Exercise 01 Solutions

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Exercise 01 Solutions: Limits

By: Andrew KerrAvailable at www.engineeringcorps.org

Answers

1. limx0 x2 sin(x) = 0

2. limx1

x= 0

3. limx0 x2 cos 1

x= 0

4. limx2 x2 + 2x + 4 = 12

5. limx0ex+1

2= 1

6. limx2x2+6x+8

x+2= 2

7. limx 6x4 3x2 15x =

8. limx8x3+9x2+2

9x3+6= 8

9

9. limx3x2+3x18

x2+5x24

= 911

10. limx 5x2

+815x3+9x2+6x = 0

Solutions

1. limx0

x2 sin(x) = 0

To evaluate this limit, we need to use Squeeze Theorem. Using our knowledge

of trigonometry, we know that the bounds of sin(x) are -1 and 1. Setting up

our inequality we get:

1 sin(x) 1

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Now, we multiply through by x2 and get:

x2 x

2 sin(x) x2

Plugging in 0 for x we get:

0 x2 sin(x) 0

So, our graph ofx2 sin(x) approaches 0 as x approaches 0. Therefore, limx0 x2 sin(x)

is equal to 0.

2. limx

1

x

= 0

For this problem, we plug straight into 1x

and we get:

1

Whenever a constant is divided by a really large number (i.e. ), the answer

is 0. Therefore, limx1

xis equal to 0.

3. limx0

x2 cos

1

x

= 0

To evaluate this limit, we need to use Squeeze Theorem. Using our knowledge

of trigonometry, we know that the bounds of cos(x) are -1 and 1. Setting up

our inequality we get:

1 cos(1

x

) 1

NOTE: Since 1x

is inside cos(x), cos(x) is still going to be between -1 and 1.

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Now, we multiply through by x2 and get:

x2 x

2 cos(1

x

) x2

Plugging in 0 for x we get:

0 x2 cos(1

x

) 0

So, our graph ofx2 cos( 1x

) approaches 0 as x approaches 0. Therefore, limx0 x2 cos 1

x

is equal to 0.

4. limx2

x2 + 2x + 4 = 12

To evaluate this limit, we can just plug 2 directly into the equation and get our

answer: 12.

5. limx0

e

x

+ 12

= 1

Our first step to evaluate this limit is to plug in zero:

e0 + 1

2

Since e0 is equal to 1, we are left with 1+12

which is just 1.

6. limx2

x2 + 6x + 8

x + 2= 2

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To solve this limit, let us plug in -2 into the problem:

(2)2 + 6(2) + 8

2 + 2

Plugging in -2 for x makes the bottom 0 which is unsolvable! But, lets factor

and see if anything can cancel:

(x + 4)(x + 2)

x + 2

The x+ 2 on the top and bottom can cancel just leaving us with limx2 x+ 4.

Plugging in -2 for x, we get 2.

***

Extra: What type of discontinuity does x2+6x+8

x+2have and where?

There is a removable discontinuity at x = 2, which is called a hole.

7.

limx 6x4

3x2

15x

=

Before solving this problem, lets factor out an x4:

limx

x4(6

3

x2

15

x3

)

Now we can easily evaluate the limit by looking at each individual term. limx x4 =

, limx3

x2 = 0, and limx

15

x3 = 0. Now, lets plug in each 0 limit into

the equation: limxx4(6 0 0) which gives us:

limx

6x4

Which is just .

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8. limx

8x3 + 9x2 + 29x3 + 6

= 89

For this problem, lets look at the degree on both the top and the bottom. On

top, we have a degree of 3 and, on the bottom, we have the same. This means

that we can just take the leading coefficients: 8 and -9. But, there is a twist.

Since the degree is odd and we are going to , we need to multiply both

coefficients by -1. Therefore, we get 89

as our answer.

9. limx3

x2 + 3x 18

x2 + 5x 24

=9

11

When we plug in 3, we get 0 in the denominator, which is not allowed. To solve

this limit, lets factor out both the numerator and the denominator and see if

anything cancels:

limx3

(x 3)(x + 6)

(x 3)(x + 8)

Since x3 is in both the numerator and the denominator, we can cancel it out.

After cancelling, we get:

limx3

x + 6

x + 8

Plugging in 3 for x, we get 911

, which is our answer.

***

Extra: What types of discontinuities does x2+3x18

x2+5x24

have and

where?

There is a hole at x = 3 and an asymptote at x = 8.

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10. limx

5x2 + 8

15x3

+ 9x2

+ 6x

= 0

To solve this limit, we look at the degrees on both the top and the bottom.

Since the degree on the top is less than the degree on the bottom, the limit is

just 0.

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Exercise 01 Solutions