# Calculus/Limits/Exercises

 ← Proofs of Some Basic Limit Rules Calculus Differentiation → Limits/Exercises

## Basic Limit Exercises

1. ${\displaystyle \lim _{x\to 2}{\Big [}4x^{2}-3x-1{\Big ]}}$
${\displaystyle 9}$
${\displaystyle 9}$
2. ${\displaystyle \lim _{x\to 5}{\Big [}x^{2}{\Big ]}}$
${\displaystyle 25}$
${\displaystyle 25}$
3. ${\displaystyle \lim _{x\to {\frac {\pi }{4}}}{\Big [}\cos ^{2}(x){\Big ]}}$
${\displaystyle 1/2}$
${\displaystyle 1/2}$
4. ${\displaystyle \lim _{x\to 1}{\Big [}5e^{x-1}-5{\Big ]}}$
${\displaystyle 0}$
${\displaystyle 0}$

## One-Sided Limits

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

5. ${\displaystyle \lim _{x\to 0^{-}}{\frac {x^{3}+x^{2}}{x^{3}+2x^{2}}}}$
${\displaystyle {\frac {1}{2}}}$
${\displaystyle {\frac {1}{2}}}$
6. ${\displaystyle \lim _{x\to 7^{-}}{\Big [}|x^{2}+x|-x{\Big ]}}$
${\displaystyle 49}$
${\displaystyle 49}$
7. ${\displaystyle \lim _{x\to -1^{+}}{\sqrt {1-x^{2}}}}$
${\displaystyle 0}$
${\displaystyle 0}$
8. ${\displaystyle \lim _{x\to -1^{-}}{\sqrt {1-x^{2}}}}$
The limit does not exist.
The limit does not exist.
9. ${\displaystyle \lim _{x\to 0^{-}}{\frac {|x|}{x}}}$
${\displaystyle -1}$
${\displaystyle -1}$
10. ${\displaystyle \lim _{x\to 1^{-}}\arcsin(x)}$
${\displaystyle {\frac {\pi }{2}}}$
${\displaystyle {\frac {\pi }{2}}}$

## Two-Sided Limits

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

11. ${\displaystyle \lim _{x\to -1}{\frac {1}{x-1}}}$
${\displaystyle -{\frac {1}{2}}}$
${\displaystyle -{\frac {1}{2}}}$
12. ${\displaystyle \lim _{x\to 4}{\frac {1}{x-4}}}$
The limit does not exist.
The limit does not exist.
13. ${\displaystyle \lim _{x\to 2}{\frac {1}{x-2}}}$
The limit does not exist.
The limit does not exist.
14. ${\displaystyle \lim _{x\to -3}{\frac {x^{2}-9}{x+3}}}$
${\displaystyle -6}$
${\displaystyle -6}$
15. ${\displaystyle \lim _{x\to 3}{\frac {x^{2}-9}{x-3}}}$
${\displaystyle 6}$
${\displaystyle 6}$
16. ${\displaystyle \lim _{x\to -1}{\frac {x^{2}+2x+1}{x+1}}}$
${\displaystyle 0}$
${\displaystyle 0}$
17. ${\displaystyle \lim _{x\to -1}{\frac {x^{3}+1}{x+1}}}$
${\displaystyle 3}$
${\displaystyle 3}$
18. ${\displaystyle \lim _{x\to 4}{\frac {x^{2}+5x-36}{x^{2}-16}}}$
${\displaystyle {\frac {13}{8}}}$
${\displaystyle {\frac {13}{8}}}$
19. ${\displaystyle \lim _{x\to 25}{\frac {x-25}{{\sqrt {x}}-5}}}$
${\displaystyle 10}$
${\displaystyle 10}$
20. ${\displaystyle \lim _{x\to 0}{\frac {|x|}{x}}}$
The limit does not exist.
The limit does not exist.
21. ${\displaystyle \lim _{x\to 2}{\frac {1}{(x-2)^{2}}}}$
${\displaystyle \infty }$
${\displaystyle \infty }$
22. ${\displaystyle \lim _{x\to 3}{\frac {\sqrt {x^{2}+16}}{x-3}}}$
The limit does not exist.
The limit does not exist.
23. ${\displaystyle \lim _{x\to -2}{\frac {3x^{2}-8x-3}{2x^{2}-18}}}$
${\displaystyle -{\frac {5}{2}}}$
${\displaystyle -{\frac {5}{2}}}$
24. ${\displaystyle \lim _{x\to 2}{\frac {x^{2}+2x+1}{x^{2}-2x+1}}}$
${\displaystyle 9}$
${\displaystyle 9}$
25. ${\displaystyle \lim _{x\to 3}{\frac {x+3}{x^{2}-9}}}$
The limit does not exist.
The limit does not exist.
26. ${\displaystyle \lim _{x\to -1}{\frac {x+1}{x^{2}+x}}}$
${\displaystyle -1}$
${\displaystyle -1}$
27. ${\displaystyle \lim _{x\to 1}{\frac {1}{x^{2}+1}}}$
${\displaystyle {\frac {1}{2}}}$
${\displaystyle {\frac {1}{2}}}$
28. ${\displaystyle \lim _{x\to 1}\left[x^{2}+5x-{\frac {1}{2-x}}\right]}$
${\displaystyle 5}$
${\displaystyle 5}$
29. ${\displaystyle \lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}}$
${\displaystyle 0}$
${\displaystyle 0}$
30. ${\displaystyle \lim _{x\to 1}{\frac {x^{2}-1}{x^{2}+2x-3}}}$
${\displaystyle {\frac {1}{2}}}$
${\displaystyle {\frac {1}{2}}}$
31. ${\displaystyle \lim _{x\to 1}{\frac {5x}{x^{2}+2x-3}}}$
The limit does not exist.
The limit does not exist.
32. ${\displaystyle \lim _{x\to -1}\ln({\sqrt {x^{2}-1}})}$
The limit does not exist.
The limit does not exist.
33. ${\displaystyle \lim _{x\to 1}\arcsin(x)}$
The limit does not exist.
The limit does not exist.

