplify the limit entirely:

Simplifying the numerator gives

Canceling out equal limits results in

Either way, the equation is correct.

Example 7Evaluate the limit, if it exists.

SolutionTo start with, we use direct substitution; that is, we put x = –3. If we do this, we find that the denominator equals zero.

To this end, we conclude that the limit does not exist.

Then again, there's a possibility we are jumping to conclusions here. The truth is, we are.

Here's one rule you should STRICTLY adhere to when evaluating the limit of a rational function: TRY TO SIMPLIFY THE

NUMERATOR OR DENOMINATOR WHEREVER POSSIBLE. In some cases, it is possible to simplify both numerator and

denominator, so that you'll end up canceling out equal values. This is what we did in example 6.

In this case, the function CANNOT be simplified any further. Hence, we can officially say that the limit DOES NOT

EXIST.

For cases like this, one way of verifying your result is by graphing the function. Since a limit does not exist when x = –3,

you should expect to see some kind of discontinuity.

If you look at the graph carefully, you'll agree that the limit is actually infinite (both positive and negative). Thus, the

actual answer should be

limx→2

x2 + x – 6

x – 2 =

limx→2 (x2 + x – 6)

limx→2

(x – 2)

(Quotient Law)

limx→2 (x + 3)(x – 2)

limx→2

(x – 2)

=

limx→2

(x + 3) limx→2

(x – 2)

limx→2

(x – 2)

(Product Law)

limx→2

(x + 3) limx→2

(x – 2)

limx→2

(x – 2)

= limx→2

(x + 3)

x2 – x + 12

x + 3lim

x→ –3

x2 – x + 12

x + 3lim

x→ –3 = ± ∞


SolutionWe start by simplifying the denominator:

Using direct substitution,

On the other hand, we can use the limit laws:

which equals

Thus,

x + 2

x2 – x – 6lim

x→ –2

x + 2

x2 – x – 6lim

x→ –2 =

x + 2

(x + 2)(x – 3)lim

x→ –2 =

1

(x – 3)lim

x→ –2

1

(x – 3)lim

x→ –2 =

1

(– 2 – 3)lim

x→ –2 =

1

–5lim

x→ –2 =

1

–5

1

(x – 3)lim

x→ –2 =

limx→ –2

(x – 3)

limx→ –2

1

=

limx→ –2

1

limx→ –2

x – limx→ –2

3

1

(– 2 – 3)lim

x→ –2 =

1

–5lim

x→ –2 =

1

–5

x + 2

x2 – x – 6lim

x→ –2 =

1

–5


SolutionHere, we see that direct substitution is not an option, for obvious reasons. Neither can we use the quotient law at this

stage. Therefore, we can try to evaluate the limit, by simplifying it first. In this case, we see that both numerator and

denominator can be factorized:

Now we can use the limit laws:

Then, using laws 7 and 8, we have

Therefore,


SolutionLike example 9, using direct substitution or the quotient rule at this stage is not a good idea. The function needs to be

simplified first. We start by expanding the numerator:

x2 + x – 2

x2 – 3x + 2limx→1

x2 + x – 2

x2 – 3x + 2limx→1 =

(x + 2)(x – 1)

(x – 2)(x – 1)limx→1 =

x + 2

x – 2limx→1

limx→1

(x – 2)

limx→1 (x + 2)

x + 2

x – 2limx→1 = (Quotient Law)

limx→1

(x – 2)

limx→1 (x + 2)

=

limx→1

x limx→1

2+

limx→1

x limx→1

2–

(Addition & Subtraction Laws)

1 + 2

1 – 2

limx→1

x limx→1

2+

limx→1

x limx→1

2–

= = – 3

x2 + x – 2

x2 – 3x + 2limx→1

= – 3

(h – 5)2 – 25

hlimh→0

You can see that we have reduced

Now we can use the limit laws to evaluate the limit

Therefore,


SolutionWe need to simplify both numerator and denominator of the function:


Therefore,

Here, I used direct substitution to evaluate the limit, after reducing the function to a simpler form. Now, try using the

limit laws to evaluate the limit.

One thing I've noticed is that, anytime you evaluate a limit, you'll always end up using laws 7, 8 and/or 9 (usually in the

final stage).

(h – 5)2 – 25

hlimh→0

=(h2 – 10h + 25) – 25

hlimh→0

=h2 – 10h + 25 – 25

hlimh→0

=h2 – 10h

hlimh→0

=h(h – 10)

hlimh→0

= limh→0

(h – 10)

(h – 5)2 – 25

hlimh→0

to limh→0

(h – 10)

limh→0

(h – 10) = limh→0

h limh→0

10

=

– (Subtraction Law)

0 – 10 (Laws 7 and 8)

= – 10

(h – 5)2 – 25

hlimh→0

= – 10

x3 – 1

x2 – 1limx→1

x3 – 1

x2 – 1limx→1

= (x – 1)(x2 + x + 1)

(x – 1)(x + 1)limx→1

= x2 + x + 1

x + 1limx→1

x2 + x + 1

x + 1limx→1

=(1)2 + (1) + 1

(1) + 1=

3

2

x3 – 1

x2 – 1limx→1

= 3

2


SolutionExpanding the numerator gives


Again, I have used direct substitution to evaluate the limit after reducing it to a simpler form (just like example 11).

Here's another task: use the limit laws to evaluate the limit. Justify each step by indicating the appropriate limit laws

you applied.


