# Calculus/Limits/Solutions

## Basic Limit Exercises

1. $\lim _{x\to 2}[4x^{2}-3x-1]$
Since this is a polynomial, two can simply be plugged in. This results in $4(4)-2(3)-1=16-6-1=\mathbf {9}$
Since this is a polynomial, two can simply be plugged in. This results in $4(4)-2(3)-1=16-6-1=\mathbf {9}$
2. $\lim _{x\to 5}x^{2}$
$5^{2}=\mathbf {25}$
$5^{2}=\mathbf {25}$
3. $\lim _{x\to {\frac {\pi }{4}}}{\Big [}\cos ^{2}(x){\Big ]}$
$\cos ^{2}\left({\frac {\pi }{4}}\right)=\left({\frac {1}{\sqrt {2}}}\right)^{2}=\mathbf {\frac {1}{2}}$
$\cos ^{2}\left({\frac {\pi }{4}}\right)=\left({\frac {1}{\sqrt {2}}}\right)^{2}=\mathbf {\frac {1}{2}}$
4. $\lim _{x\to 1}{\Big [}5e^{x-1}-5{\Big ]}$
$5e^{1-1}-5=5e^{0}-5=5-5=\mathbf {0}$
$5e^{1-1}-5=5e^{0}-5=5-5=\mathbf {0}$

## One-Sided Limits

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

5. $\lim _{x\to 0^{-}}{\frac {x^{3}+x^{2}}{x^{3}+2x^{2}}}$
Factor as ${\frac {x^{2}}{x^{2}}}\cdot {\frac {x+1}{x+2}}$  . In this form we can see that there is a removable discontinuity at $x=0$  and that the limit is $\mathbf {\frac {1}{2}}$
Factor as ${\frac {x^{2}}{x^{2}}}\cdot {\frac {x+1}{x+2}}$  . In this form we can see that there is a removable discontinuity at $x=0$  and that the limit is $\mathbf {\frac {1}{2}}$
6. $\lim _{x\to 7^{-}}[|x^{2}+x|-x]$
$|7^{2}+7|-7=\mathbf {49}$
$|7^{2}+7|-7=\mathbf {49}$
7. $\lim _{x\to -1^{+}}{\sqrt {1-x^{2}}}$
${\sqrt {1-x^{2}}}$  is defined if $x^{2}<1$ , so the limit is ${\sqrt {1-1^{2}}}=\mathbf {0}$
${\sqrt {1-x^{2}}}$  is defined if $x^{2}<1$ , so the limit is ${\sqrt {1-1^{2}}}=\mathbf {0}$
8. $\lim _{x\to -1^{-}}{\sqrt {1-x^{2}}}$
${\sqrt {1-x^{2}}}$  is not defined if $x^{2}>1$  , so the limit does not exist.
${\sqrt {1-x^{2}}}$  is not defined if $x^{2}>1$  , so the limit does not exist.
9. $\lim _{x\to 0^{-}}{\frac {|x|}{x}}$
If approaching from the left, then $|x|=-x$ . Thus, $\lim _{x\to 0^{-}}{\frac {|x|}{x}}={\frac {-x}{x}}=\mathbf {-1}$
If approaching from the left, then $|x|=-x$ . Thus, $\lim _{x\to 0^{-}}{\frac {|x|}{x}}={\frac {-x}{x}}=\mathbf {-1}$
10. $\lim _{x\to 1^{-}}\arcsin(x)$
$\arcsin(x)=\sin ^{-1}(x)$  is defined from $x\in \left[-1,1\right]$ , so the limit is $\arcsin(1)=\mathbf {\frac {\pi }{2}}$ .
$\arcsin(x)=\sin ^{-1}(x)$  is defined from $x\in \left[-1,1\right]$ , so the limit is $\arcsin(1)=\mathbf {\frac {\pi }{2}}$ .

