
Try these problems.

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

Try to factor either the numerator or the denominator.

Try to factor either the numerator or the denominator.

Multiply by $\frac {\sqrt {4-n}+3}{\sqrt {4-n}+3}$.

Multiply by $\frac {\sqrt {-z-3}+1}{\sqrt {-z-3}+1}$.

Multiply by $\frac {(x-5) (x+3)}{(x-5) (x+3)}$.

Multiply by $\frac {(\theta -2) (\theta +3)}{(\theta -2) (\theta +3)}$.
Let $A(x) = \frac {1}{x +4}$. Compute

Let Does $\lim _{x \to 2} g(x)$ exist? If it does, give its value. Otherwise write DNE.

Note that, close to $x=2$, the rule for $g(x)$ is $x^3+1$.

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

Close to $x=-4$, $S$ has the rule $\frac {-x}{x}$.

Consider: A good way to compute this limit would be to use limit lawsindeterminate formsthe Squeeze Theoremthe Intermediate Value Theorem .

List two functions $g$ and $h$ such that for all $t$ except for $t=\answer {0}$ on some interval containing $t=0$.
Compute:
By the Squeeze Theorem: