Using the graph of $f,$ use limit notation to describe the behavior of $f(x)$ as follows:

a$.\underset{x{0}^{-}}{lim}f(x)=1$

b$.\underset{x{0}^{}}{lim}f(x)=3$

c$.\underset{x0}{lim}f(x)=$ does not exist

d$.f(0)=1x=0,f(0)=1$

Using the graph of $f$ above, use limit notation to describe the behavior of $f(x)$ as follows:

a$.\underset{x{1}^{-}}{lim}f(x)=2$

e$.\underset{x{3}^{-}}{lim}f(x)=$

b$.\underset{x{1}^{}}{lim}f(x)=1$

f$.\underset{x{3}^{}}{lim}f(x)=$

c$.\underset{x1}{lim}f(x)=$

g$.\underset{x3}{lim}f(x)=$

d$.f(1)=$

h$.f(3)=$

Find each limit$.$

a$.\underset{x0}{lim}\frac{2x}{3x1}=$

b$.\underset{x-1}{lim}\frac{x^2-1}{x1}=$

c$.\underset{x1}{lim}\frac{x^2-1}{x1}=$

Find each value for the given piecewise function.

Let $f(x)=\{\begin{array}{ccc}x& 1if& x2\\ (x-2{)}^2& if& x>2\end{array}$

a$.\underset{x{2}^{-}}{lim}f(x)=$

c$.\underset{x2}{lim}f(x)=$

b$.\underset{x{2}^{}}{lim}f(x)=$

d$.f(2)=$Use limit notation to describe the behavior of $f$ at each zero of the denominator; find all vertical asymptotes.

$f(x)=\frac{x4}{x^{2}2x-8}$

Use limit notation to find the horizontal asymptotes of each function.

a$.f(x)=\frac{2x^3-5x^{2}9}{5x^3-6x7}$

b$.f(x)=\frac{-4x^{4}1}{3x^2-7x3}$

c$.f(x)=\frac{2x^3-6x^{2}7}{3x^{6}5x^5-11x^{3}5}$

State where the function is continuous using inequality and interval notation.