## Two-Sided Limits

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

11. $\lim _{x\to -1}{\frac {1}{x-1}}$
$\mathbf {-{\frac {1}{2}}}$
$\mathbf {-{\frac {1}{2}}}$
12. $\lim _{x\to 4}{\frac {1}{x-4}}$
{\begin{aligned}&\lim _{x\to 4^{-}}{\frac {1}{x-4}}=-\infty \\&\lim _{x\to 4^{+}}{\frac {1}{x-4}}=\infty \end{aligned}}
The limit does not exist.
{\begin{aligned}&\lim _{x\to 4^{-}}{\frac {1}{x-4}}=-\infty \\&\lim _{x\to 4^{+}}{\frac {1}{x-4}}=\infty \end{aligned}}
The limit does not exist.
13. $\lim _{x\to 2}{\frac {1}{x-2}}$
$\lim _{x\to 2^{-}}{\frac {1}{x-2}}=-\infty$
$\lim _{x\to 2^{+}}{\frac {1}{x-2}}=+\infty$
The limit does not exist.
$\lim _{x\to 2^{-}}{\frac {1}{x-2}}=-\infty$
$\lim _{x\to 2^{+}}{\frac {1}{x-2}}=+\infty$
The limit does not exist.
14. $\lim _{x\to -3}{\frac {x^{2}-9}{x+3}}$
$\lim _{x\to -3}{\frac {(x+3)(x-3)}{x+3}}=\lim _{x\to -3}x-3=-3-3=\mathbf {-6}$
$\lim _{x\to -3}{\frac {(x+3)(x-3)}{x+3}}=\lim _{x\to -3}x-3=-3-3=\mathbf {-6}$
15. $\lim _{x\to 3}{\frac {x^{2}-9}{x-3}}$
$\lim _{x\to 3}{\frac {(x-3)(x+3)}{x-3}}=\lim _{x\to 3}x+3=3+3=\mathbf {6}$
$\lim _{x\to 3}{\frac {(x-3)(x+3)}{x-3}}=\lim _{x\to 3}x+3=3+3=\mathbf {6}$
16. $\lim _{x\to -1}{\frac {x^{2}+2x+1}{x+1}}$
$\lim _{x\to -1}{\frac {(x+1)(x+1)}{x+1}}=\lim _{x\to -1}x+1=-1+1=\mathbf {0}$
$\lim _{x\to -1}{\frac {(x+1)(x+1)}{x+1}}=\lim _{x\to -1}x+1=-1+1=\mathbf {0}$
17. $\lim _{x\to -1}{\frac {x^{3}+1}{x+1}}$
$\lim _{x\to -1}{\frac {(x^{2}-x+1)(x+1)}{x+1}}=\lim _{x\to -1}x^{2}-x+1=(-1)^{2}-(-1)+1=1+1+1=\mathbf {3}$
$\lim _{x\to -1}{\frac {(x^{2}-x+1)(x+1)}{x+1}}=\lim _{x\to -1}x^{2}-x+1=(-1)^{2}-(-1)+1=1+1+1=\mathbf {3}$
18. $\lim _{x\to 4}{\frac {x^{2}+5x-36}{x^{2}-16}}$
$\lim _{x\to 4}{\frac {(x-4)(x+9)}{(x-4)(x+4)}}=\lim _{x\to 4}{\frac {x+9}{x+4}}={\frac {4+9}{4+4}}=\mathbf {\frac {13}{8}}$
$\lim _{x\to 4}{\frac {(x-4)(x+9)}{(x-4)(x+4)}}=\lim _{x\to 4}{\frac {x+9}{x+4}}={\frac {4+9}{4+4}}=\mathbf {\frac {13}{8}}$
19. $\lim _{x\to 25}{\frac {x-25}{{\sqrt {x}}-5}}$
$\lim _{x\to 25}{\frac {({\sqrt {x}}-5)({\sqrt {x}}+5)}{{\sqrt {x}}-5}}=\lim _{x\to 25}{\sqrt {x}}+5)={\sqrt {25}}+5)=5+5=\mathbf {10}$
$\lim _{x\to 25}{\frac {({\sqrt {x}}-5)({\sqrt {x}}+5)}{{\sqrt {x}}-5}}=\lim _{x\to 25}{\sqrt {x}}+5)={\sqrt {25}}+5)=5+5=\mathbf {10}$
20. $\lim _{x\to 0}{\frac {\left|x\right|}{x}}$
$\lim _{x\to 0^{-}}{\frac {\left|x\right|}{x}}=\lim _{x\to 0^{-}}{\frac {-x}{x}}=\lim _{x\to 0^{-}}-1=-1$
$\lim _{x\to 0^{+}}{\frac {\left|x\right|}{x}}=\lim _{x\to 0^{+}}{\frac {x}{x}}=\lim _{x\to 0^{+}}1=1$
The limit does not exist.
$\lim _{x\to 0^{-}}{\frac {\left|x\right|}{x}}=\lim _{x\to 0^{-}}{\frac {-x}{x}}=\lim _{x\to 0^{-}}-1=-1$