## Limits to Infinity

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

34. ${\displaystyle \lim _{x\to \infty }{\frac {-x+\pi }{x^{2}+3x+2}}}$
${\displaystyle 0}$
${\displaystyle 0}$
35. ${\displaystyle \lim _{x\to -\infty }{\frac {x^{2}+2x+1}{3x^{2}+1}}}$
${\displaystyle {\frac {1}{3}}}$
${\displaystyle {\frac {1}{3}}}$
36. ${\displaystyle \lim _{x\to -\infty }{\frac {3x^{2}+x}{2x^{2}-15}}}$
${\displaystyle {\frac {3}{2}}}$
${\displaystyle {\frac {3}{2}}}$
37. ${\displaystyle \lim _{x\to -\infty }{\Big [}3x^{2}-2x+1{\Big ]}}$
${\displaystyle \infty }$
${\displaystyle \infty }$
38. ${\displaystyle \lim _{x\to \infty }{\frac {2x^{2}-32}{x^{3}-64}}}$
${\displaystyle 0}$
${\displaystyle 0}$
39. ${\displaystyle \lim _{x\to \infty }6}$
${\displaystyle 6}$
${\displaystyle 6}$
40. ${\displaystyle \lim _{x\to \infty }{\frac {3x^{2}+4x}{x^{4}+2}}}$
${\displaystyle 0}$
${\displaystyle 0}$
41. ${\displaystyle \lim _{x\to -\infty }{\frac {2x+3x^{2}+1}{2x^{2}+3}}}$
${\displaystyle {\frac {3}{2}}}$
${\displaystyle {\frac {3}{2}}}$
42. ${\displaystyle \lim _{x\to -\infty }{\frac {x^{3}-3x^{2}+1}{3x^{2}+x+5}}}$
${\displaystyle -\infty }$
${\displaystyle -\infty }$
43. ${\displaystyle \lim _{x\to \infty }{\frac {x^{2}+2}{x^{3}-2}}}$
${\displaystyle 0}$
${\displaystyle 0}$
44. ${\displaystyle \lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}}$
${\displaystyle 0}$
${\displaystyle 0}$
45. ${\displaystyle \lim _{x\to -\infty }x^{2}\cos \left({\frac {1}{x}}\right)}$
${\displaystyle \infty }$
${\displaystyle \infty }$
46. ${\displaystyle \lim _{x\to \infty }{\frac {\sin(\arctan(x))}{\arctan(-x)}}}$
${\displaystyle -{\frac {2}{\pi }}}$
${\displaystyle -{\frac {2}{\pi }}}$

## Limits of Piecewise Functions

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

48. Consider the function

${\displaystyle f(x)={\begin{cases}(x-2)^{2}&{\text{if }}x<2\\x-3&{\text{if }}x\geq 2\end{cases}}}$
a. ${\displaystyle \lim _{x\to 2^{-}}f(x)}$
${\displaystyle 0}$
${\displaystyle 0}$
b. ${\displaystyle \lim _{x\to 2^{+}}f(x)}$
${\displaystyle -1}$
${\displaystyle -1}$
c. ${\displaystyle \lim _{x\to 2}f(x)}$
The limit does not exist
The limit does not exist

