Calculus/Limits/Exercises

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	Limits/Exercises

Basic Limit Exercises

1.

lim_{x \to 2} [4 x^{2} - 3 x - 1]

9

9

2.

lim_{x \to 5} [x^{2}]

25

25

3.

lim_{x \to \frac{π}{4}} [\cos^{2} (x)]

1 / 2

1 / 2

4.

lim_{x \to 1} [5 e^{x - 1} - 5]

0

0

Solutions

One-Sided Limits

Evaluate the following limits or state that the limit does not exist.

5.

lim_{x \to 0^{-}} \frac{x^{3} + x^{2}}{x^{3} + 2 x^{2}}

\frac{1}{2}

\frac{1}{2}

6.

lim_{x \to 7^{-}} [| x^{2} + x | - x]

49

49

7.

lim_{x \to - 1^{+}} \sqrt{1 - x^{2}}

0

0

8.

lim_{x \to - 1^{-}} \sqrt{1 - x^{2}}

The limit does not exist.

9.

lim_{x \to 0^{-}} \frac{| x |}{x}

- 1

- 1

10.

lim_{x \to 1^{-}} \arcsin (x)

\frac{π}{2}

\frac{π}{2}

Solutions

Two-Sided Limits

Evaluate the following limits or state that the limit does not exist.

11.

lim_{x \to - 1} \frac{1}{x - 1}

- \frac{1}{2}

- \frac{1}{2}

12.

lim_{x \to 4} \frac{1}{x - 4}

The limit does not exist.

13.

lim_{x \to 2} \frac{1}{x - 2}

The limit does not exist.

14.

lim_{x \to - 3} \frac{x^{2} - 9}{x + 3}

- 6

- 6

15.

lim_{x \to 3} \frac{x^{2} - 9}{x - 3}

6

6

16.

lim_{x \to - 1} \frac{x^{2} + 2 x + 1}{x + 1}

0

0

17.

lim_{x \to - 1} \frac{x^{3} + 1}{x + 1}

3

3

18.

lim_{x \to 4} \frac{x^{2} + 5 x - 36}{x^{2} - 16}

\frac{13}{8}

\frac{13}{8}

19.

lim_{x \to 25} \frac{x - 25}{\sqrt{x} - 5}

10

10

20.

lim_{x \to 0} \frac{| x |}{x}

The limit does not exist.

21.

lim_{x \to 2} \frac{1}{(x - 2)^{2}}

\infty

\infty

22.

lim_{x \to 3} \frac{\sqrt{x^{2} + 16}}{x - 3}

The limit does not exist.

23.

lim_{x \to - 2} \frac{3 x^{2} - 8 x - 3}{2 x^{2} - 18}

- \frac{5}{2}

- \frac{5}{2}

24.

lim_{x \to 2} \frac{x^{2} + 2 x + 1}{x^{2} - 2 x + 1}

9

9

25.

lim_{x \to 3} \frac{x + 3}{x^{2} - 9}

The limit does not exist.

26.

lim_{x \to - 1} \frac{x + 1}{x^{2} + x}

- 1

- 1

27.

lim_{x \to 1} \frac{1}{x^{2} + 1}

\frac{1}{2}

\frac{1}{2}

28.

lim_{x \to 1} [x^{2} + 5 x - \frac{1}{2 - x}]

5

5

29.

lim_{x \to 0} \frac{x^{2}}{x^{2} + 2 x - 3}

0

0

30.

lim_{x \to 1} \frac{x^{2} - 1}{x^{2} + 2 x - 3}

\frac{1}{2}

\frac{1}{2}

31.

lim_{x \to 1} \frac{5 x}{x^{2} + 2 x - 3}

The limit does not exist.

32.

lim_{x \to - 1} \ln (\sqrt{x^{2} - 1})

The limit does not exist.

33.

lim_{x \to 1} \arcsin (x)

The limit does not exist.

Solutions

Limits to Infinity

Evaluate the following limits or state that the limit does not exist.

34.

lim_{x \to \infty} \frac{- x + π}{x^{2} + 3 x + 2}

0

0

35.

lim_{x \to - \infty} \frac{x^{2} + 2 x + 1}{3 x^{2} + 1}

\frac{1}{3}

\frac{1}{3}

36.

lim_{x \to - \infty} \frac{3 x^{2} + x}{2 x^{2} - 15}

\frac{3}{2}

\frac{3}{2}

37.

lim_{x \to - \infty} [3 x^{2} - 2 x + 1]

\infty

\infty

38.

lim_{x \to \infty} \frac{2 x^{2} - 32}{x^{3} - 64}

0

0

39.

lim_{x \to \infty} 6

6

6

40.

lim_{x \to \infty} \frac{3 x^{2} + 4 x}{x^{4} + 2}

0

0

41.

lim_{x \to - \infty} \frac{2 x + 3 x^{2} + 1}{2 x^{2} + 3}

\frac{3}{2}

\frac{3}{2}

42.

lim_{x \to - \infty} \frac{x^{3} - 3 x^{2} + 1}{3 x^{2} + x + 5}

- \infty

- \infty

43.

lim_{x \to \infty} \frac{x^{2} + 2}{x^{3} - 2}

0

0

44.

lim_{x \to \infty} \frac{\sin (\frac{1}{x})}{x}

0

0

45.

lim_{x \to - \infty} x^{2} \cos (\frac{1}{x})

\infty

\infty

46.

lim_{x \to \infty} \frac{\sin (\arctan (x))}{\arctan (- x)}

- \frac{2}{π}

- \frac{2}{π}

Solutions

Limits of Piecewise Functions

Evaluate the following limits or state that the limit does not exist.

48. Consider the function

f (x) = {\begin{matrix} (x - 2)^{2} & if x < 2 \\ x - 3 & if x \geq 2 \end{matrix}

a.

lim_{x \to 2^{-}} f (x)

0

0

b.

lim_{x \to 2^{+}} f (x)

- 1

- 1

c.

lim_{x \to 2} f (x)

The limit does not exist

49. Consider the function

g (x) = {\begin{matrix} - 2 x + 1 & if x \leq 0 \\ x + 1 & if 0 < x < 4 \\ x^{2} + 2 & if x \geq 4 \end{matrix}

a.

lim_{x \to 4^{+}} g (x)

18

18

b.

lim_{x \to 4^{-}} g (x)

5

5

c.

lim_{x \to 0^{+}} g (x)

1

1

d.

lim_{x \to 0^{-}} g (x)

1

1

e.

lim_{x \to 0} g (x)

1

1

f.

lim_{x \to 1} g (x)

2

2

50. Consider the function

h (x) = {\begin{matrix} 2 x - 3 & if x < 2 \\ 8 & if x = 2 \\ - x + 3 & if x > 2 \end{matrix}

a.

lim_{x \to 0} h (x)

- 3

- 3

b.

lim_{x \to 2^{-}} h (x)

1

1

c.

lim_{x \to 2^{+}} h (x)

1

1

d.

lim_{x \to 2} h (x)

1

1

Solutions

Intermediate Value Theorem

51. Use the intermediate value theorem to show that there exists a value

f (x) = 3

for

f (x) = \sqrt{x} + \frac{1}{\sqrt{x}}

from

1 \leq x \leq 9

. If you cannot use the intermediate value theorem to show this, explain why.

Notice

f (x)

is continuous from

x \in [1, 9]

. Ergo, the intermediate value theorem applies. For all

y \in [2, \frac{10}{3}]

, there exists a

c \in (1, 9)

so that

f (c) = 3

.

◼

Notice

f (x)

is continuous from

x \in [1, 9]

. Ergo, the intermediate value theorem applies. For all

y \in [2, \frac{10}{3}]

, there exists a

c \in (1, 9)

so that

f (c) = 3

.

◼

52. Use the intermediate value theorem to show that there exists an

x = μ

so that

f (μ) = \frac{π}{4}

for

f (x) = \frac{\sin (x) - \cos (x)}{\ln x}

from

π \leq x \leq \frac{9}{4} π

. If you cannot use the intermediate value theorem to show this, explain why.

Notice

f (x)

is continuous from

x \in [π, \frac{9}{4} π]

. Ergo, the intermediate value theorem applies.

f (π) = \frac{1}{\ln (π)}

f (\frac{9}{4} π) = 0

It is known the following is true: $1 \geq \frac{π}{4} \geq 0$ . From there, we can directly argue the following:

\begin{matrix} \frac{1}{\ln (π)} \geq 1 & \Rightarrow \frac{1}{\ln (π)} \geq 1 \geq \frac{π}{4} \geq 0 \\ \Rightarrow \frac{1}{\ln (π)} \geq \frac{π}{4} \geq 0 \end{matrix}

By the intermediate value theorem, if

f (x)

is continuous from

x \in [π, \frac{9}{4} π]

, then there exists an

x = μ

so that

f (\frac{9}{4} π) \leq f (μ) \leq f (π)

for

μ \in [π, \frac{9}{4} π]

.

◼

Notice

f (x)

is continuous from

x \in [π, \frac{9}{4} π]

. Ergo, the intermediate value theorem applies.

f (π) = \frac{1}{\ln (π)}

f (\frac{9}{4} π) = 0

It is known the following is true: $1 \geq \frac{π}{4} \geq 0$ . From there, we can directly argue the following:

\begin{matrix} \frac{1}{\ln (π)} \geq 1 & \Rightarrow \frac{1}{\ln (π)} \geq 1 \geq \frac{π}{4} \geq 0 \\ \Rightarrow \frac{1}{\ln (π)} \geq \frac{π}{4} \geq 0 \end{matrix}

By the intermediate value theorem, if

f (x)

is continuous from

x \in [π, \frac{9}{4} π]

, then there exists an

x = μ

so that

f (\frac{9}{4} π) \leq f (μ) \leq f (π)

for

μ \in [π, \frac{9}{4} π]

.

◼

53. Use the intermediate value theorem to show that there exists a value

x = Γ

so that

f (Γ) = 2

for

f (x) = \sqrt{- 2 x + 4} + 5 \ln (x^{2})

from

- 1 \leq x \leq 1

. If you cannot use the intermediate value theorem to show this, explain why.

Notice

f (x)

is not continuous for

x = 0

since

lim_{x \to 0} f (x)

is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.

Notice

f (x)

is not continuous for

x = 0

since

lim_{x \to 0} f (x)

is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.

Solutions

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