# Calculus/Limits/Solutions
## Basic Limit Exercises

$\lim _{x\to 2}[4x^{2}-3x-1]$
$4(4)-2(3)-1=16-6-1=\mathbf {9}$
$4(4)-2(3)-1=16-6-1=\mathbf {9}$
$\lim _{x\to 5}x^{2}$
$5^{2}=\mathbf {25}$
$5^{2}=\mathbf {25}$
$\lim _{x\to {\frac {\pi }{4}}}{\Big\cos ^{2}(x){\Big ]}$
$\cos ^{2}\left({\frac {\pi }{4}}\right)=\left({\frac {1}{\sqrt {2}}}\right)^{2}=\mathbf {\frac {1}{2}}$
$\cos ^{2}\left({\frac {\pi }{4}}\right)=\left({\frac {1}{\sqrt {2}}}\right)^{2}=\mathbf {\frac {1}{2}}$
$\lim _{x\to 1}{\Big5e^{x-1}-5{\Big ]}$
$5e^{1-1}-5=5e^{0}-5=5-5=\mathbf {0}$
$5e^{1-1}-5=5e^{0}-5=5-5=\mathbf {0}$

## One-Sided Limits

Evaluate the following limits or state that the limit does not exist.
$\lim _{x\to 0^{-}}{\frac {x^{3}+x^{2}}{x^{3}+2x^{2}}}$
${\frac {x^{2}}{x^{2}}}\cdot {\frac {x+1}{x+2}}$
$x=0$
$\mathbf {\frac {1}{2}}$
${\frac {x^{2}}{x^{2}}}\cdot {\frac {x+1}{x+2}}$
$x=0$
$\mathbf {\frac {1}{2}}$
$\lim _{x\to 7^{-}}[|x^{2}+x|-x]$
$|7^{2}+7|-7=\mathbf {49}$
$|7^{2}+7|-7=\mathbf {49}$
$\lim _{x\to -1^{+}}{\sqrt {1-x^{2}}}$
${\sqrt {1-x^{2}}}$
$x^{2}<1$
${\sqrt {1-1^{2}}}=\mathbf {0}$
${\sqrt {1-x^{2}}}$
$x^{2}<1$
${\sqrt {1-1^{2}}}=\mathbf {0}$
$\lim _{x\to -1^{-}}{\sqrt {1-x^{2}}}$
${\sqrt {1-x^{2}}}$
$x^{2}>1$
${\sqrt {1-x^{2}}}$
$x^{2}>1$
$\lim _{x\to 0^{-}}{\frac {|x|}{x}}$
$|x|=-x$
$\lim _{x\to 0^{-}}{\frac {|x|}{x}}={\frac {-x}{x}}=\mathbf {-1}$
$|x|=-x$
$\lim _{x\to 0^{-}}{\frac {|x|}{x}}={\frac {-x}{x}}=\mathbf {-1}$
$\lim _{x\to 1^{-}}\arcsin(x)$
$\arcsin(x)=\sin ^{-1}(x)$
$x\in \left[-1,1\right]$
$\arcsin(1)=\mathbf {\frac {\pi }{2}}$
$\arcsin(x)=\sin ^{-1}(x)$
$x\in \left[-1,1\right]$
$\arcsin(1)=\mathbf {\frac {\pi }{2}}$

