Exam 3: Limits and Continuity

Find the limit. - $\lim _ { x \rightarrow } \sqrt { 8 + \cos ^ { 2 } x }$

Find the limit. -f $\lim _ { x \rightarrow 2 } \frac { f ( x ) } { x } = 3$ , find $\lim _ { x \rightarrow - 2 } f ( x )$ .

$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$ - $\lim _ { x \rightarrow } \frac { \tan 4 x } { x }$

Find the average rate of change of the function over the given interval. - $y= - 3 x ^ { 2 } - x , [ 5,6 ]$

Graph the rational function. Include the graphs and equations of the asymptotes. - $y = \frac { 1 } { x + 1 }$

Sketch the graph of a function y = f(x) that satisfies the given conditions. - $f ( 0 ) = 4 , f ( 1 ) = - 4 , f ( - 1 ) = - 4 , \lim _ { x \rightarrow \infty } f ( x ) = 0 \text {. }$

Use a CAS to plot the function near the point x0 being approached. From your plot guess the value of the limit. -Provide a short sentence that summarizes the general limit principle given by the formal notation $\lim _ { x \rightarrow a } [ f ( x ) \pm g ( x ) ] = \lim _ { x \rightarrow a } f ( x ) \pm \lim _ { x \rightarrow a } g ( x ) = L \pm M$ , given that $\lim _ { x \rightarrow a } f ( x ) = L$ and $\lim _ { x \rightarrow a } g ( x ) = M$ .

Find all points where the function is discontinuous. -

Use the table of values of f to estimate the limit. - $\text { Let } f(x)=x^{2}+8 x-2, \text { find } \lim _{x \rightarrow 2} f(x)$ 1.9 1.99 1.999 2.001 2.01 2.1 ()

Find the limit. - $\lim _ { x \rightarrow 0 } \frac { x ^ { 2 } - 8 x + 15 } { x ^ { 3 } - 9 x }$

Determine if the given function can be extended to a continuous function at x = 0. If so, approximate the extended function's value at x = 0 (rounded to four decimal places if necessary). If not, determine whether the function can be continuously extended from the left or from the right and provide the values of the extended functions at x = 0. Otherwise write "no continuous extension." - $f ( x ) = ( 1 + 2 x ) ^ { 1 / x }$

Provide an appropriate response. -The inequality $1 - \frac { x ^ { 2 } } { 2 } < \frac { \sin x } { x } < 1$ holds when $x$ is measured in radians and $| x | < 1$ . Find $\lim _ { x \rightarrow \theta } \frac { \sin x } { x }$ if it exists.

A function $\mathrm { f } ( \mathrm { x } )$ , a point $\mathrm { c }$ , the limit of $\mathrm { f } ( \mathrm { x } )$ as $x$ approaches $\mathrm { c }$ , and a positive number $\varepsilon$ is given. Find a number $\delta > 0$ such that for all $x , 0 < | \mathrm { x } - \mathrm { c } | < \delta \Rightarrow | \mathrm { f } ( \mathrm { x } ) - \mathrm { L } | < \varepsilon$ . - $f ( x ) = 8 x - 9 , L = 15 , c = 3 \text {, and } \varepsilon = 0.01$

Find the limit. - $\lim _ { x \rightarrow } \sqrt { x ^ { 2 } + 18 x } - x$

Provide an appropriate response. -Given $\lim _ { x \rightarrow \theta ^ { - } } f ( x ) = L _ { 1 } , \lim _ { x \rightarrow \theta ^ { + } } f ( x ) = L _ { r }$ , and $L _ { 1 } \neq L _ { r }$ , which of the following statements is true? I. $\lim _ { x \rightarrow - } f ( x ) = L _ { 1 }$ II. $\lim _ { x \rightarrow \theta } f ( x ) = L _ { r }$ III. $\lim _ { x \rightarrow 0 } f ( x )$ does not exist.

Find numbers a and b, or k, so that f is continuous at every point. - $f ( x ) = \left\{ \begin{array} { l l } x ^ { 2 } , & \text { if } x \leq 4 \\ k x , & \text { if } x > 4 \end{array} \right.$

Use a CAS to plot the function near the point x0 being approached. From your plot guess the value of the limit. - $\lim _ { x \rightarrow 0 } \frac { \sqrt { 9 + 2 x } - 3 } { x }$

Provide an appropriate response. - $\text { If } f ( x ) = 2 x ^ { 3 } - 5 x + 5 \text {, show that there is at least one value of } c \text { for which } f ( x ) \text { equals } \pi$

Find all points where the function is discontinuous. -

Find the limit and determine if the function is continuous at the point being approached. - $\lim _ { x \rightarrow 2 \pi } \sin \left( \frac { 9 \pi } { 2 } \cos ( \tan x ) \right)$