Let $f$ be defined on the interval $\left [0,2\right ]$, and no where else, whose graph is:

Find

(a)

$\lim _{x\to 1^-} f(x)\begin {prompt} = \answer {2}\end {prompt}$

(b)

$\lim _{x\to 1^+} f(x)\begin {prompt} = \answer {2}\end {prompt}$

(c)

$\lim _{x\to 1} f(x)\begin {prompt} = \answer {2}\end {prompt}$

(d)

$f({1})\begin {prompt} = \answer {1}\end {prompt}$

(e)

$\lim _{x\to 0^-} f(x)\begin {prompt} = \answer {DNE}\end {prompt}$

(f)

$\lim _{x\to 0^+} f(x)\begin {prompt} = \answer {1}\end {prompt}$

$$\lim _{v\to -5 } \frac {v^2+6 v+5}{v+5}\begin {prompt} = \answer {-4}\end {prompt}$$

$$\lim _{z\to -4 } \frac {z^2+6 z+8}{z+4}\begin {prompt} = \answer {-2}\end {prompt}$$

$$\lim _{n\to -5 } \frac {\sqrt {4-n}-3}{n+5}\begin {prompt} = \answer {-\frac {1}{6}}\end {prompt}$$

$$\lim _{z\to -4 } \frac {\sqrt {-z-3}-1}{z+4}\begin {prompt} = \answer {-\frac {1}{2}}\end {prompt}$$

$$\lim _{x\to -1 } \frac {\frac {1}{x+3}+\frac {3}{x-5}}{x+1} \begin {prompt}=\answer {-\frac {1}{3}}\end {prompt}$$

$$\lim _{\theta \to 4 } \frac {\frac {1}{\theta -2}-\frac {7}{2 (\theta +3)}}{\theta -4} \begin {prompt}=\answer {-\frac {5}{28}}\end {prompt}$$

$$\lim _{ x\to 2 } \frac {A(x)-A(2)}{x-2} \begin {prompt}=\answer {-\frac {1}{36}}\end {prompt}$$

Let $$g(x) = \begin {cases} \frac {x^3 - 8}{x-2} &\text {if $x<1$,} \\ x^3+1 &\text {if $x>1$.} \end {cases}$$ Does $\lim _{x \to 2} g(x)$ exist? If it does, give its value. Otherwise write DNE. $$\lim _{x \to 2} g(x) = \answer {9}$$

Let $S(x) = \frac {|x|}{x}$. Does $\lim _{x \to -4} S(x)$ exist? If it does, give its value. Otherwise write DNE. $$\lim _{x \to -4} S(x) = \answer {-1}$$

Consider: $$\lim _{t\to 0} \left (t \cos \left (\frac {3}{t}\right )\right )$$ A good way to compute this limit would be to use limit lawsindeterminate formsthe Squeeze Theoremthe Intermediate Value Theorem .