SolutionThere are two possible approaches to solving this limit:

➔ Expressing the numerator as a difference of squares, and canceling out the common factors:

OR

➔ Rationalize the denominator:

(1 + h)4 – 1

hlimh→0

(1 + h)4 – 1

hlimh→0 =

(1 + 4h + 6h2 + 4h3 + h4) – 1

hlimh→0

=4h + 6h2 + 4h3 + h4 – 1 + 1

hlimh→0

= 4h + 6h2 + 4h3 + h4

hlimh→0

= h(4 + 6h + 4h2 + h3)

hlimh→0

= limh→0

(4 + 6h + 4h2 + h3)

limh→0

(4 + 6h + 4h2 + h3) = 4 + 6(0) + 4(0)2 + (0)3 = 4

9 – t

3 – √t limt→9

9 – t

3 – √t limt→9

=(3)2 – (√t)2

3 – √t limt→9

=(3 – √t)(3 + √t)

3 – √t limt→9

= limt→9

(3 + √t)

9 – t

3 – √t limt→9

=9 – t

3 – √t limt→9

3 + √t

3 + √t × =

(9 – t)(3 + √t)

(3 – √t )(3 + √t)limt→9

= (9 – t)(3 + √t)

(9 – t)limt→9

= limt→9

(3 + √t)

Either way, we end up with one limit:

which can easily be evaluated. Using the limit laws:

Therefore,


SolutionUnderstand this: when evaluating the limit of a rational function, there is often the need to simplify it. The aim of

simplifying the limit is to reduce it to a relatively simpler form, or perhaps its simplest possible form so that the

appropriate limit laws can be applied. Here, we factorize both numerator and denominator:

At this point, it will do absolutely no harm to perform direct substitution:

Try using the limit laws to evaluate the limit.


SolutionWe cannot use direct substitution, nor the quotient law, for obvious reasons. Neither can we simplify the numerator by

factorization. So, how do we simplify this limit?

limt→9

(3 + √t)

limt→9

(3 + √t) = limt→9

3 + limt→9

√t (Addition Law)

= limt→9

3 + √limt→9

t (Root Law)

= 3 + √9 (Laws 7 and 8)

= 6

9 – t

3 – √t limt→9

= limt→9

(3 + √t) = 6

t2 + t – 6

t2 – 4 lim t→2

t2 + t – 6

t2 – 4 lim t→2

= (t + 3)(t – 2)

(t + 2)(t – 2)lim t→2

= (t + 3)

(t + 2)lim t→2

(t + 3)

(t + 2)lim t→2

=2 + 3

2 + 2=

54

√2 – t – √2

tlim t→0

When attempting to evaluate the limit of a rational function whose numerator or denominator (or perhaps both) is

irrational, one good approach is rationalizing the irrational expression first. This process should simplify the function

considerably. This is demonstrated in example 13. In this case, we rationalize the numerator:

which gives

After simplifying the numerator, we end up with

Now that the limit is simplified considerably, we can use the limit laws to evaluate it. Thus,

If you choose to rationalize the answer, then you'd have

Thus,

√2 – t – √2 t

lim t→0

× √2 – t + √2

√2 – t + √2

lim t→0

(2 – t) + (√2 × √2 – t) – (√2 × √2 – t) – 2

t(√2 – t + √2 )= lim

t→0 (2 – t) + (√2(2 – t)) – (√2(2 – t)) – 2

t(√2 – t + √2 )

lim t→0

2 – t – 2

t(√2 – t + √2 )= lim

t→0 – t

t(√2 – t + √2 )= lim

t→0 – 1

(√2 – t + √2 )

lim t→0

– 1

(√2 – t + √2 )=

lim t→0

–1

lim t→0

(√2 – t + √2 )

(Quotient Law)

=

lim t→0

–1

lim t→0

√2 – t + lim t→0

√2

(Addition Law)

=

lim t→0

–1

+ lim t→0

2√lim t→0

(2 – t)√(Root Law)

– 1

√2 – 0 + √2 (Laws 7 & 8)=

= – 1

√2 + √2 =

– 1

2√2

– 1

2√2 ×

2√2

2√2 =

– 2√2

8 =

– √2

4

√2 – t – √2

tlim t→0

= lim t→0

– 1

(√2 – t + √2 )=

– 1

2√2 =

– √2

4


SolutionAs usual, we simplify the function. We start by expressing the numerator as a difference of squares:

We split the numerator even further to get

which leaves us with


Alternatively, you can use the limit laws.


SolutionAt a glance, it would seem that the subtraction law would be the best option. On the contrary, it would be the worst:

Clearly, it won't work because both denominators will evaluate to zero, and we know what that means!

A better option is to evaluate the expression in the bracket first:

limx→2

x4 – 16x – 2

limx→2

x4 – 16x – 2

= limx→2

(x2)2 – (4)2

x – 2 = lim

x→2

(x2 – 4)(x2 + 4)x – 2

limx→2

(x – 2)(x + 2)(x2 + 4)x – 2

=limx→2

(x2 – 4)(x2 + 4)x – 2

limx→2 (x + 2)(x2 + 4) = lim

x→2 (x3 + 4x + 2x2 + 8)

limx→2

(x3 + 4x + 2x2 + 8) = (2)3 + 4(2) + 2(2)2 + 8

=

(2)3 + 4(2) + 2(2)2 + 8

32

limx→1

1 x – 1

– 2 x2 – 1

limx→1

1 x – 1

– 2 x2 – 1 = lim

x→1

1 x – 1

– 2 x2 – 1

limx→1

limx→1

1 x – 1

– 2 x2 – 1 = lim

x→1

(x2 – 1) – 2(x – 1)(x – 1)(x2 – 1)

= limx→1

x2 – 1 – 2x + 2(x – 1)(x – 1)(x + 1)

= limx→1

x2 – 2x + 1(x – 1)(x – 1)(x + 1)

= limx→1

(x – 1)(x – 1)(x – 1)(x – 1)(x + 1)

= limx→1

1x + 1

So, by simply evaluating the bracketed expression, we have reduced

This further proves the importance of simplifying a function before attempting to evaluate its limit. It makes the task a

whole lot easier. Using direct substitution,

Thus,

Using direct substitution, the limit evaluates to ½. Now try using the limit laws to evaluate the limit.