$\lim _{x\to 0^{+}}{\frac {\left|x\right|}{x}}=\lim _{x\to 0^{+}}{\frac {x}{x}}=\lim _{x\to 0^{+}}1=1$
The limit does not exist.
21. $\lim _{x\to 2}{\frac {1}{(x-2)^{2}}}$
As $x$  approaches $2$ , the denominator will be a very small positive number, so the whole fraction will be a very large positive number. Thus, the limit is $\mathbf {\infty }$ .
As $x$  approaches $2$ , the denominator will be a very small positive number, so the whole fraction will be a very large positive number. Thus, the limit is $\mathbf {\infty }$ .
22. $\lim _{x\to 3}{\frac {\sqrt {x^{2}+16}}{x-3}}$
As $x$  approaches $3$ , the numerator goes to 5 and the denominator goes to 0. Depending on whether you approach $3$  from the left or the right, the denominator will be either a very small negative number, or a very small positive number. So the limit from the left is $-\infty$  and the limit from the right is $+\infty$ . Thus, the limit does not exist.
As $x$  approaches $3$ , the numerator goes to 5 and the denominator goes to 0. Depending on whether you approach $3$  from the left or the right, the denominator will be either a very small negative number, or a very small positive number. So the limit from the left is $-\infty$  and the limit from the right is $+\infty$ . Thus, the limit does not exist.
23. $\lim _{x\to -2}{\frac {3x^{2}-8x-3}{2x^{2}-18}}$
${\frac {3(-2)^{2}-8(-2)-3}{2(-2)^{2}-18}}={\frac {3(4)+16-3}{2(4)-18}}={\frac {12+16-3}{8-18}}={\frac {25}{-10}}=\mathbf {-{\frac {5}{2}}}$
${\frac {3(-2)^{2}-8(-2)-3}{2(-2)^{2}-18}}={\frac {3(4)+16-3}{2(4)-18}}={\frac {12+16-3}{8-18}}={\frac {25}{-10}}=\mathbf {-{\frac {5}{2}}}$
24. $\lim _{x\to 2}{\frac {x^{2}+2x+1}{x^{2}-2x+1}}$
${\frac {2^{2}+2(2)+1}{2^{2}-2(2)+1}}={\frac {4+4+1}{4-4+1}}={\frac {9}{1}}=\mathbf {9}$
${\frac {2^{2}+2(2)+1}{2^{2}-2(2)+1}}={\frac {4+4+1}{4-4+1}}={\frac {9}{1}}=\mathbf {9}$
25. $\lim _{x\to 3}{\frac {x+3}{x^{2}-9}}$
$\lim _{x\to 3}{\frac {x+3}{(x+3)(x-3)}}=\lim _{x\to 3}{\frac {1}{x-3}}$
$\lim _{x\to 3^{-}}{\frac {1}{x-3}}=-\infty$
$\lim _{x\to 3^{+}}{\frac {1}{x-3}}=+\infty$
The limit does not exist.
$\lim _{x\to 3}{\frac {x+3}{(x+3)(x-3)}}=\lim _{x\to 3}{\frac {1}{x-3}}$
$\lim _{x\to 3^{-}}{\frac {1}{x-3}}=-\infty$
$\lim _{x\to 3^{+}}{\frac {1}{x-3}}=+\infty$
The limit does not exist.
26. $\lim _{x\to -1}{\frac {x+1}{x^{2}+x}}$
$\lim _{x\to -1}{\frac {x+1}{x(x+1)}}=\lim _{x\to -1}{\frac {1}{x}}={\frac {1}{-1}}=\mathbf {-1}$
$\lim _{x\to -1}{\frac {x+1}{x(x+1)}}=\lim _{x\to -1}{\frac {1}{x}}={\frac {1}{-1}}=\mathbf {-1}$
27. $\lim _{x\to 1}{\frac {1}{x^{2}+1}}$
${\frac {1}{1^{2}+1}}={\frac {1}{1+1}}=\mathbf {\frac {1}{2}}$
${\frac {1}{1^{2}+1}}={\frac {1}{1+1}}=\mathbf {\frac {1}{2}}$
28. $\lim _{x\to 1}x^{3}+5x-{\frac {1}{2-x}}$