49. Consider the function

${\displaystyle g(x)={\begin{cases}-2x+1&{\text{if }}x\leq 0\\x+1&{\text{if }}0
a. ${\displaystyle \lim _{x\to 4^{+}}g(x)}$
${\displaystyle 18}$
${\displaystyle 18}$
b. ${\displaystyle \lim _{x\to 4^{-}}g(x)}$
${\displaystyle 5}$
${\displaystyle 5}$
c. ${\displaystyle \lim _{x\to 0^{+}}g(x)}$
${\displaystyle 1}$
${\displaystyle 1}$
d. ${\displaystyle \lim _{x\to 0^{-}}g(x)}$
${\displaystyle 1}$
${\displaystyle 1}$
e. ${\displaystyle \lim _{x\to 0}g(x)}$
${\displaystyle 1}$
${\displaystyle 1}$
f. ${\displaystyle \lim _{x\to 1}g(x)}$
${\displaystyle 2}$
${\displaystyle 2}$

50. Consider the function

${\displaystyle h(x)={\begin{cases}2x-3&{\text{if }}x<2\\8&{\text{if }}x=2\\-x+3&{\text{if }}x>2\end{cases}}}$
a. ${\displaystyle \lim _{x\to 0}h(x)}$
${\displaystyle -3}$
${\displaystyle -3}$
b. ${\displaystyle \lim _{x\to 2^{-}}h(x)}$
${\displaystyle 1}$
${\displaystyle 1}$
c. ${\displaystyle \lim _{x\to 2^{+}}h(x)}$
${\displaystyle 1}$
${\displaystyle 1}$
d. ${\displaystyle \lim _{x\to 2}h(x)}$
${\displaystyle 1}$
${\displaystyle 1}$

## Intermediate Value Theorem

51. Use the intermediate value theorem to show that there exists a value ${\displaystyle f(x)=3}$ for ${\displaystyle f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}}$ from ${\displaystyle 1\leq x\leq 9}$. If you cannot use the intermediate value theorem to show this, explain why.
Notice ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[1,9\right]}$. Ergo, the intermediate value theorem applies. For all ${\displaystyle y\in \left[2,{\frac {10}{3}}\right]}$, there exists a ${\displaystyle c\in (1,9)}$ so that ${\displaystyle f(c)=3}$. ${\displaystyle \blacksquare }$
Notice ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[1,9\right]}$. Ergo, the intermediate value theorem applies. For all ${\displaystyle y\in \left[2,{\frac {10}{3}}\right]}$, there exists a ${\displaystyle c\in (1,9)}$ so that ${\displaystyle f(c)=3}$. ${\displaystyle \blacksquare }$
52. Use the intermediate value theorem to show that there exists an ${\displaystyle x=\mu }$ so that ${\displaystyle f(\mu )={\frac {\pi }{4}}}$ for ${\displaystyle f(x)={\frac {\sin(x)-\cos(x)}{\ln {x}}}}$ from ${\displaystyle \pi \leq x\leq {\frac {9}{4}}\pi }$. If you cannot use the intermediate value theorem to show this, explain why.
Notice ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}$. Ergo, the intermediate value theorem applies.
${\displaystyle f(\pi )={\frac {1}{\ln(\pi )}}}$
${\displaystyle f\left({\frac {9}{4}}\pi \right)=0}$

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

{\displaystyle {\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}}
By the intermediate value theorem, if ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}$, then there exists an ${\displaystyle x=\mu }$ so that ${\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}$ for ${\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}$. ${\displaystyle \blacksquare }$
Notice ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}$. Ergo, the intermediate value theorem applies.
${\displaystyle f(\pi )={\frac {1}{\ln(\pi )}}}$
${\displaystyle f\left({\frac {9}{4}}\pi \right)=0}$

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

{\displaystyle {\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}}
By the intermediate value theorem, if ${\displaystyle f(x)}$ is continuous from ${\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}$, then there exists an ${\displaystyle x=\mu }$ so that ${\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}$ for ${\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}$. ${\displaystyle \blacksquare }$
53. Use the intermediate value theorem to show that there exists a value ${\displaystyle x=\Gamma }$ so that ${\displaystyle f(\Gamma )=2}$ for ${\displaystyle f(x)={\sqrt {-2x+4}}+5\ln \left(x^{2}\right)}$ from ${\displaystyle -1\leq x\leq 1}$. If you cannot use the intermediate value theorem to show this, explain why.
Notice ${\displaystyle f(x)}$ is not continuous for ${\displaystyle x=0}$ since ${\displaystyle \lim _{x\to 0}f(x)}$ is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.
Notice ${\displaystyle f(x)}$ is not continuous for ${\displaystyle x=0}$ since ${\displaystyle \lim _{x\to 0}f(x)}$ is unbounded. Ergo, the intermediate value theorem cannot be used to solve this problem.