# Calculus/Limits/Exercises
## Basic Limit Exercises

$\lim _{x\to 2}{\Big4x^{2}-3x-1{\Big ]}$
$9$
$9$
$\lim _{x\to 5}{\Bigx^{2}{\Big ]}$
$25$
$25$
$\lim _{x\to {\frac {\pi }{4}}}{\Big\cos ^{2}(x){\Big ]}$
$1/2$
$1/2$
$\lim _{x\to 1}{\Big5e^{x-1}-5{\Big ]}$
$0$
$0$
Solutions

## One-Sided Limits

Evaluate the following limits or state that the limit does not exist.
$\lim _{x\to 0^{-}}{\frac {x^{3}+x^{2}}{x^{3}+2x^{2}}}$
${\frac {1}{2}}$
${\frac {1}{2}}$
$\lim _{x\to 7^{-}}{\Big|x^{2}+x|-x{\Big ]}$
$49$
$49$
$\lim _{x\to -1^{+}}{\sqrt {1-x^{2}}}$
$0$
$0$
$\lim _{x\to -1^{-}}{\sqrt {1-x^{2}}}$
$\lim _{x\to 0^{-}}{\frac {|x|}{x}}$
$-1$
$-1$
$\lim _{x\to 1^{-}}\arcsin(x)$
${\frac {\pi }{2}}$
${\frac {\pi }{2}}$
Solutions

## Two-Sided Limits

Evaluate the following limits or state that the limit does not exist.
$\lim _{x\to -1}{\frac {1}{x-1}}$
$-{\frac {1}{2}}$
$-{\frac {1}{2}}$
$\lim _{x\to 4}{\frac {1}{x-4}}$
$\lim _{x\to 2}{\frac {1}{x-2}}$
$\lim _{x\to -3}{\frac {x^{2}-9}{x+3}}$
$-6$
$-6$
$\lim _{x\to 3}{\frac {x^{2}-9}{x-3}}$
$6$
$6$
$\lim _{x\to -1}{\frac {x^{2}+2x+1}{x+1}}$
$0$
$0$
$\lim _{x\to -1}{\frac {x^{3}+1}{x+1}}$
$3$
$3$
$\lim _{x\to 4}{\frac {x^{2}+5x-36}{x^{2}-16}}$
${\frac {13}{8}}$
${\frac {13}{8}}$
$\lim _{x\to 25}{\frac {x-25}{{\sqrt {x}}-5}}$
$10$
$10$
$\lim _{x\to 0}{\frac {|x|}{x}}$
$\lim _{x\to 2}{\frac {1}{(x-2)^{2}}}$
$\infty$
$\infty$
$\lim _{x\to 3}{\frac {\sqrt {x^{2}+16}}{x-3}}$
$\lim _{x\to -2}{\frac {3x^{2}-8x-3}{2x^{2}-18}}$
$-{\frac {5}{2}}$
$-{\frac {5}{2}}$
$\lim _{x\to 2}{\frac {x^{2}+2x+1}{x^{2}-2x+1}}$
$9$
$9$
$\lim _{x\to 3}{\frac {x+3}{x^{2}-9}}$
$\lim _{x\to -1}{\frac {x+1}{x^{2}+x}}$
$-1$
$-1$
$\lim _{x\to 1}{\frac {1}{x^{2}+1}}$
${\frac {1}{2}}$
${\frac {1}{2}}$
$\lim _{x\to 1}\left[x^{2}+5x-{\frac {1}{2-x}}\right]$
$5$
$5$
$\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}$
$0$
$0$
$\lim _{x\to 1}{\frac {x^{2}-1}{x^{2}+2x-3}}$
${\frac {1}{2}}$
${\frac {1}{2}}$
$\lim _{x\to 1}{\frac {5x}{x^{2}+2x-3}}$
$\lim _{x\to -1}\ln({\sqrt {x^{2}-1}})$
$\lim _{x\to 1}\arcsin(x)$
Solutions

## Limits to Infinity

Evaluate the following limits or state that the limit does not exist.
$\lim _{x\to \infty }{\frac {-x+\pi }{x^{2}+3x+2}}$
$0$
$0$
$\lim _{x\to -\infty }{\frac {x^{2}+2x+1}{3x^{2}+1}}$
${\frac {1}{3}}$
${\frac {1}{3}}$
$\lim _{x\to -\infty }{\frac {3x^{2}+x}{2x^{2}-15}}$
${\frac {3}{2}}$
${\frac {3}{2}}$
$\lim _{x\to -\infty }{\Big3x^{2}-2x+1{\Big ]}$
$\infty$
$\infty$
$\lim _{x\to \infty }{\frac {2x^{2}-32}{x^{3}-64}}$
$0$
$0$
$\lim _{x\to \infty }6$
$6$
$6$
$\lim _{x\to \infty }{\frac {3x^{2}+4x}{x^{4}+2}}$
$0$
$0$
$\lim _{x\to -\infty }{\frac {2x+3x^{2}+1}{2x^{2}+3}}$
${\frac {3}{2}}$
${\frac {3}{2}}$
$\lim _{x\to -\infty }{\frac {x^{3}-3x^{2}+1}{3x^{2}+x+5}}$
$-\infty$
$-\infty$
$\lim _{x\to \infty }{\frac {x^{2}+2}{x^{3}-2}}$
$0$
$0$
$\lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}$
$0$
$0$
$\lim _{x\to -\infty }x^{2}\cos \left({\frac {1}{x}}\right)$
$\infty$
$\infty$
$\lim _{x\to \infty }{\frac {\sin(\arctan(x))}{\arctan(-x)}}$
$-{\frac {2}{\pi }}$
$-{\frac {2}{\pi }}$
Solutions