SolutionWe simplify the numerator:

Using the limit laws,

limx→1

1 x – 1

– 2 x2 – 1 to lim

x→1

1x + 1

limx→1

1x + 1

=1

1 + 1 =

12

limx→1

1 x – 1

– 2 x2 – 1 = lim

x→1

1x + 1

=12

limh→0

(3 + h)–1 – 3–1 h

limh→0

(3 + h)–1 – 3–1 h

= limh→0

13 + h

1 3

–

h

= limh→0

3 – (3 + h) 3(3 + h)

h

= limh→0

3 – 3 – h 9 + 3h

h= lim

h→0

– h 9 + 3h

h= lim

h→0– h

(9 + 3h) 1

h ×

= limh→0

– 1

(9 + 3h)

limh→0

– 1

(9 + 3h) =

limh→0

– 1

limh→0

(9 + 3h)

(Quotient Law)

=

limh→0

– 1

+ limh→0

9 limh→0

h 3

(Addition & Constant Multiple Laws)

Using laws 7 and 8, we have

Therefore,


SolutionFirst, we simplify the function:

Next, we multiply the denominator by a minus sign. This allows us to cancel out the common factors:

Using limit laws,

Thus,

In the next set of examples (20 – 22), we learn how to evaluate limits of absolute functions.

– 1

9 + 3(0) =

– 1

9

limh→0

(3 + h)–1 – 3–1 h

=– 1

9

1 2

–

x – 2

1 x lim

x→2

1 2

–

x – 2

1 x lim

x→2= lim

x→2

2 – x 2x

x – 2 = lim

x→2

2 – x 2x

× 1x – 2

= limx→2

2 – x (x – 2)(2x)

limx→2

2 – x

– [(x – 2)(2x)] = lim

x→2 2 – x

(–x + 2)(–2x) = lim

x→2 2 – x

(2 – x)(–2x) = lim

x→21

–2x

limx→2

1

–2x =

limx→2

1

limx→2

x –2

(Quotient & Constant Multiple Laws)

= 1

–2(2) =

1

–4 (Laws 7 and 8)

1 2

–

x – 2

1 x lim

x→2 = 1

–4


SolutionFrom the definition of an absolute function,

Let's assume

Therefore,

which equals

In other words,

➔ f(x) = 1 for x ≥ 0 (i)

➔ f(x) = –1 for x < 0 (ii)

The task here is to evaluate the limit of f as x → –4 from the left. This means that we consider case (ii) only. Thus, since

f(x) = –1 for x < 0, then

You can verify the answer by graphing f.


limx→–4–

|x + 4| x + 4

|x| = x if x ≥ 0

–x if x < 0

f(x) = |x + 4| x + 4

|x + 4| x + 4

=

x + 4 x + 4

– (x + 4) x + 4

if x ≥ 0

if x < 0

|x + 4| x + 4

= 1 if x ≥ 0

–1 if x < 0

limx→–4– f(x) = –1 (direct substitution; Law 7)

lim x→1.5

2x2 – 3x |2x – 3|

SolutionFirst, we simplify the function (if possible):

From the definition of an absolute function,

which equals

This means that

➔ g(x) = x for x ≥ 0 (i)

➔ g(x) = – x for x < 0 (ii)

So,

➔ For case (i).

➔ For case (ii).

Clearly,

In other words, we are saying that


lim x→1.5

2x2 – 3x |2x – 3|

=lim

x→1.5 x(2x – 3)|2x – 3|

g(x) = = x(2x – 3)|2x – 3|

x(2x – 3)2x – 3

x(2x – 3)– (2x – 3)

if x ≥ 0

if x < 0

x if x ≥ 0

–x if x < 0 g(x) = =

x(2x – 3)

|2x – 3|

lim x→1.5+

g(x) = 1.5 (Law 8)

lim x→1.5–

g(x) = –1.5 (Law 8)

lim x→1.5+

g(x) lim x→1.5–

g(x) ≠

lim x→1.5

2x2 – 3x |2x – 3|

DOES NOT EXIST

lim x→0 x

1|x|1

–

SolutionFollowing the definition of an absolute function,

This becomes

So, to evaluate

we consider two cases (just as we did in example 21 above):

➔ h(x) = x for x ≥ 0 (i)

➔ h(x) = 2/x for x < 0 (ii)

Now, since h(x) = x for x ≥ 0 (case (i)), then

On the other hand, there's case (ii), which says that h(x) = 2/x for x < 0. Therefore, by direct substitution,

This means a limit DOES NOT EXIST for case (ii). In summary,

Since these one sided limits are not equal, it follows that

Example 23Let

Determine if

exists, and sketch the graph of F.

h(x) =x1

|x|1

– =

x1

x1

–

x1

x1 – –

if x ≥ 0

if x < 0

0 if x ≥ 0

if x < 0 h(x) =

x2

lim x→0

x1

|x|1

–

lim x→0+

h(x) = 0 (Law 7)

lim x→0–

h(x) = 02

lim x→0+

h(x) = 0 and lim x→0–

h(x) DOES NOT EXIST

lim x→0 x

1|x|1

– DOES NOT EXIST

x2 – 1

|x – 1|F(x) =

lim x→1

F(x)

SolutionBelow is a graph of F:

To evaluate the limit, the first step to be taken is to rewrite F using the definition of the absolute function:

Thus,

OR

Now that F has been broken down into a simplified piecewise defined function, we can evaluate the limit by considering

two simple cases:

➔ F(x) = x + 1 for x ≥ 0 (i)

➔ F(x) = – x – 1 for x < 0 (ii)

x2 – 1

|x – 1|F(x) = =

(x + 1)(x – 1)x – 1

if x ≥ 0

if x < 0 (x + 1)(x – 1)

–(x – 1)

(x + 1) if x ≥ 0

–(x + 1) if x < 0 F(x) =

x + 1 if x ≥ 0

–x – 1 if x < 0 F(x) =

CASE 1: F(x) = x + 1 for x ≥ 0

Thus,

CASE 2: F(x) = – x – 1 for x < 0

Thus,

At this point, it is obvious that

This means that

The graph of F above clearly verifies this result.

Example 24The signum (or sign) function, denoted by sgn, is defined by

(a) Sketch the graph of this function.

(b) Find each of the following limits or explain why it does not exist.