## Two-Sided Limits

Evaluate the following limits or state that the limit does not exist.
$\lim _{x\to -1}{\frac {1}{x-1}}$
$\mathbf {-{\frac {1}{2}}}$
$\mathbf {-{\frac {1}{2}}}$
$\lim _{x\to 4}{\frac {1}{x-4}}$
${\begin{aligned}&\lim _{x\to 4^{-}}{\frac {1}{x-4}}=-\infty \\&\lim _{x\to 4^{+}}{\frac {1}{x-4}}=\infty \end{aligned}}$
${\begin{aligned}&\lim _{x\to 4^{-}}{\frac {1}{x-4}}=-\infty \\&\lim _{x\to 4^{+}}{\frac {1}{x-4}}=\infty \end{aligned}}$
$\lim _{x\to 2}{\frac {1}{x-2}}$
$\lim _{x\to 2^{-}}{\frac {1}{x-2}}=-\infty$
$\lim _{x\to 2^{+}}{\frac {1}{x-2}}=+\infty$
$\lim _{x\to 2^{-}}{\frac {1}{x-2}}=-\infty$
$\lim _{x\to 2^{+}}{\frac {1}{x-2}}=+\infty$
$\lim _{x\to -3}{\frac {x^{2}-9}{x+3}}$
$\lim _{x\to -3}{\frac {(x+3)(x-3)}{x+3}}=\lim _{x\to -3}x-3=-3-3=\mathbf {-6}$
$\lim _{x\to -3}{\frac {(x+3)(x-3)}{x+3}}=\lim _{x\to -3}x-3=-3-3=\mathbf {-6}$
$\lim _{x\to 3}{\frac {x^{2}-9}{x-3}}$
$\lim _{x\to 3}{\frac {(x-3)(x+3)}{x-3}}=\lim _{x\to 3}x+3=3+3=\mathbf {6}$
$\lim _{x\to 3}{\frac {(x-3)(x+3)}{x-3}}=\lim _{x\to 3}x+3=3+3=\mathbf {6}$
$\lim _{x\to -1}{\frac {x^{2}+2x+1}{x+1}}$
$\lim _{x\to -1}{\frac {(x+1)(x+1)}{x+1}}=\lim _{x\to -1}x+1=-1+1=\mathbf {0}$
$\lim _{x\to -1}{\frac {(x+1)(x+1)}{x+1}}=\lim _{x\to -1}x+1=-1+1=\mathbf {0}$
$\lim _{x\to -1}{\frac {x^{3}+1}{x+1}}$
$\lim _{x\to -1}{\frac {(x^{2}-x+1)(x+1)}{x+1}}=\lim _{x\to -1}x^{2}-x+1=(-1)^{2}-(-1)+1=1+1+1=\mathbf {3}$
$\lim _{x\to -1}{\frac {(x^{2}-x+1)(x+1)}{x+1}}=\lim _{x\to -1}x^{2}-x+1=(-1)^{2}-(-1)+1=1+1+1=\mathbf {3}$
$\lim _{x\to 4}{\frac {x^{2}+5x-36}{x^{2}-16}}$
$\lim _{x\to 4}{\frac {(x-4)(x+9)}{(x-4)(x+4)}}=\lim _{x\to 4}{\frac {x+9}{x+4}}={\frac {4+9}{4+4}}=\mathbf {\frac {13}{8}}$
$\lim _{x\to 4}{\frac {(x-4)(x+9)}{(x-4)(x+4)}}=\lim _{x\to 4}{\frac {x+9}{x+4}}={\frac {4+9}{4+4}}=\mathbf {\frac {13}{8}}$
$\lim _{x\to 25}{\frac {x-25}{{\sqrt {x}}-5}}$
$\lim _{x\to 25}{\frac {({\sqrt {x}}-5)({\sqrt {x}}+5)}{{\sqrt {x}}-5}}=\lim _{x\to 25}{\sqrt {x}}+5)={\sqrt {25}}+5)=5+5=\mathbf {10}$
$\lim _{x\to 25}{\frac {({\sqrt {x}}-5)({\sqrt {x}}+5)}{{\sqrt {x}}-5}}=\lim _{x\to 25}{\sqrt {x}}+5)={\sqrt {25}}+5)=5+5=\mathbf {10}$
$\lim _{x\to 0}{\frac {\left|x\right|}{x}}$
$\lim _{x\to 0^{-}}{\frac {\left|x\right|}{x}}=\lim _{x\to 0^{-}}{\frac {-x}{x}}=\lim _{x\to 0^{-}}-1=-1$
$\lim _{x\to 0^{+}}{\frac {\left|x\right|}{x}}=\lim _{x\to 0^{+}}{\frac {x}{x}}=\lim _{x\to 0^{+}}1=1$
$\lim _{x\to 0^{-}}{\frac {\left|x\right|}{x}}=\lim _{x\to 0^{-}}{\frac {-x}{x}}=\lim _{x\to 0^{-}}-1=-1$
$\lim _{x\to 0^{+}}{\frac {\left|x\right|}{x}}=\lim _{x\to 0^{+}}{\frac {x}{x}}=\lim _{x\to 0^{+}}1=1$
$\lim _{x\to 2}{\frac {1}{(x-2)^{2}}}$
$x$
$2$
$\mathbf {\infty }$
$x$
$2$
$\mathbf {\infty }$
$\lim _{x\to 3}{\frac {\sqrt {x^{2}+16}}{x-3}}$
$x$
$3$
$3$
$-\infty$
$+\infty$
$x$
$3$
$3$
$-\infty$
$+\infty$
$\lim _{x\to -2}{\frac {3x^{2}-8x-3}{2x^{2}-18}}$
${\frac {3(-2)^{2}-8(-2)-3}{2(-2)^{2}-18}}={\frac {3(4)+16-3}{2(4)-18}}={\frac {12+16-3}{8-18}}={\frac {25}{-10}}=\mathbf {-{\frac {5}{2}}}$
