1.5 Limits

Goal: to understand limits.def: If f(x) becomes arbitrarily close to a single number L as x approaches c from either side, then

lim ( )x cf x L

Ex.

2 1( )

1

xf x

x

Graph this in your calculator

2

1

-1

-2

-4 -2 2 4

f x = x2-1

x-1

2 1( )

1

xf x

x

It looks like f(1) = 2, is this true?

• Look at the table.

• Reset the table and look closely as x gets closer to 1.

• Is f(1) = 2?

• No!!!!

What do you think

2

1

1lim

1x

x

x

In this case it is 2 since we can say that f(x) becomes arbitrarily close to 2 as x approaches 1 from either side (positive or negative)

will be?

Huh? Go back to the graph for a second and think about

the function (domain in particular)

2

1

-1

-2

-4 -2 2 4

f x = x2-1

x-1

2 1( )

1

xf x

x

We know that x ≠ 1, the function is undefined at x = 1

That would mean that we actually have a hole in our graph at x = 1

2

1

-1

-2

-4 -2 2 4

f x = x2-1

x-1

As we approach (imagine you are riding on the function) from the negative (left) side we get closer to the function having a value of 2

2

1

-1

-2

-4 -2 2 4

f x = x2-1

x-1

As we approach (again, imagine you are riding on the function) from the positive (right) side we also get closer to the function having a value of 2

2

1

-1

-2

-4 -2 2 4

f x = x2-1

x-1

Since we are traveling towards a value of 2 for the function from both sides, we can say that the limit of the function as x–› 1 is 2

Is there another way to find this value? Sure, we can use the table in your calculator

Set your table to start at x = 1 Change your ∆x setting to .1 to start Your table should look something like this:

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4

1.5 1.6 1.7 1.8 1.9 ER 2.1 2.2 2.3 2.4

Now let’s move in closer Again set your table to start at x = 1 Change your ∆x setting to .01 Your table should look something like this:

.95 .96 .97 .98 .99 1 1.01 1.02 1.03 1.04

1.95 1.96 1.97 1.98 1.99 ER 2.01 2.02 2.03 2.04

We can probably guess pretty accurately that the limit is equal to 2

0.995 0.996 0.997 0.998 0.999 1 1.001 1.002 1.003 1.004

1.995 1.996 1.997 1.998 1.999 ER 2.001 2.002 2.003 2.004

One last time even closer Again set your table to start at x = 1 Change your ∆x setting to .001 Your table should look something like this:

Limit rules:A limit only exists if both the left side and the right side approach the same value!!!! THIS is VERY IMPORTANT!!!!!

Ex. 1

( )1

xf x

x

does

1

1lim

1x

x

x

exist?

Graph 1

( )1

xf x

x

Look at the table as x gets closer to 1 What value is the function approaching from the left side? -1 What is value is the function approaching from the right side? 1

in your calculator.

Since the left side limit ≠ right side limit, we say that the limit does not exist.

Ex. 1

( )0 1

x xf x

x

find 1

lim ( )xf x

Graph the function first, remember how to graph a piecewise function.

8

6

4

2

-2

-4

-6

-8

-15 -10 -5 5 10 15

1( )

0 1

x xf x

x

Ex.

Find

3

1

4 5lim

1x

x x

x

How can I figure this one out? Graph? Yes Table? Yes What do you think the limit is? 7 What if I forgot my calculator today? Or my

batteries died? How did people do limits before there were calculators?

Direct Substitution:When the domain does not create difficulty for us (undefined, etc.), we can substitute what x approaches directly into the function.

lim ( ) ( )x cf x f c

Go back to our first example.

You must have figured by now that there was a reason we went back and reviewed all of our

algebra skills. Simplify first!

2

1

1lim

1x

x

x

2 1 11( ) 1, 1

1 1

x xxf x x x

x x

Now Direct Substitution works, we can substitute x = 1 into the function to find the value the function is approaching.

1

lim( 1) 1 1 2x

x

Remember this is not necessarily solving for the actual value of the function, it is looking for what value the function is approaching as x -> that specific number. Often x is approaching a number that will make the function undefined!

Properties and Operations with Limits:Let b & c be real numbers and let n be a positive integer.

limx cb b

2lim5 5x

1.

ex.

If you are taking the limit of a constant the limit is just that constant.

limx cx c

4lim 4xx

ex.

If you are taking the limit of a lone variable just substitute in what x is approaching for the x in the equation.

2.

lim n n

x cx c

2 2

3lim 3 9xx

ex.

If you are taking the limit of a variable raised to a positive power, just substitute in what x is approaching for the x in the equation and simplify.

3.

lim n n

x cx c

3 3

8lim 8 2x

x

ex.

If you are taking the limit of a variable under a radical, just substitute in what x is approaching for the x in the equation and simplify. Be careful that when your index is even then c must be positive!

4.

Operations with Limits:

Let b and c be real numbers, let n be a positive integer and let

f and g be functions with the following limits:

lim ( )x cf x L

and lim ( )

x cg x k

1. Scalar Multiple

lim ( )x c

bf x bL

ex.

4lim10 10 4 40x

x

Ex. Find 2

1lim 2 4x

x x

2

1 1 1lim2 lim lim4x x x

x x

22(1) (1) 4

5

lim ( ) ( )x c

f x g x L k

2. Sum or Difference

lim ( ) ( )x c

f x g x L k

3. Product

lim ( ) 5x cf x

Ex.

and lim ( ) 6x cg x

lim ( ) ( ) 5 6 30x c

f x g x

lim ( ) 12x cf x

Ex.

and lim ( ) 6x cg x

4. Quotient( )

lim 0( )x c

f x Lk

g x k

( ) 12lim 2

( ) 6x c

f x

g x

The Limit of a Polynomial:If p is a polynomial function and c is any real number then

lim ( ) ( )x cp x p c

Can you think of a method that allows us to substitute values into polynomials very easily?

Synthetic Substitution!

Ex.

Find 4 3 2

1lim 7 6 2 7x

x x x

7 6 2 0 7

7 1 1 1

1 7 1 1 1 6

The Replacement Theorem:Let a be a real number and let f(x)=g(x) for all x≠c. If the limit of g(x) exists as x->c then the limit of f(x) also exists and

lim ( ) lim ( )x c x cf x g x

This is just the theorem that actually allows us to simplify before we take limits.

Ex.

Find

3

2

8lim

2x

x

x

2

2

2 ( 2 4)lim

( 2)x

x x x

x

2

2lim 2 4xx x

Direct substitution won’t work here. We would be dividing by zero.

4 4 4 12

Here is a place where it is very helpful to remember how to factor the difference of cube. Again, that is why we reviewed it:-)

Ex.

Find

2

3

12lim

3x

x x

x

3

3 ( 4)lim

( 3)x

x x

x

3lim 4xx

3 4 7

One Sided Limits:When the limit doesn’t exist

(since the left and right side are approaching at very different values) we look at one sided limits.

Graph:

2( )

2

xf x

x

Does the limit as x->2 exist?

2

2lim

2x

x

x

2

2lim

2x

x

x

Left side (see the little negative sign)

Right side (see the little positive sign)

4

2

-2

-4

-6

-5 5

f x = x-2

x-2

2

2lim 1

2x

x

x

2

2lim 1

2x

x

x

Existence of a Limit:If f is a function and c and L are

real numbers then

i.f.f. both the left and right hand limits are equal to L.

So:

lim ( )x cf x L

lim ( ) lim ( )x c x c

f x L and f x L

For the limit to exist

2

0

1 0lim

2 1 0x

x x

x x

Ex.

2

0lim 1x

x

0

lim 2 1x

x

and

0 1 1 2(0) 1 1 and

So the limit exits and is equal to 1

Unbounded Behavior:When the function approaches

+∞ or -∞ as x->c, we say the function is unbounded. The limit does not exist.

Graph:

3( )

2f x

x

20

15

10

5

-5

-10

-15

-20 -10 10 20 30

f x = 3

x-2

2

3lim

2x x Ex.

and

and

So the limit does not exits

2

3lim

2x x 2

3lim

2x x

Unbounded Behavior:When the function approaches

+∞ or -∞….

Think about that statement for a minute.Remember,

+∞ and -∞ are not numbers, when we say something is approaching +∞ or -∞, we are not giving an exact numerical answer.That is why the function is unbounded and the limit does not exist

Special Cases with Radicals

What about functions where direct substitution doesn’t work?

If you can’t divide out factors, try to rationalize the numerator to get a larger expression in the denominator.

To rationalize the numerator, you use the same strategies that you use to rationalize the denominator, but this time you are actually trying to put the radical into the denominator!

0

1 1limx

x

x

0

1 1 1 1lim

1 1x

x x

x x

0

1 1lim

1 1x

x

x x

Ex.

0lim

1 1x

x

x x

0

1lim

1 1x x

1 1 1 1

1 1 21 10 1 1

0

4 2limx

x

x

0

4 2 4 2lim

4 2x

x x

x x

0

4 4lim

4 2x

x

x x

Ex.

0lim

4 2x

x

x x

0

1lim

4 2x x

1 1 1 1

2 2 44 20 4 2

Homework Problems:

1. 2.

3. 4.