SolutionBased on the definition of the signum function, we have this graph:

lim x→1+

F(x) = 1 + 1 = 2 (Direct Substitution)

lim x→1–

F(x) = –1 – 1 = –2 (Direct Substitution)

lim x→1+

F(x) lim x→1–

F(x) ≠

lim x→1

F(x) DOES NOT EXIST

sgn x =

–1 if x < 0

0 if x = 0

1 if x > 0

(i) lim x→0

sgn x (ii) lim x→0

|sgn x|

The graph is quite straightforward, and will be used to evaluate the limits in question:

To evaluate this limit, we need to approach it from both sides, that is, find the left and right-hand limits. From

the graph, we see that

Since the two one-sided limits are not equal, it follows that

This one is a little tricky, but recall the definition of an absolute function:

Using this definition, then

which results in

Graphically, we have

(i) lim x→0

sgn x

lim x→0+

sgn x = 1 and lim x→0–

sgn x = –1

lim x→0

sgn x DOES NOT EXIST

(ii) lim x→0

|sgn x|

x if x ≥ 0

– x if x < 0 |x| =

|sgn x| =

–(–1) if x < 0

0 if x = 0

1 if x > 0

|sgn x| =

1 if x < 0

0 if x = 0

1 if x > 0

From the graph,

which means

Again, always remember that, when dealing with absolute functions, always apply this rule:

On my graphing calculator (CASIO Fx-9750G PLUS), the absolute function |x| is input as abs x. The abs function is one

of the many built in functions of a typical graphing calculator. Consult your calculator's manual for more info on how to

use this function, both for calculations and graphing.

Example 25Let

Sketch the graph of h, and evaluate each of the following limits, if it exists:

SolutionThis is another piecewise defined function, defined by three individual functions on three different domains:

➔ CASE 1: h(x) = x if x < 0

➔ CASE 2: h(x) = x2 if 0 ≤ x ≤ 2

➔ CASE 3: h(x) = 8 – x if x > 2

The values of the given limits depends solely on at least one of these cases.

Here, we want to find the value of the limit for values of x that are greater than zero. This means we consider case 2, as

it deals only with values of x greater than zero but less than/equal to 2. Thus, by direct substitution,

To solve this limit, we condense it into two simpler limits:

lim x→0+

|sgn x| = 1 and lim x→0–

|sgn x| = 1

lim x→0

|sgn x| = 1

x if x ≥ 0

– x if x < 0 |x| =

h(x) =

x if x < 0

x2 if x = 0

8 – x if x > 0

(i) lim x→0+

h(x) (ii) lim x→0

h(x) (iii) lim x→1

h(x) (iv) lim x→2

h(x)

(i) lim x→0+

h(x)

lim x→0+

h(x) = lim x→0+

(x2) = 02 = 0

(ii) lim x→0

h(x)

lim x→0–

h(x)

lim x→0+

h(x)

(This limit points to CASE 1)

(This limit points to CASE 2)

From CASE 1, h(x) = x if x < 0. Therefore,

For CASE 2, h(x) = x2 if 0 ≤ x ≤ 2. Thus,

Since, both cases yield the same result, we conclude that

Here, we are to evaluate the limit of h as x approaches 1. Clearly, this would fit into CASE 2, as described by its domain,

0 ≤ x ≤ 2. Besides, don't forget that h(x) = x2 for 0 ≤ x ≤ 2. Therefore, by direct substitution,

Therefore,

To solve this limit, we consider two cases:

and

From CASE 2, h(x) = x2 for 0 ≤ x ≤ 2. Therefore,

and, from CASE 3, h(x) = 8 – x for x > 2. Therefore,

In summary,

Clearly, this is an indication that

In the following example, we take a look at a special function.

lim x→0–

h(x) = lim x→0–

x = 0

lim x→0+

h(x) = lim x→0+

(x2) = 02 =

lim x→0–

h(x) =

0

0

(iii) lim x→1

h(x)

lim x→1

h(x) = lim x→1

(x2) = 12 = 1

lim x→1

h(x) = 1

(iv) lim x→2

h(x)

lim x→2–

h(x) (which fits into CASE 2)

lim x→2+

h(x) (which fits into CASE 3)

lim x→2–

h(x) = lim x→2–

(x2) = 22 = 4

lim x→2+

h(x) = lim x→2+

(8 – x) = 8 – 2 = 6

lim x→2–

h(x) = 4 lim x→2+

h(x) = 6and

DOES NOT EXISTlim x→2

h(x)

Example 26In this example, we'll look at the greatest integer function. By definition, the greatest integer function, written as y = [x]

, is the largest integer that is less than or equal to x. For example, [3] = 3, [3.6] = 3, [4.7] = 4, [π] = 3, and so on. The

greatest integer function is graphed below:

Using this definition, evaluate

From the graph, if we trace the lines, we see that [x] → –2 as x → –2 from the right. Therefore,

To solve this limit, we need to determine the values of

From (i), we see that

Also, from the graph, we see that

The left and right hand limits are not equal, which means that

lim x→ –2+

[x] (i)

lim x→ –2+

[x] = – 2

lim x→ –2

[x] (ii)

lim x→ –2+

[x] lim x→ –2–

[x] and

lim x→ –2+

[x] = – 2

lim x→ –2–

[x] = – 3

lim x→ –2

[x] DOES NOT EXIST

(A)

Like (ii), we need to evaluate the left and right hand limits:

From the graph,

which means

If n is an integer, evaluate

One good approach to solving this problem is by determining the values of these limits, using selected values of n. In

this case, we select n such that –10 ≤ x ≤ 10 . Thus we have the following table:

n n

–10 –11 1 0

–9 –10 2 1

–8 –9 3 2

–7 –8 4 3

–6 –7 5 4

–5 –6 6 5

–4 –5 7 6

–3 –4 8 7

–2 –3 9 8

–1 –2 10 9

n n

–10 –10 1 1

–9 –9 2 2

–8 –8 3 3

–7 –7 4 4

–6 –6 5 5

–5 –5 6 6

–4 –4 7 7

–3 –3 8 8

–2 –2 9 9

–1 –1 10 10

lim x→–2.4

[x] (iii)

lim x→–2.4+

[x] lim x→–2.4–

[x] and

lim x→–2.4+

[x] = – 3 and lim x→–2.4–

[x] = – 3

lim x→–2.4

[x] = – 3

(B)

lim x→n–

[x] lim x→n+

[x] and

lim x→n–

[x] lim x→n–

[x]

lim x→n+

[x] lim x→n+

[x]

From the first table, you can see that each value of

is ALWAYS 1 less than its corresponding value of n. Thus, we say

that

Using this same technique, we evaluate

and we get the second table below.

lim x→n–

[x]

lim x→n–

[x] = n – 1

lim x→n+

[x]

The pattern is simply too clear: each value of

is equal to its corresponding value of n. This means

lim x→n+

[x]

lim x→n+

[x] = n

From (B) above, we saw that, for any integer value of n (positive and negative),

and

Since the left and right hand limits are not equal, it follows that, for any integer n,

So now, we've seen that the limit does not exist for integer values. For what values of n does it exist then? Let's take

one more look at the graph:

Observe the graph carefully, and you'll see that a limit does not exist for integer values. So, let's consider another

category of numbers on this graph: non-integers. Take a number like 1.5. This number will have a left and right hand

limit (which is 1). The same will apply to any non-integer. You can test this theory yourself. In the end, we find that the

limit

exists only for only non-integers. In other words, the limit exists if and only if n is a non-integer

(C)

lim x→n–

[x] = n – 1

lim x→n+

[x] = n

(left hand limit)

(right hand limit)

lim x→n

[x] DOES NOT EXIST

lim x→n

[x]

Example 27In the theory of relativity, the Lorentz contraction formula

expresses the length L of an object as a function of its velocity v with respect to an observer, where Lo is the length of

the object at rest and c is the speed of light. Find

and interpret the result. Why is a left hand limit necessary?

SolutionThe function in question is

and we're given the following information:

L = Object length

V = Velocity of object

Lo = Length of object at rest

c = Speed of light (approx. 3.0 ×105km/s)

The question is asking us to find the value of the limit

Verbally speaking, we've been asked to determine what happens to the length of an object as its velocity approaches

that of light. Mathematically, we want to know what happens to L as v→c from the left. Thus, from a logical perspective,

if v→c– (which means v approaches c, but remains less than c), then

As v gets closer and closer to c, v2/c2 gets closer and closer to 1. Thus, we can assume that, at some point,

which means

Once again, it is assumed that we get to a point where

So,

L √ c2v2

1 –= Lo

lim v→c–

L

L √ c2v2

1 –= Lo

lim v→c–

L

c2v2

1→ (i)

c2v2

1= (ii)

would approach zero√ c2v2

1 – (iii)

√ c2v2

1 – = 0 (iv)

lim v→c–

L = lim v→c– √ c2

v21 –Lo

= lim v→c– √ 1 – 1Lo From (i) and (ii)

= lim v→c– √ 0Lolim

v→c–

Therefore,

So, what is the significance of this result?

First, don't forget that the function we're dealing with is a “contraction formula”. With that in mind, the value of the

limit can be taken to mean that the length of the object approaches zero (i.e, the object “shrinks in length”) as its speed

approaches the velocity of light.

Here's another interpretation of the result, taken from Cramster:

“As an object approaches the speed of light from velocities necessarily less than that of light, the object cotracts in the

direction of motion to a limiting length of zero”.

Example 28Is there a number a such that

exists? If so, find the value of a and the value of the limit.

SolutionTake a close look at the limit:

Clearly, we can't use the quotient rule or direct substitution, because then, it would mean the limit does not exist (as the

denominator would evaluate to zero). For now, let's assume a limit exists. We'll represent the limit by L. Therefore,

Next, we try to simplify the limit. We start by factorizing the denominator:

You'll find that, at this point, any method (quotient rule or direct substitution, although I prefer the latter) just might

work. However, there's a problem: the expression (x + 2) in the denominator prohibits us from moving any further. We

therefore partially eliminate the expression by taking it to the other side of the equation in (ii):

Now, we can evaluate the limit. Here we use direct substitution (i.e, we put x = –2)

= lim v→c–

(0)Lolim

v→c– = lim

v→c– Lo 0 From (iv)

= 0

lim v→c–

L = 0

3x2 + ax + a + 3

x2 + x – 2 lim

X→ –2

3x2 + ax + a + 3

x2 + x – 2 lim

X→ –2

3x2 + ax + a + 3

x2 + x – 2 lim

X→ –2 = L (i)

3x2 + ax + a + 3

(x + 2) (x – 1) lim

X→ –2 = L (ii)

3x2 + ax + a + 3

(x – 1) lim

X→ –2 = (x + 2)L (iii)

http://www.cramster.com/

Solving for a gives a = 15.

Now that we know the value of a, we can find the value of L. To do that, we put a = 15 in (i) above:

We simplify the numerator and denominator to give

Canceling out common factors leaves

Finally, we use direct substitution to evaluate what's left of the limit (or you can use the limit laws. My advice:Use

whatever method you're convenient with unless otherwise specified). In other words, we put x = –2:

Thus, L = –1

Example 29If

prove that

Solutionf is a function defined by two different functions in two different domains:

➔ f(x) = x2 if x is rational (I)

➔ f(x) = 0 if x is irrational (II)

Let's consider the first case:

3(–2)2 + a(–2) + a + 3

(–2 – 1) = (–2 + 2)L

= 12 – 2a + a + 3

–3 = 0

= 15 – a

–3 = 0

3x2 + 15x + 15 + 3

x2 + x – 2 lim

X→ –2 = L

3x2 + 15x + 18

x2 + x – 2 lim

X→ –2 = L=

3((x + 2)(x + 3)) (x + 2) (x – 1)

limX→ –2

= L

=3(x + 3)

(x – 1) lim

X→ –2

3x + 9

x – 1 lim

X→ –2 = L

3(–2) + 9

–2 – 1 = –6 + 9

–2 – 1 = –1

x2 if x is rational

0 if x is irrationalf(x) =

lim x→0

f(x) = 0

f(x) = x2 for any rational value of x. Therefore,

Next, let's look at the second case:

f(x) = 0 for any irrational value of x. Thus,

We see that, for both scenarios, the limits evaluate to zero. Thus, for ANY value of x, rational or irrational,

Example 30Show by means of an example that

may exist even though neither

exists.

SolutionTo solve this problem, we apply one of the basic principles of problem solving – working backwards.

In this regard, a feasible solution to this problem will be to come up with a function that can be broken down into two

simpler functions whose limits do not exist at a specified number a. This is perhaps the toughest part.

Here's a simple rational function I've come up with:

This is the point where we start working backwards. Suppose we break h into two functions, m and n such that

h(x) = m(x) + n(x)

where

Now, let's put a = – 3. We see that the limits

do NOT EXIST, because the denominators of both limits evaluate to zero; however, if we combine m and n, we get a

different result:

lim x→0

f(x) = 02lim x→0

(x2) = = 0 (Law 9)

lim x→0

f(x) = 0lim x→0

0 = (Law 7)

lim x→0

f(x) = 0

lim x→a

[f(x) + g(x)]

lim x→a

f(x) nor lim x→a

g(x)

x2 + 5x + 6

x +3 h(x) =

x2 + 5x

x +3 m(x) = and 6

x +3 n(x) =

lim x→ –3

m(x) and lim x→ –3

n(x)

lim x→ –3

m(x) + lim x→ –3

n(x) = lim x→ –3

[m(x) + n(x)]

= x2 + 5x

x +3 +

6

x +3 lim

x→ –3

Thus, even though the limits

did not exist, their sum yielded a value. The same applies to products of limits, as the next example will show.

Try to practice this example over and over by deriving your own functions and proving the theorem.

Example 31Show by means of an example that

may exist even though neither

exists.

SolutionWe apply the “working backwards” technique we applied in Example 30 above. Here we use a relatively simple rational

function:

which is then split into two simpler functions:

and put a = 0. Thus we have two limits

both of which do not exist because the denominators evaluate to zero. But if we combine the limits of f and g via

multiplication, we get

= x2 + 5x + 6

x +3 lim

x→ –3 = lim

x→ –3 (x +2)(x +3 )

x +3

= lim x→ –3

(x + 2)

= – 3 + 2

= – 1

(Direct Substitution)

lim x→ –3

m(x) and lim x→ –3

n(x)

lim x→a

[f(x)g(x)]

lim x→a

f(x) nor lim x→a

g(x)

2√x(3x2 – 15x)

x√x(4 + 3x)h(x) =

2√x

x(4 + 3x) f(x) = and 3x2 – 15x

√x g(x) =

lim x→0

f(x) and lim x→0

g(x)

lim x→0

[f(x)g(x)] = lim x→0

f(x) × lim x→0

g(x)

= lim x→0

2√x

x(4 + 3x)

3x2 – 15x

√x ×

= lim x→0

2√x

x(4 + 3x)

3x2 – 15x

√x ×

Canceling out common factors leaves

You can see this limit can be evaluated without any problems. We use direct substitution:

Again, we see that even though the limits

did not exist, their product yielded a value.

Example 32Let f(x) = x – [x]

(a) Sketch the graph of f.

(b) If n is an integer, evaluate

(c) For what values does

exist?

SolutionObserve that f contains a relatively familiar function, the greatest integer function y = [x]. We've examined this function

in Example 26.

To solve these problems, we need to apply what we've learnt about the function y = [x]. First, we have to graph f, and to

do that, we need two vital pieces of information:

➔ How f is defined for integer values

➔ How f is defined for non-integer values

Let's say we want to graph f for –5 ≤ x ≤ 5.

In the first instance, we find that, for any integer x,

f(x) = x – [x]

= x – x = 0

This means that f(x) = 0 if x is an integer.

To determine how f is defined for non-integers, we select a few domains and see how f is defined in each domain, like

this:

➔ In the domain –5 < x < –4 for example, [x] = –5 and therefore f(x) = x – [–5] = x – (–5) = x + 5

➔ In the domain –4 < x < –3, [x] = –4 and therefore f(x) = x – [–4] = x – (–4) = x + 4

and so on. You'll understand the pattern better when its tabulated:

= lim x→0

2√x(3x2 – 15x)

x√x(4 + 3x) = lim

x→0

2x√x(3x – 15)

x√x(4 + 3x)

lim x→0

2(3x – 15)

(4 + 3x)

6(0) – 30

4 + 3(0)=

= lim x→0

6x – 30

4 + 3x

–30

4 = –7.5

lim x→0

f(x) and lim x→0

g(x)

lim x→n–

f(x) and lim x→n+

f(x)

lim x→a

f(x)

The solution to this example has been adapted from Cramster


Domain of x [x] f(x) = x – [x]

–5 < x < –4 –5 f(x) = x + 5

–4 < x < –3 –4 f(x) = x + 4

–3 < x < –2 –3 f(x) = x + 3

–2 < x < –1 –2 f(x) = x + 2

–1 < x < 0 –1 f(x) = x + 1

0 < x < 1 0 f(x) = x

1 < x < 2 1 f(x) = x – 1

2 < x < 3 2 f(x) = x – 2

3 < x < 4 3 f(x) = x – 3

4 < x < 5 4 f(x) = x – 4

(c)

In summary, we've seen from (a) and (b) that, for any integer n,

Using this table and the fact that f(x) = 0 for any integer x, we have the graph below.

(b) From the graph below, it is quite obvious that, for instance, f(x)→1 as x→ –5–, f(x)→1 as x→ 3–,f(x)→1 as x→ 5–, and so on, as illustrated by the upward- facing arrows in the graph.Thus, for any integer n,

On the other hand, we also see from the graph that, for example,f(x)→0 as x→ –5+, f(x)→0 as x→ 3+,f(x)→0 as x→ 5+, and so on, as illustrated by the downward- facing arrows in the graph.Thus, for any integer n,

lim x→n–

f(x) = 1

lim x→n+

f(x) = 0

y = [x]

lim x→n–

f(x) = 1 and lim x→n+

f(x) = 0

Since the two one-sided limits are not equal, it follows that the limit

does exists if a is an integer. However, if you look carefully at the graph, you'll see that a limit exists for non-integers,

and this implies that the limit exists if and only if a is a non-integer.

Example 33If f(x) = [x] + [– x], show that

exists but is not equal to f(2).

SolutionIf you read the question carefully, you'll see that, to solve the problem, we have to determine the values of

➔

➔ f(2)

The question is asking us to establish the fact that

If f(x) = [x] + [– x], then

f(2) = [2] + [– 2]

f(2) = 2 + (– 2) = 2 – 2

= 0

Thus, f(2) = 0.

