# Limit Worksheets

Limit Worksheets
• Page 1
1.
Evaluate $\underset{x\to 7}{\mathrm{lim}}$ .
 a. 10 b. 8 c. 9 d. 7

#### Solution:

limx7 8x -56x - 7

= limx78(x - 7)x - 7

= limx7 8

= 8.

2.
Find the value of $\underset{x\to 7}{\mathrm{lim}}$ .
 a. - $\frac{1}{5}$ b. - $\frac{1}{7}$ c. does not exist

#### Solution:

limx7 x - 7x2 - 14x + 49

= limx7 x - 7(x - 7)2

= limx7 1x - 7

= ± ∞

So , the limit does not exist.

3.
What is the value of $\underset{x\to 5}{\mathrm{lim}}$ ?
 a. - 5 b. $\frac{5}{4}$ c. $\frac{5}{6}$

#### Solution:

limx5 5(x - 5)x2 - 6x + 5

= limx5 5(x - 5)(x - 1)(x - 5)
[Factor x2 - 6 x + 5.]

= limx5 5x - 1

= 55 - 1

= 5 / 4

4.
Find the value of $\underset{x\to 0}{\mathrm{lim}}$.
 a. 2 b. 3 c. -1

#### Solution:

limx02x - 3x2x

= limx0x(2 - 3x)x
[Factor 2x - 3x2.]

= limx0 (2 - 3x)

= 2 - 3(0)

= 2.

5.
Evaluate $\underset{x\to 3}{\mathrm{lim}}$ .
 a. 6 b. 3 c. ∞

#### Solution:

limx3 x2 - 9x - 3

= limx3 (x - 3)(x + 3)x - 3
[Factor x2 - 9.]

= limx3 (x + 3)

= (3 + 3) = 6.

6.
Evaluate $\underset{x\to 1}{\mathrm{lim}}$.
 a. $\frac{1}{\sqrt{82}}$ b. does not exist c. ∞

#### Solution:

limx1x2 + 81 -82x - 1

= limx1 (x2 + 81-82)(x2 + 81 +82)(x - 1)(x2 + 81 +82)
[Rationalize the numerator.]

= limx1 x2 - 1(x - 1)(x2 + 81 +82)

= limx1 (x +1)(x - 1)(x - 1)(x2 + 81 +82)
[Factor (x2 - 1).]

= limx1 x + 1(x2 + 81 +82)
[Cancel the common factor.]

= 2282 = 182.

7.
Find the equation of the tangent to the curve $f$ ($x$) = $x$2 such that $f$ (8) = 64.
 a. $y$ - 16$x$ - 64 = 0 b. $y$ = 16$x$ + 64 c. 16$x$ - $y$ - 64 = 0 d. None of the above

#### Solution:

f (x) = x2, f (8) = 64.

limh0f (8 + h) - f (8)h = limh0(8 + h )2 - 64h = limh0 64 + 16h +h2 - 64h

= limh016 + h =16.

So, the slope of the tangent line to the curve at the point (8, 64) = 16.

The equation of the tangent line to the curve at the point (8, 64) is y - 64 = 16 (x - 8) or y = 16x - 64.
[Use slope point form.]

8.
Find the equation of the tangent to the curve $f$ ($x$) = 1 - 2$x$ + $x$2 ; $f$ (-4) = 25.
 a. $y$ - 2$x$ = 0 b. $y$ + $x$ = 0 c. $y$ + 10 $x$ = -15 d. $y$ - 10$x$ = 0

#### Solution:

f (x) = 1 - 2x + x2, f (-4) = 25

f (-4 + h) - f (-4)h = 1 - 2(-4 + h) +(-4 + h)2 - 25h = h - 10

limh0 f (-4 + h) - f (-4)h = limh0 ( h - 10) = - 10

So, the slope of the tangent to the curve at the point (-4, 25) = - 10

The equation of the tangent line to the curve at the point (-4, 25) is y - 25 = - 10(x-(-4)) y + 10x = -15.
[Use slope point form.]

9.
Find the equation of the tangent to the curve $f$ ($x$) = $\sqrt{x}$, $f$ (8) = $\sqrt{8}$.
 a. $y$ = $x$ + $\frac{\sqrt{8}}{2}$ b. $y$ = $\frac{x}{2\sqrt{8}}$ + $\frac{\sqrt{8}}{2}$ c. $y$ = $x$ - 1 d. $y$ = $x$ - $\frac{1}{2}$

#### Solution:

f (x) = x; f (8) = 8

f (8 + h) - f (8)h = 8 + h -8 h

= 8 + h -8h . 8 + h +88 + h +8

= hh(8 + h +8)

= 18 + h +8

limh0 f (1 + h) - f (1)h = limh0 18 + h +8 = 128

So , the slope of the tangent line to the curve at the point (8,8) = 128 .

The equation of the tangent line to the curve at the point (8,8) is y - 8 = 128(x - 8) or y = x28 + 82.
[Use slope point form.]

10.
Which of the following is the tangent to the curve $f$ ($x$) = $x$5 + 4 such that $f$ (0) = 4?
 a. $y$ + $x$ = 1 b. $y$ = 0 c. $y$ = $x$ d. $y$ = 4

#### Solution:

f (x) = x5 + 4; f (0) = 4

f (0 + h) - f (0)h = (h5 + 4) - (0 + 4)h = h4

limh0 f (0 + h) - f (0)h = limh0 h4 = 0

So, The slope of the tangent line to the curve at the poiunt (0,4) = 0.

The equation of the tangent line to the curve at the point (0,4) is y - 4 = 0(x - 0) y = 4
[Use slope point form.]