Finding Limits AlgebraicallyChapter 2: Limits and Continuity

What you’ll learn about

• Finding a limit algebraically (or analytically) • Using properties of limits• Finding a limit using the sandwich (or squeeze) theorem• Two special limits• Limits of piecewise functions

Direct SubstitutionIn the last lesson, we learned that the limit of f(x) as x approaches c doesn’t depend on the actual value of f at x = c. It often happens, however, that the limit is actually f(c). In these cases, the limit can be evaluated by direct substitution. In other words,

(just substitute c for x)For example, This can be verified by examining the function’s graph and table of values near x = 2.These well-behaved functions are considered continuous at c. We will examine this concept more in a later lesson. Now let’s look at some properties of limits that will help us when performing direct substitution.

Properties of Limits

If , , , and are real numbers and

lim and lim , then

1. : lim

The limit of the sum of two functions is the sum of their limits.

2. : lim

The limit

x c x c

x c

x c

L M c k

f x L g x M

Sum Rule f x g x L M

DifferenceRule f x g x L M

of the difference of two functions is the difference

of their limits.

Product Rule:

Constant Multiple Rule:

Properties of Limits continued

Properties of Limits continued

6. : If and are integers, 0, then

lim

provided that is a real number.

The limit of a rational power of a function is that power of the

limit of the function, provided the latte

rrss

x c

r

s

Power Rule r s s

f x L

L

r is a real number.

Other properties of limits:

lim

limx c

x c

k k

x c

Polynomial and Rational FunctionsTrying to evaluate functions using just these rules can be a grueling process (I’m not going to make you do that) so it is a little easier to just combine a bunch of those rules to evaluate polynomial and rational functions.

11 0

11 0

1. If ... is any polynomial function and

is any real number, then

lim ...

2. If and are polynomials and is any real number, then

lim , prov

n nn n

n nn nx c

x c

f x a x a x a

c

f x f c a c a c a

f x g x c

f x f c

g x g c

ided that 0.g c

Now let’s look at a few examples using these new properties and rules.

Example LimitsUse the rules from the previous slides to find the following limits using direct substitution.

a)

22

5lim 4 -2 6 4 2 6 4 25 15 0 6 100 0 95 1 6 6x

x x

b)

= = = 2

Evaluating LimitsAs with polynomials, limits of many familiar functions can be found by substitution at points where they are defined. This includes trigonometric, exponential, and logarithmic functions, and composites of these functions.

Now lets try an example:

Let’s look at the graph first:

0

1 sinFind lim

cosx

x

x

Solve graphically:

1 sinThe graph of suggests that the limit exists and is 1.

cos

xf x

x

If we can, we always want to confirm this algebraically:

0

00

Confirm Analytically:

lim 1 sin 1 sin 01 sinFind lim

cos lim cos cos 0

1 0 1

1

x

xx

xx

x x

Evaluating Limits cont. Now let’s try another example:

0

5Find lim

x x

Confirm Analytically :

We can't use substitution in this example because when is relaced by 0,

the denominator becomes 0 and the function is undefined.

This would suggest that we rely on the graph to s

x

ee that the

limit does not exist.

Dividing Out TechniqueSometimes direct substitution fails even when the limit exists. This is one technique to help find the limit in that case. Let’s use it in an example

Find: In this case here we can’t use direct substitution because if we try to we will get . Let’s see what the graph looks like.

There appears to be a limit as x

One approach we can try to do is factor our original function and see if we can cancel anything out that may allow us to do direct substitution.

Now we can easily find the limit using direct substitution (

Rationalizing TechniqueAnother technique that we can try when direct substitution fails but a limit appears to exist is to rationalize the numerator or denominator and see if that helps us to make a cancellation that allows for us to use direct substitution.

Find Now just like in the last problem, direct substitution won’t work because we get the form . If you look at the graph, it appears that there is a limit as x approaches 1. There is no dividing out that we can do, so let’s try rationalizing the numerator and see what happens.