In fact, we see that, for ANY integer x, we see that f(x) = 0.

To determine the value of

we apply the technique used to solve Example 32: evaluate f for different domains. Here, we choose the domain such

that –2 < x < 3. Like Example 32, we use a table:

Domain of x f(x) = [x] + [–x]

–2 < x < –1 f(x) = – 2 + 2 = – 1

–1 < x < 0 f(x) = – 1 + 0 = – 1

0 < x < 1 f(x) = 0 + (–1) = – 1

1 < x < 2 f(x) = 1 + (–2) = – 1

2 < x < 3 f(x) = 2 + (–3) = – 1

From the table, we see that, for any non-integer x, f(x) = –1.

Using this information and the fact that f(x) = 0 for integer values of x, we have the following graph:

lim x→a

f(x)

lim x→2

f(x)

lim x→2

f(x)

lim x→2

f(x) f(2) ≠

lim x→2

f(x)

From the graph of f, it is quite clear that f(2) = 0, but

Thus, we have shown that


SolutionObserve that the numerator and denominator are irrational. Thus, this limit can be evaluated by rationalizing either.

Here, we rationalize the denominator first:

We need to cancel out the denominator, and doing this requires the rearrangement of the terms in the numerator:

Then we factorize the numerator:

lim x→2

f(x) = –1

lim x→2

f(x) f(2) ≠

lim x→1

√x – x2

1 – √x

lim x→1

√x – x2

1 – √x

1 + √x

1 + √x × = lim

x→1

(√x – x2)(1 + √x)

(1 – √x)(1 + √x)

= lim x→1

√x + x – x2 – x2√x

1 – x

lim x→1

√x + x – x2 – x2√x

1 – x= lim

x→1

√x – x2√x + x – x2

1 – x

Therefore,

Now, let's evaluate the limit by rationalizing the numerator:

After rationalizing the numerator, you can see that the limit hasn't been simplified much. In cases like this, we

rationalize again using the original denominator:

Crossing out common factors gives

At this point, the limit has been simplified well enough for us to evaluate.

lim x→1

√x – x2√x + x – x2

1 – x= lim

x→1

√x(1 – x2) + x(1 – x)

1 – x

= lim x→1

√x(1 – x)(1 + x) + x(1 – x)

1 – x

= lim x→1

x(1 – x)

1 – x

√x(1 – x)(1 + x)

1 – x+

= lim x→1

(√x(1 + x) + x )

= lim x→1

(√x + x√x + x)

= √(1) + (1)√(1) + (1)

= 3

lim x→1

√x – x2

1 – √x = 3

lim x→1

√x – x2

1 – √x

√x + x2

√x + x2 × = lim

x→1

(√x – x2)(√x + x2)

(1 – √x)(√x + x2)

= lim x→1

x – x4

(1 – √x)(√x + x2)= lim

x→1

x(1 – x3)

(1 – √x)(√x + x2)

lim x→1

×x(1 – x3)

(1 – √x)(√x + x2)

1 + √x

1 + √x = lim

x→1

x(1 – x3)(1 + √x)

(1 – √x)(1 + √x)(√x + x2)

= lim x→1

–x(x3 – 1)(1 + √x)

(1 – x)(√x + x2)

= lim x→1

–x(x3 – 1)(1 + √x)

–(x – 1)(√x + x2)

= lim x→1

x(x – 1)(x2 + x + 1)(1 + √x)

(x – 1)(√x + x2)

lim x→1

x(x2 + x + 1)(1 + √x)

(√x + x2)


You'll agree that the second method is relatively complicated, but I want you to realize that sometimes, you need to

rationalize a relatively complicated rational limit more than once before you can move any further.

Example 35The figure shows a fixed circle C1 with equation (x – 1)2 + y2 = 1 and a shrinking circle C2 with radius r and center the

origin. P is the point (0, r), Q is the upper point of intersection of the two circles, and R is the point of intersection of the

line PQ and the x-axis. What happens to R as C2 shrinks, that is, as r → 0+ ?

SolutionThe figure shows two circles, C1, whose equation is (x – 1)2 + y2 = 1. This equation can be rewritten to look like a

standard circle equation:

(x – 1)2 + y2 = 1

= x2 – 2x + 1 + y2 = 1

= x2 – 2x + y2 = 1 – 1

= x2 – 2x + y2 = 0

= x2 + y2 = 2x (i)

There's a second circle C2 whose equation is not given. Nonetheless, we know that it has a radius r and its center is the

origin (0,0). Thus we use the standard circle equation:

x2 + y2 = r2 (ii)

lim x→1

x(x2 + x + 1)(1 + √x)

(√x + x2)=

(1)((1)2 + (1) + 1)(1 + √(1))

(√(1) + (1)2)

=(1)(3)(2)

(2)

= 3

The solutions to example 34 have been adapted from Cramster

y

R

C1

C2

0

QP

x

Note: Figure not drawn to scale


to represent C2.

The task here is to determine what happens to R as r → 0+ . Try to imagine for a moment what the figure would look like

when C2 starts to shrink! You'll find that the line PQ starts to shrink too, and this means that the point R begins to move

as well. To be more specific, the x-coordinate of R will approach a certain value. But what is the limit? Well, we're about

to find out.

We start by solving (i) and (ii) simultaneously:

x2 + y2 = 2x (i)

x2 + y2 = r2 (ii)

When we solve the two equations, we find that r2 = 2x. Solving for x gives

This will be the x-coordinate of the point Q (where C1 and C2 intersect). To get the y-coordinate, we plug (iii) into (ii) to

give

Therefore,

So,

This is the y-coordinate of Q. Therefore,

and from the graph, the coordinate of P = (0, r).

