Limit Examples

8/10/2019 Limit Examples

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Limit examples

Example 1

Evaluatelimx 4

x 2

x 2 4If we try direct substitution, we end up with 160 (i.e., a non-zero constant over zero), sowell get either + or as we approach 4. We then need to check left- and right-handlimits to see which one it is, and to make sure the limits are equal from both sides.

Left-hand limit:lim

x 4x 2

(x 4)(x + 4)

As x 4 , the function is negative since (+)2

( )(+) = ( ), so the left-hand limit is .

Right-hand limit:

limx 4+

x 2

(x 4)(x + 4)

As x 4+ , the function is positive since (+)2

(+)(+) = (+), so the right-hand limit is + .

Since the left- and right-hand limits are not equal,

limx 4

x 2

x 2 4 DNE

Example 2

Evaluatelim

x 3

x 2 2x 3x 2 + 6 x + 9

If we try direct substitution, we end up with 120 , so well get either + or as weapproach -3. As in the last example, we need to check left- and right-hand limits to seewhich one it is, and to make sure the limits are equal from both sides.

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Left-hand limit:lim

x 3(x 3)(x + 1)

(x + 3) 2

As x 3 , the function is positive since ( )( )( )2 = (+)(+) = (+), so the left-hand limit

is + .

Right-hand limit:

limx 3+

(x 3)(x + 1)(x + 3) 2

As x 3+ , the function is positive since ( )( )(+) 2 = (+)(+) = (+), so the right-hand limit

is also + .

Since the left- and right-hand limits are the same,

limx 4

x 2 2x 3x 2 + 6 x + 9

=

Example 3

Evaluatelim

x 0+

2sin(x )

First of all, we note that direct substitution fails (we get 20 ). There are a couple of differentways we can look at this problem. For either one, we observe that as x 0+ , sin(x ) also

goes to zero from values greater than zero (i.e., sin( x ) 0+ ): So, limx 0+

2sin(x )

is either +

or . From what we observed above, we know the function will be (+)(+) = (+), so the limitis + .

The other way we can approach this is to replace sin( x ) with another variable that goesto the same value as sin( x ) when we take the limit. Since sin( x ) 0+ as x 0+ , then

limx 0+

2sin(x )

= limt 0+

2t

( which still = ).

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Example 4

Evaluatelim

x

x cos(x )x

We are eventually going to use the Squeeze Theorem on this example. There are a couple of ways to approach this; the part of the function being squeezed will be different in each case,but the end result is the same.

1. We rst rewrite the function by dividing the numerator through by x , and then uselimit laws to split into two separate limits:

limx

x cos(x )x

= limx

1 cos(x )

x= lim

x 1 lim

x

cos(x )x

= 1 limx

cos(x )x

We now use the Squeeze Theorem on the remaining limit:

We know that 1 cos(x ) 1

Since x + , x is positive, dividing this inequality through by x wont change theinequalities:

1x

cos(x )x

1x

So,

limx

1x

limx

cos(x )x

limx

1x

The outer limits are both 0, so by the Squeeze Theorem,

limx

cos(x )x

= 0 ,

and thuslim

x

x cos(x )x

= 1 0 = 1

2. In this second approach, we start into the Squeeze Theorem right away:

1 cos(x ) 1

So

1

cos(x )

1,which is the same as

1 cos(x ) 1.

Adding x to all sides gives us

x 1 x cos(x ) x + 1 ,

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To show this one formally, we rst note that as x , then

x 2

as well, so

x2

and3 x 2

also. So, we can replace the 3 x 2 in the exponent with another variable (say, t ) that goesto without changing the limit, i.e.,

limx

e3x 2 = lim

t e

t (= 0 by properties mentioned in class).

Example 7

Evaluatelim

x

1e x

In this example, we rst rewrite the limit as

limx

e x

,

which is + from properties mentioned in class.

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Limit Examples

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