$1^{3}+5(1)-{\frac {1}{2-1}}=1+5-{\frac {1}{1}}=6-1=\mathbf {5}$
$1^{3}+5(1)-{\frac {1}{2-1}}=1+5-{\frac {1}{1}}=6-1=\mathbf {5}$
29. $\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}$
The numerator evaluates to $0$  while the denominator evaluates to $-3$ . Hence, $\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}=\mathbf {0}$ .
The numerator evaluates to $0$  while the denominator evaluates to $-3$ . Hence, $\lim _{x\to 0}{\frac {x^{2}}{x^{2}+2x-3}}=\mathbf {0}$ .
30. $\lim _{x\to 1}{\frac {x^{2}-1}{x^{2}+2x-3}}$
$\lim _{x\to 1}{\frac {(x-1)(x+1)}{(x-1)(x+3)}}=\lim _{x\to 1}{\frac {x+1}{x+3}}={\frac {1+1}{1+3}}={\frac {2}{4}}=\mathbf {\frac {1}{2}}$
$\lim _{x\to 1}{\frac {(x-1)(x+1)}{(x-1)(x+3)}}=\lim _{x\to 1}{\frac {x+1}{x+3}}={\frac {1+1}{1+3}}={\frac {2}{4}}=\mathbf {\frac {1}{2}}$
31. $\lim _{x\to 1}{\frac {5x}{x^{2}+2x-3}}$
Notice that as $x$  approaches $1$ , the numerator approaches $5$  while the denominator approaches $0$ . However, if you approach from below, the denominator is negative, and if you approach from above, the denominator is positive. So the limits from the left and right will be $-\infty$  and $+\infty$  respectively. Thus, the limit does not exist.
Notice that as $x$  approaches $1$ , the numerator approaches $5$  while the denominator approaches $0$ . However, if you approach from below, the denominator is negative, and if you approach from above, the denominator is positive. So the limits from the left and right will be $-\infty$  and $+\infty$  respectively. Thus, the limit does not exist.
32. $\lim _{x\to -1}\ln \left({\sqrt {1-x^{2}}}\right)$
Notice how $\lim _{x\to -1}\ln \left({\sqrt {1-x^{2}}}\right)=\ln \left(\lim _{x\to -1}{\sqrt {x^{2}-1}}\right)$ .
$\ln \left(\lim _{x\to -1^{+}}{\sqrt {x^{2}-1}}\right)=\ln({\text{DNE}})={\text{DNE}}$
$\ln \left(\lim _{x\to -1^{-}}{\sqrt {x^{2}-1}}\right)=\ln \left(0^{+}\right)=\infty$
Thus, the limit does not exist.
Notice how $\lim _{x\to -1}\ln \left({\sqrt {1-x^{2}}}\right)=\ln \left(\lim _{x\to -1}{\sqrt {x^{2}-1}}\right)$ .
$\ln \left(\lim _{x\to -1^{+}}{\sqrt {x^{2}-1}}\right)=\ln({\text{DNE}})={\text{DNE}}$
$\ln \left(\lim _{x\to -1^{-}}{\sqrt {x^{2}-1}}\right)=\ln \left(0^{+}\right)=\infty$
Thus, the limit does not exist.
33. $\lim _{x\to 1}\arcsin(x)$
Notice that
$\lim _{x\to 1^{+}}\arcsin(x)$  does not exist since the domain of $\arcsin(x)$  is $x\in \left[-1,1\right]$
$\lim _{x\to 1^{-}}\arcsin(x)={\frac {\pi }{2}}$
Thus, the limit does not exist.
Notice that
$\lim _{x\to 1^{+}}\arcsin(x)$  does not exist since the domain of $\arcsin(x)$  is $x\in \left[-1,1\right]$
$\lim _{x\to 1^{-}}\arcsin(x)={\frac {\pi }{2}}$
Thus, the limit does not exist.

## Limits to Infinity

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

34. $\lim _{x\to \infty }{\frac {-x+\pi }{x^{2}+3x+2}}$
This rational function is bottom-heavy, so the limit is $\mathbf {0}$ .
This rational function is bottom-heavy, so the limit is $\mathbf {0}$ .
35. $\lim _{x\to -\infty }{\frac {x^{2}+2x+1}{3x^{2}+1}}$
This rational function has evenly matched powers of $x$  in the numerator and denominator, so the limit will be the ratio of the coefficients, i.e. $\mathbf {\frac {1}{3}}$ .
This rational function has evenly matched powers of $x$  in the numerator and denominator, so the limit will be the ratio of the coefficients, i.e. $\mathbf {\frac {1}{3}}$ .
36. $\lim _{x\to -\infty }{\frac {3x^{2}+x}{2x^{2}-15}}$
Balanced powers in the numerator and denominator, so the limit is the ratio of the coefficients, i.e. $\mathbf {\frac {3}{2}}$ .
Balanced powers in the numerator and denominator, so the limit is the ratio of the coefficients, i.e. $\mathbf {\frac {3}{2}}$ .
37. $\lim _{x\to -\infty }3x^{2}-2x+1$
This is a top-heavy rational function, where the exponent of the ratio of the leading terms is $2$ . Since it is even, the limit will be $\mathbf {\infty }$ .
This is a top-heavy rational function, where the exponent of the ratio of the leading terms is $2$ . Since it is even, the limit will be $\mathbf {\infty }$ .
38. $\lim _{x\to \infty }{\frac {2x^{2}-32}{x^{3}-64}}$
Bottom-heavy rational function, so the limit is $\mathbf {0}$ .
Bottom-heavy rational function, so the limit is $\mathbf {0}$ .
39. $\lim _{x\to \infty }6$
This is a rational function, as can be seen by writing it in the form ${\frac {6x^{0}}{1x^{0}}}$ . Since the powers of $x$  in the numerator and denominator are evenly matched, the limit will be the ratio of the coefficients, i.e. $\mathbf {6}$ .
This is a rational function, as can be seen by writing it in the form ${\frac {6x^{0}}{1x^{0}}}$ . Since the powers of $x$  in the numerator and denominator are evenly matched, the limit will be the ratio of the coefficients, i.e. $\mathbf {6}$ .
40. $\lim _{x\to \infty }{\frac {3x^{2}+4x}{x^{4}+2}}$
Bottom-heavy, so the limit is $\mathbf {0}$ .
Bottom-heavy, so the limit is $\mathbf {0}$ .
41. $\lim _{x\to -\infty }{\frac {2x+3x^{2}+1}{2x^{2}+3}}$
Evenly matched highest powers of $x$  in the numerator and denominator, so the limit will be the ratio of the corresponding coefficients, i.e. $\mathbf {\frac {3}{2}}$ .
Evenly matched highest powers of $x$  in the numerator and denominator, so the limit will be the ratio of the corresponding coefficients, i.e. $\mathbf {\frac {3}{2}}$ .
42. $\lim _{x\to -\infty }{\frac {x^{3}-3x^{2}+1}{3x^{2}+x+5}}$
Top-heavy rational function, where the exponent of the ratio of the leading terms is $1$ , so the limit is $\mathbf {-\infty }$ .
Top-heavy rational function, where the exponent of the ratio of the leading terms is $1$ , so the limit is $\mathbf {-\infty }$ .
43. $\lim _{x\to \infty }{\frac {x^{2}+2}{x^{3}-2}}$
Bottom-heavy, so the limit is $\mathbf {0}$ .
Bottom-heavy, so the limit is $\mathbf {0}$ .
44. $\lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}$
$\lim _{x\to \infty }{\frac {1}{x}}=0$ , so $\lim _{x\to \infty }\sin \left({\frac {1}{x}}\right)=0$ . From this, $\lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}={\frac {0}{\infty }}$ . Because $\lim _{x\to \infty }{\frac {n}{x}}=0$  for any $n\in \left(-\infty ,\infty \right)$ , $\lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}=\mathbf {0}$ .
$\lim _{x\to \infty }{\frac {1}{x}}=0$ , so $\lim _{x\to \infty }\sin \left({\frac {1}{x}}\right)=0$ . From this, $\lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}={\frac {0}{\infty }}$ . Because $\lim _{x\to \infty }{\frac {n}{x}}=0$  for any $n\in \left(-\infty ,\infty \right)$ , $\lim _{x\to \infty }{\frac {\sin \left({\frac {1}{x}}\right)}{x}}=\mathbf {0}$ .
45. $\lim _{x\to -\infty }x^{2}\cos \left({\frac {1}{x}}\right)$
$\lim _{x\to -\infty }x^{2}\cos \left({\frac {1}{x}}\right)=(-\infty )^{2}\cos(0)=\infty \cdot 1=\mathbf {\infty }$
$\lim _{x\to -\infty }x^{2}\cos \left({\frac {1}{x}}\right)=(-\infty )^{2}\cos(0)=\infty \cdot 1=\mathbf {\infty }$
46. $\lim _{x\to \infty }{\frac {\sin(\arctan(x))}{\arctan(-x)}}$
Notice that $\lim _{x\to \infty }\arctan(x)={\frac {\pi }{2}}$ . Because $\arctan(-x)=-\arctan(x)$ , $\lim _{x\to \infty }{\frac {\sin(\arctan(x))}{\arctan(-x)}}={\frac {\sin \left({\frac {\pi }{2}}\right)}{-{\frac {\pi }{2}}}}={\frac {1}{-{\frac {\pi }{2}}}}=\mathbf {-{\frac {2}{\pi }}}$
Notice that $\lim _{x\to \infty }\arctan(x)={\frac {\pi }{2}}$ . Because $\arctan(-x)=-\arctan(x)$ , $\lim _{x\to \infty }{\frac {\sin(\arctan(x))}{\arctan(-x)}}={\frac {\sin \left({\frac {\pi }{2}}\right)}{-{\frac {\pi }{2}}}}={\frac {1}{-{\frac {\pi }{2}}}}=\mathbf {-{\frac {2}{\pi }}}$

## Limits of Piecewise Functions

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

48. Consider the function

$f(x)={\begin{cases}(x-2)^{2}&{\mbox{if }}x<2\\x-3&{\mbox{if }}x\geq 2.\end{cases}}$
a. $\lim _{x\to 2^{-}}f(x)$
$(2-2)^{2}=\mathbf {0}$
$(2-2)^{2}=\mathbf {0}$
b. $\lim _{x\to 2^{+}}f(x)$
$2-3=\mathbf {-1}$
$2-3=\mathbf {-1}$
c. $\lim _{x\to 2}f(x)$
Since the limits from the left and right don't match, the limit does not exist.
Since the limits from the left and right don't match, the limit does not exist.

49. Consider the function

$g(x)={\begin{cases}-2x+1&{\mbox{if }}x\leq 0\\x+1&{\mbox{if }}0
a. $\lim _{x\to 4^{+}}g(x)$
$4^{2}+2=16+2=\mathbf {18}$
$4^{2}+2=16+2=\mathbf {18}$
b. $\lim _{x\to 4^{-}}g(x)$
$4+1=\mathbf {5}$
$4+1=\mathbf {5}$
c. $\lim _{x\to 0^{+}}g(x)$
$0+1=\mathbf {1}$
$0+1=\mathbf {1}$
d. $\lim _{x\to 0^{-}}g(x)$
$-2(0)+1=\mathbf {1}$
$-2(0)+1=\mathbf {1}$
e. $\lim _{x\to 0}g(x)$
Since the left and right limits match, the overall limit is also $\mathbf {1}$ .
Since the left and right limits match, the overall limit is also $\mathbf {1}$ .
f. $\lim _{x\to 1}g(x)$
$1+1=\mathbf {2}$
$1+1=\mathbf {2}$

50. Consider the function

$h(x)={\begin{cases}2x-3&{\mbox{if }}x<2\\8&{\mbox{if }}x=2\\-x+3&{\mbox{if }}x>2.\end{cases}}$
a. $\lim _{x\to 0}h(x)$
$2(0)-3=\mathbf {-3}$
$2(0)-3=\mathbf {-3}$
b. $\lim _{x\to 2^{-}}h(x)$
$2(2)-3=4-3=\mathbf {1}$
$2(2)-3=4-3=\mathbf {1}$
c. $\lim _{x\to 2^{+}}h(x)$
$-(2)+3=\mathbf {1}$
$-(2)+3=\mathbf {1}$
d. $\lim _{x\to 2}h(x)$
Since the limits from the right and left match, the overall limit is $\mathbf {1}$ . Note that in this case, the limit at 2 does not match the function value at 2, so the function is discontinuous at this point, hence the function is nondifferentiable at this point as well.
Since the limits from the right and left match, the overall limit is $\mathbf {1}$ . Note that in this case, the limit at 2 does not match the function value at 2, so the function is discontinuous at this point, hence the function is nondifferentiable at this point as well.

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

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

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}}
By the intermediate value theorem, if $f(x)$  is continuous from $x\in \left[\pi ,{\frac {9}{4}}\pi \right]$ , then there exists an $x=\mu$  so that $f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )$  for $\mu \in \left[\pi ,{\frac {9}{4}}\pi \right]$ . $\blacksquare$
53. Use the intermediate value theorem to show that there exists a value $x=\Gamma$  so that $f(\Gamma )=2$  for $f(x)={\sqrt {-2x+4}}+5\ln \left(x^{2}\right)$  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.