Now that we have the coordinates of two points P and Q, we can compute the slope m:

r2

2x = (iii)

r2

2

2+ y2 = r2

= r4

4+ y2 = r2

y2 = r2 r4

4–

y2 =4r2 – r4

4=

r2(4 – r2) 4

y = √r2(4 – r2) 4 √r2

4(4 – r2)=

y =r2 √(4 – r2)

Coordinate of point Q = r2 √(4 – r2)

r2

2,

Slope of PQ =

r2 √(4 – r2) – r

r2

2 – 0

=

r2 √(4 – r2) – r

r2

2

×

=

2r2

12 √(4 – r2) – 1r=

=

12 √(4 – r2) – 12

r

=√(4 – r2) – 2

r

Therefore,

We have the slope of PQ, which means we can compute the equation of the line. Using the slope-intercept equation of a

line, we have

y = mx + c

where m = slope of PQ

c = y-intercept (which in this case, equals r)

Thus the equation can be rewritten as

y = mx + r

Recall the problem, we want to know what happens to the point R (x, 0) as r → 0+ . We see that if r → 0+ , then the point

R begins to shift. To be more specific, the x-coordinate of R changes. But what point does it approach? That's what we

really want to know.

Therefore, we'll need an equation that explicitly describes the x-coordinate of R. In other words, we express (iv) in

terms of r, which is done by making x the subject of (iv):

We know that R = (x,y) = (x,0). Putting y = 0 in (iv) gives

mx + c = 0

∴

Plugging in the value of m into (iv) gives

Which equals

We are getting close!!

Once again, recall the problem – we want to know what happens to the point R (x, 0) as r → 0+ . The answer can be found

by evaluating the limit

Since the denominator is irrational, we rationalize it. This yields

=m√(4 – r2) – 2

r

(iv)

=x –rm (v)

=x√(4 – r2) – 2

r

– r

√(4 – r2) – 2

– r2=x (vi)

lim r→0+ √(4 – r2) – 2

– r2

lim r→0+ √(4 – r2) – 2

– r2

×√(4 – r2) + 2

√(4 – r2) + 2

=

√(4 – r2) – 2 √(4 – r2) + 2

√(4 – r2) + 2– r2

lim r→0+


Thus,

In other words, as r → 0+ , the x-coordinate of R approaches 4. Here's a better interpretation: as r → 0+, the point R

approaches the point (4,0) on the x-axis.

Example 36Evaluate the limit

SolutionI've tried to evaluate this limit using many methods. I came to understand that the best (and easiest) way around it is to

“double rationalize”! See how it's done:

which equals

=√(4 – r2) + 2– r2

4 – r2 – 4

= √(4 – r2) + 2– r2

– r2

lim r→0+

lim r→0+

= lim r→0+ √(4 – r2) + 2

lim r→0+ √(4 – r2) + 2 = √ 4 – (0)2 + 2

= √4 + 2 =

2 + 2 =

4=

lim r→0+ √(4 – r2) – 2

– r2

= 4

The solution to example 35 has been adapted from Cramster

√ 6 – x – 2

√ 3 – x – 1

lim x→2

√ 6 – x – 2

√ 3 – x – 1

lim x→2

× √ 6 – x + 2

√ 6 – x + 2× √ 3 – x + 1

√ 3 – x + 1

√ 6 – x – 2

√ 3 – x – 1

√ 6 – x + 2 √ 3 – x + 1

√ 6 – x + 2 √ 3 – x + 1

lim x→2


Crossing out common factors gives

The function is still irrational, but at this point, it does not matter because it can be evaluated That's the whole point of

rationalization – to simplify a function far enough so that it can be evaluated as easily as possible.

So now, we use direct substitution:

Therefore,

This tutorial is concluded in Part II, where we look at a very important Theorem: The Sandwich Theorem.

Before you exit, try the following exercises. If you need help, you know how to reach me.

EXERCISES(1) Given that

find the limits that exist. If the limit does not exist, explain why.

=(6 – x – 4) √ 3 – x + 1

(3 – x – 1) √ 6 – x + 2

lim x→2

=(2 – x) √ 3 – x + 1

√ 6 – x + 2

lim x→2 (2 – x)

lim x→2

√ 3 – x + 1

√ 6 – x + 2

lim x→2

√ 3 – x + 1

√ 6 – x + 2=

√ 3 – 2 + 1

√ 6 – 2 + 2

=√1 + 1√4 + 2

= 12

√ 6 – x – 2

√ 3 – x – 1

lim x→2

=12

The solution to example 36 has been adapted from Cramster

f(x) = –3 limx→a

g(x) = 0 limx→a

h(x) = 8 limx→a

(a) [f(x) + h(x)] limx→a

(b) [f(x)]2 lim x→a

(c) [h(x)]1/3 limx→a

(d) limx→a

1f(x)

(e) limx→a

f(x)h(x)

(f) limx→a

g(x)f(x)

(g) limx→a

f(x)g(x)

(h) limx→a

2f(x)h(x) – f(x)


(2) Evaluate the limits and justify each step by indicating the appropriate limit laws.

(3)(a) Estimate the value of

by graphing the function

(b) Make a table of values for f(x) for x close to 0 and guess the value of the limit.

(c) Use the limit laws to prove that your guess is correct .

(4)(a) Use a graph of

to estimate the value of limx→0 f(x) to two decimal places

(b) Use a table of values of f(x) to estimate the limit to four decimal places

(c) Use the limit laws to find the exact value of the limit.

(5) Let

Evaluate each limit, if it exists.

calculus4engineeringstudents.com

(a) limx→–1

x – 2 x2 + 4x – 3

(b) √16 – x2 limx→4–

(c) limx→–3

x2 – x – 12 x + 3

(d) limh→0

(1 + h)4 – 1 h

(e) limh→0

(2 – h )3 – 8 h

(f) limx→2

x2 + x – 6 x2 – 4

(g) limx→9

x2 – 81 √x – 3

(h) limx→0

1

x√1 + x

1

x – (i) lim

x→16

4 – √x x – 16

(j) limh→0

1 – √1 – h2 h

(k) limx→0

√x + 2 – √2xx2 – 2x

(l) (√x – 9 + [x + 1])

limx→9+

limx→0 √1 + 3x – 1

x

√1 + 3x – 1

xf(x) =

√3 + x – √3x

f(x) =

(m) |x + 4|limx→– 4

(n) limx→2

|x – 2|

x – 2

f(x) =

√–x if x < 0

3 – x if 0 ≤ x ≤ 3

(x – 3)2 if x > 3

limx→0

f(x) (a) (b) limx→3

f(x)

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EvaluatingLimits(Limit Laws)