${\frac {3(-2)^{2}-8(-2)-3}{2(-2)^{2}-18}}={\frac {3(4)+16-3}{2(4)-18}}={\frac {12+16-3}{8-18}}={\frac {25}{-10}}=\mathbf {-{\frac {5}{2}}}$
$\lim _{x\to 2}{\frac {x^{2}+2x+1}{x^{2}-2x+1}}$
${\frac {2^{2}+2(2)+1}{2^{2}-2(2)+1}}={\frac {4+4+1}{4-4+1}}={\frac {9}{1}}=\mathbf {9}$
${\frac {2^{2}+2(2)+1}{2^{2}-2(2)+1}}={\frac {4+4+1}{4-4+1}}={\frac {9}{1}}=\mathbf {9}$
$\lim _{x\to 3}{\frac {x+3}{x^{2}-9}}$
$\lim _{x\to 3}{\frac {x+3}{(x+3)(x-3)}}=\lim _{x\to 3}{\frac {1}{x-3}}$
$\lim _{x\to 3^{-}}{\frac {1}{x-3}}=-\infty$
$\lim _{x\to 3^{+}}{\frac {1}{x-3}}=+\infty$
$\lim _{x\to 3}{\frac {x+3}{(x+3)(x-3)}}=\lim _{x\to 3}{\frac {1}{x-3}}$
$\lim _{x\to 3^{-}}{\frac {1}{x-3}}=-\infty$
$\lim _{x\to 3^{+}}{\frac {1}{x-3}}=+\infty$
$\lim _{x\to -1}{\frac {x+1}{x^{2}+x}}$
$\lim _{x\to -1}{\frac {x+1}{x(x+1)}}=\lim _{x\to -1}{\frac {1}{x}}={\frac {1}{-1}}=\mathbf {-1}$
$\lim _{x\to -1}{\frac {x+1}{x(x+1)}}=\lim _{x\to -1}{\frac {1}{x}}={\frac {1}{-1}}=\mathbf {-1}$
$\lim _{x\to 1}{\frac {1}{x^{2}+1}}$
${\frac {1}{1^{2}+1}}={\frac {1}{1+1}}=\mathbf {\frac {1}{2}}$
${\frac {1}{1^{2}+1}}={\frac {1}{1+1}}=\mathbf {\frac {1}{2}}$
$\lim _{x\to 1}x^{3}+5x-{\frac {1}{2-x}}$
$1^{3}+5(1)-{\frac {1}{2-1}}=1+5-{\frac {1}{1}}=6-1=\mathbf {5}$
$1^{3}+5(1)-{\frac {1}{2-1}}=1+5-{\frac {1}{1}}=6-1=\mathbf {5}$
$\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}$
$0$
$-3$
$\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}=\mathbf {0}$
$0$
$-3$
$\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}=\mathbf {0}$
$\lim _{x\to 1}{\frac {x^{2}-1}{x^{2}+2x-3}}$
$\lim _{x\to 1}{\frac {(x-1)(x+1)}{(x-1)(x+3)}}=\lim _{x\to 1}{\frac {x+1}{x+3}}={\frac {1+1}{1+3}}={\frac {2}{4}}=\mathbf {\frac {1}{2}}$
$\lim _{x\to 1}{\frac {(x-1)(x+1)}{(x-1)(x+3)}}=\lim _{x\to 1}{\frac {x+1}{x+3}}={\frac {1+1}{1+3}}={\frac {2}{4}}=\mathbf {\frac {1}{2}}$
$\lim _{x\to 1}{\frac {5x}{x^{2}+2x-3}}$
$x$
$1$
$5$
$0$
$-\infty$
$+\infty$
$x$
$1$
$5$
$0$
$-\infty$
$+\infty$
$\lim _{x\to -1}\ln \left({\sqrt {1-x^{2}}}\right)$
$\lim _{x\to -1}\ln \left({\sqrt {1-x^{2}}}\right)=\ln \left(\lim _{x\to -1}{\sqrt {x^{2}-1}}\right)$
    $\ln \left(\lim _{x\to -1^{+}}{\sqrt {x^{2}-1}}\right)=\ln({\text{DNE}})={\text{DNE}}$
    $\ln \left(\lim _{x\to -1^{-}}{\sqrt {x^{2}-1}}\right)=\ln \left(0^{+}\right)=\infty$
$\lim _{x\to -1}\ln \left({\sqrt {1-x^{2}}}\right)=\ln \left(\lim _{x\to -1}{\sqrt {x^{2}-1}}\right)$
    $\ln \left(\lim _{x\to -1^{+}}{\sqrt {x^{2}-1}}\right)=\ln({\text{DNE}})={\text{DNE}}$
    $\ln \left(\lim _{x\to -1^{-}}{\sqrt {x^{2}-1}}\right)=\ln \left(0^{+}\right)=\infty$
$\lim _{x\to 1}\arcsin(x)$
    $\lim _{x\to 1^{+}}\arcsin(x)$ does not exist since the domain of $\arcsin(x)$ is $x\in \left[-1,1\right]$
    $\lim _{x\to 1^{-}}\arcsin(x)={\frac {\pi }{2}}$
    $\lim _{x\to 1^{+}}\arcsin(x)$ does not exist since the domain of $\arcsin(x)$ is $x\in \left[-1,1\right]$
    $\lim _{x\to 1^{-}}\arcsin(x)={\frac {\pi }{2}}$

## Limits to Infinity

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

## Limits of Piecewise Functions

Evaluate the following limits or state that the limit does not exist.
48. Consider the function
    $f(x)={\begin{cases}(x-2)^{2}&{\mbox{if }}x<2\\x-3&{\mbox{if }}x\geq 2.\end{cases}}$
$\lim _{x\to 2^{-}}f(x)$
$(2-2)^{2}=\mathbf {0}$
$(2-2)^{2}=\mathbf {0}$
$\lim _{x\to 2^{+}}f(x)$
$2-3=\mathbf {-1}$
$2-3=\mathbf {-1}$
$\lim _{x\to 2}f(x)$
49. Consider the function
    $g(x)={\begin{cases}-2x+1&{\mbox{if }}x\leq 0\\x+1&{\mbox{if }}0<x<4\\x^{2}+2&{\mbox{if }}x\geq 4.\end{cases}}$
$\lim _{x\to 4^{+}}g(x)$
$4^{2}+2=16+2=\mathbf {18}$
$4^{2}+2=16+2=\mathbf {18}$
$\lim _{x\to 4^{-}}g(x)$
$4+1=\mathbf {5}$
$4+1=\mathbf {5}$
$\lim _{x\to 0^{+}}g(x)$
$0+1=\mathbf {1}$
$0+1=\mathbf {1}$
$\lim _{x\to 0^{-}}g(x)$
$-2(0)+1=\mathbf {1}$
$-2(0)+1=\mathbf {1}$
$\lim _{x\to 0}g(x)$
$\mathbf {1}$
$\mathbf {1}$
$\lim _{x\to 1}g(x)$
$1+1=\mathbf {2}$
$1+1=\mathbf {2}$
50. Consider the function
    $h(x)={\begin{cases}2x-3&{\mbox{if }}x<2\\8&{\mbox{if }}x=2\\-x+3&{\mbox{if }}x>2.\end{cases}}$
$\lim _{x\to 0}h(x)$
$2(0)-3=\mathbf {-3}$
$2(0)-3=\mathbf {-3}$
$\lim _{x\to 2^{-}}h(x)$
$2(2)-3=4-3=\mathbf {1}$
$2(2)-3=4-3=\mathbf {1}$
$\lim _{x\to 2^{+}}h(x)$
$-(2)+3=\mathbf {1}$
$-(2)+3=\mathbf {1}$
$\lim _{x\to 2}h(x)$
$\mathbf {1}$
$\mathbf {1}$

## Intermediate Value Theorem

$f(x)=3$
$f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}$
$1\leq x\leq 9$
$f(x)$
$x\in \left[1,9\right]$
$y\in \left[2,{\frac {10}{3}}\right]$
$c\in (1,9)$
$f(c)=3$
$f(x)$
$x\in \left[1,9\right]$
$y\in \left[2,{\frac {10}{3}}\right]$
$c\in (1,9)$
$f(c)=3$
$x=\mu$
$f(\mu )={\frac {\pi }{4}}$
$f(x)={\frac {\sin(x)-\cos(x)}{\ln {x}}}$
$\pi \leq x\leq {\frac {9}{4}}\pi$
$f(x)$
$x\in \left[\pi ,{\frac {9}{4}}\pi \right]$
    $f(\pi )={\frac {1}{\ln(\pi )}}$
    $f\left({\frac {9}{4}}\pi \right)=0$
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:
    ${\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}$
$f(x)$
$x\in \left[\pi ,{\frac {9}{4}}\pi \right]$
$x=\mu$
$f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )$
$\mu \in \left[\pi ,{\frac {9}{4}}\pi \right]$
$\blacksquare$
$x=\Gamma$
$f(\Gamma )=2$
$f(x)={\sqrt {-2x+4}}+5\ln \left(x^{2}\right)$
$-1\leq x\leq 1$
$f(x)$
$x=0$
$\lim _{x\to 0}f(x)$
$f(x)$
$x=0$
$\lim _{x\to 0}f(x)$
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