## Limits of Piecewise Functions

Evaluate the following limits or state that the limit does not exist.
48. Consider the function
    $f(x)={\begin{cases}(x-2)^{2}&{\text{if }}x<2\\x-3&{\text{if }}x\geq 2\end{cases}}$
$\lim _{x\to 2^{-}}f(x)$
$0$
$0$
$\lim _{x\to 2^{+}}f(x)$
$-1$
$-1$
$\lim _{x\to 2}f(x)$
49. Consider the function
    $g(x)={\begin{cases}-2x+1&{\text{if }}x\leq 0\\x+1&{\text{if }}0<x<4\\x^{2}+2&{\text{if }}x\geq 4\end{cases}}$
$\lim _{x\to 4^{+}}g(x)$
$18$
$18$
$\lim _{x\to 4^{-}}g(x)$
$5$
$5$
$\lim _{x\to 0^{+}}g(x)$
$1$
$1$
$\lim _{x\to 0^{-}}g(x)$
$1$
$1$
$\lim _{x\to 0}g(x)$
$1$
$1$
$\lim _{x\to 1}g(x)$
$2$
$2$
50. Consider the function
    $h(x)={\begin{cases}2x-3&{\text{if }}x<2\\8&{\text{if }}x=2\\-x+3&{\text{if }}x>2\end{cases}}$
$\lim _{x\to 0}h(x)$
$-3$
$-3$
$\lim _{x\to 2^{-}}h(x)$
$1$
$1$
$\lim _{x\to 2^{+}}h(x)$
$1$
$1$
$\lim _{x\to 2}h(x)$
$1$
$1$
Solutions

## Intermediate Value Theorem

$f(x)=3$
$f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}$
$1\leq x\leq 9$
$f(x)$
$x\in \left[1,9\right]$
$y\in \left[2,{\frac {10}{3}}\right]$
$c\in (1,9)$
$f(c)=3$
$\blacksquare$
$f(x)$
$x\in \left[1,9\right]$
$y\in \left[2,{\frac {10}{3}}\right]$
$c\in (1,9)$
$f(c)=3$
$\blacksquare$
$x=\mu$
$f(\mu )={\frac {\pi }{4}}$
$f(x)={\frac {\sin(x)-\cos(x)}{\ln {x}}}$
$\pi \leq x\leq {\frac {9}{4}}\pi$
$f(x)$
$x\in \left[\pi ,{\frac {9}{4}}\pi \right]$
    $f(\pi )={\frac {1}{\ln(\pi )}}$
    $f\left({\frac {9}{4}}\pi \right)=0$
It is known the following is true: $1\geq {\frac {\pi }{4}}\geq 0$. From there, we can directly argue the following:
    ${\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}$
$f(x)$
$x\in \left[\pi ,{\frac {9}{4}}\pi \right]$
$x=\mu$
$f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )$
$\mu \in \left[\pi ,{\frac {9}{4}}\pi \right]$
$\blacksquare$
$f(x)$
$x\in \left[\pi ,{\frac {9}{4}}\pi \right]$
    $f(\pi )={\frac {1}{\ln(\pi )}}$
    $f\left({\frac {9}{4}}\pi \right)=0$
It is known the following is true: $1\geq {\frac {\pi }{4}}\geq 0$. From there, we can directly argue the following:
    ${\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}$
$f(x)$
$x\in \left[\pi ,{\frac {9}{4}}\pi \right]$
$x=\mu$
$f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )$
$\mu \in \left[\pi ,{\frac {9}{4}}\pi \right]$
$\blacksquare$
$x=\Gamma$
$f(\Gamma )=2$
$f(x)={\sqrt {-2x+4}}+5\ln \left(x^{2}\right)$
$-1\leq x\leq 1$
$f(x)$
$x=0$
$\lim _{x\to 0}f(x)$
$f(x)$
$x=0$
$\lim _{x\to 0}f(x)$
Solutions

## External Links

* Limits 1 exercise at Khan Academy
* Limits 2 exercise at Khan Academy
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