√𝑥+1 −1𝑥

=(√𝑥+1 −1𝑥 )( √𝑥+1+1

√𝑥+1+1 )= (𝑥+1 ) −1𝑥 (√𝑥+1+1)

= 𝑥𝑥 (√𝑥+1+1)

=¿

1

√𝑥+1+1Now we can easily find the limit using direct substitution.

lim𝑥→ 0

1

√𝑥+1+1=

12

Recap – Finding Limits AlgebraicallyA few important points on the last two examples that we just did.• What we are actually doing when we divide out or rationalize is find a new function. This

function is technically different from our original function because it has a different domain. But what you will notice if you look at the graphs or tables of values of our original and new function is that they have the exact same values at every point EXCEPT at the point(s) where our original function is undefined. While in practice this doesn’t change your approach to the problem, it is important to note that while the functions aren’t technically equal at EVERY point (because the original function is undefined at some point(s)), their LIMITS ARE EQUAL AT EVERY POINT.

• As a general rule, when you try direct substitution and you get , this is called indeterminate form and a limit exists at that point. You should then try rationalizing or dividing out to make the denominator ≠ 0 and then try direct substitution again.

• If you try direct substitution and you get where c is a non-zero constant, then a limit will NOT exist at that point.

A General Strategy for Finding Limits1) Try direct substitution. Direct substitution will work for most types of

functions. For piecewise functions, if you are looking for the limit at an x value where the rule for the function changes, use direct substitution on the left and right rules at that point and see if they are equal. Remember, in order for a limit to exist, both of the one-sided limits must be equal to each other.

2) If you can’t evaluate the limit of f(x) at a point c using direct substitution, try dividing out or rationalizing to find a new function g that agrees with f for all x other than x = c. Choose a g where you can find the limit at x = c using direct substitution.

3) Always confirm or reinforce your conclusion using a graph or table of values or both if you can. It doesn’t matter if you check before or after you try direct substitution, but it’s very important to make sure that your answer makes sense. If you don’t have calculator access, you just have to trust your algebra.

The Sandwich (or Squeeze) TheoremIf we cannot find a limit directly, we may be able to find

it indirectly with the Sandwich Theorem. The theorem

refers to a function whose values are sandwiched between

the values of two other func

f

tions, and .

If and have the same limit as then has that limit too.

g h

g h x c f

If for all in some interval about , and

lim =lim = ,

then

lim =

x c x c

x c

g x f x h x x c c

g x h x L

f x L

Sandwich Theorem ExampleFind In this case here, we can’t use direct substitution and we can’t

divide out or rationalize anything to help with that .

Now we know that because of the range of sine. Therefore We also know that -1 again because of the range of sine. Therefore (-1)

This leads to the following: -

This means that our function is sandwiched between - and for all values of .

Let’s try some limits: (-) )

By direct substitution: 0 0

which means that the only possible answer is 0!

Take a look at a graph of the three functions we just worked with in this example. Notice that the function with the in it is “sandwiched” between the other functions for all x values.

Two Special Trigonometric LimitsThere are two special trig limits that you need to memorize. The first one is much more likely to appear on tests in this class and on the AP exam but they are both worth memorizing.

On your own, you should verify both of these graphically and/or numerically.

Example With a Piecewise FunctionFind where

Remember that we can only say that a limit exists at a given x value if both of the one-sided limits exist and are equal at that x value. In this case, we will use direct substitution in both of the given definitions of f(x) since x=2 is the value where the definition of f(x) changes.

lim𝑥→ 2−

𝑓 (𝑥 )=4 (2 )−1=7 lim𝑥→ 2+¿ 𝑓 (𝑥 )=22+3=7¿

¿

Since both one sided limits are equal at x = 2, we can say that .

In symbols we would say:

therefore

The symbols representation in the box above is what is expected of you on my exams and on the AP exam for Free-Response questions as justification.

Summary• For most functions where the denominator ≠ 0 at the x value where you are trying to find a

limit, direct substitution is the most accurate way to find a limit. The properties of limits will also help to simplify more difficult limits with direct substitution.

• If direct substitution gives an answer of the form then try the methods of factoring out or rationalizing to find a new function where direct substitution will work.

• If none of the above methods work, the Sandwich Theorem may help.• If direct substitution gives an answer of the form where c is a non-zero constant, then the

limit will not exist at that point. • For piecewise functions, be sure that both one-sided limits are equal at the point where you

are trying to find the limit if it is at an x value where the rules of the function change. • Memorize and be able to use the following two special properties: