Practice for Exam One

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Practice for Exam One.

Compute the left and right limits of $\frac {\sage {p2fdisptop}}{\sage {p2fdispbot}}$ as $x$ approaches $\sage {p2c7}$ .

$\lim \limits _{x\rightarrow \sage {p2c7}^+} \frac {\sage {p2fdisptop}}{\sage {p2fdispbot}} = \answer {\sage {p2ans1}}$

$\lim \limits _{x\rightarrow \sage {p2c7}^-} \frac {\sage {p2fdisptop}}{\sage {p2fdispbot}} = \answer {\sage {p2ans2}}$

Given the previous one sided limit values, determine if the general limit at $\sage {p2c7}$ exists.

The Limit: $\lim \limits _{x\rightarrow \sage {p2c7}} \frac {\sage {p2fdisptop}}{\sage {p2fdispbot}}$ ...

Exists and equals $\sage {p2ans1}$ . Exists and equals $\sage {p2ans2}$ . Fails to exists

Remember that, in order for a limit to exist, both the left and right limits must exist and be equal.

If you know that $\lim \limits _{x \to \sage {p3target}} f(x) = \sage {p3c1}$ and $\lim \limits _{x \to \sage {p3target}} g(x) = \sage {p3c2}$ , then evaluate the following limit:

$\lim \limits _{x \to \sage {p3target}} \left ( \sage {p3c3} f(x) + \sage {p3c4}g(x) \right ) = \answer {\sage {p3ans}}$

Remember that you can use limit laws to rewrite $\lim _{x \to \sage {p3target}} \left ( \sage {p3c3} f(x) + \sage {p3c4}g(x) \right ) = \lim _{x \to \sage {p3target}}( \sage {p3c3}) \cdot \lim _{x \to \sage {p3target}}(f(x)) + \lim _{x \to \sage {p3target}}(\sage {p3c4}) \cdot \lim _{x \to \sage {p3target}}(g(x))$

If $f(x)$ is a continuous real-valued function, and you know that $f(\sage {p4start})=\sage {p4c1}$ and $f(\sage {p4end})=\sage {p4c2}$ , must it be true that $f$ attains the value $\sage {p4c3}$ at some $x$ value between $\sage {p4start}$ and $\sage {p4end}$ ? If yes, enter $1$ into the following box. If no, enter $0$ . $\answer {\sage {p4ans}}$

Consider the following piecewise function: $f(x) = \begin {cases} \sage {p5f1} & \sage {p5l1}\leq x \leq \sage {p5r1} \\ \sage {p5f2} & \sage {p5r1} < x \leq \sage {p5r2} \\ \sage {p5f3} & \sage {p5r2} < x \leq \sage {p5r3} \end {cases}$

Determine if $f(x)$ is continuous at $x = \sage {p5r1}$ . If it is, enter $1$ in the following box, otherwise enter $0$ . $\answer {\sage {p5contFlip}}$ .

To evaluate, find the row that has the input in the listed domain span. Once you find the correct row, plug the input in as the x-value into the function in that row to find the value of the piecewise function at that input.

Compute the following limit: (If the limit does not exist, enter DNE) $\lim \limits _{x\rightarrow \sage {p6target}} \sage {p6f1} = \answer {\sage {p6ans}}$

Continuous functions are nice for limits, since (by definition) the limit of a continuous function is just the value of that function!

Compute the following limit: (If the limit does not exist, enter DNE) $\lim \limits _{x\rightarrow \sage {p7c3}}\frac {\sage {p7ftop}}{\sage {p7fbot}} = \answer {\sage {p7ans}}$

If the limit has a 0/0 form, then you should try factoring it to cancel any common terms and see if you can resolve the indeterminate form. If it resolves to a C/0 form for some non-zero C, that tells you it is a vertical asymptote; to tell if the limit exists you need to use a sign chart!

Compute the following limit: (If the limit does not exist, enter DNE) $\lim \limits _{x\rightarrow \sage {p8c3}}\frac {\sage {p8ftop}}{\sage {p8fbot}} = \answer {\sage {p8ans}}$

Let $f(x) = \frac {\sage {p9ftop}}{\sage {p9fbot}}$ . Compute:

$\lim \limits _{x\rightarrow \infty }f(x) = \answer {\sage {p9ansR}}$
$\lim \limits _{x\rightarrow -\infty }f(x) = \answer {\sage {p9ansL}}$

Limits at infinity involve both right and left limits. The key to computing limits at infinity is manipulating the functions so that there aren’t more than one infinitely large term appearing anywhere when you try to compute the limit.

Consider the function $f(x) = \frac {\sage {p10ftop}}{\sage {p10fbot}}$ . What is the sum of the $x$ -values that have vertical asymptotes? $\answer {\sage {p10ans}}$ .

Vertical Asymptotes occur when you have something of the form non-zero over zero, when trying to evaluate a limit. In particular, they occur when at least one of the one-sided limits goes to infinity.

Let $f(x) = \frac {\sage {p11ftop}}{\sage {p11fbot}}$ . Determine if $f(x)$ has any horizontal asymptotes.

$f(x)$ has a right horizontal asymptote of $\answer {\sage {p11ansR}}$ (enter $\infty$ if $f(x)$ is positive and has no horizontal asymptote, or $-\infty$ if $f(x)$ is negative and has no horizontal asymptote).
$\lim \limits _{x\rightarrow -\infty }f(x) = \answer {\sage {p11ansL}}$ (enter $\infty$ if $f(x)$ is positive and has no horizontal asymptote, or $-\infty$ if $f(x)$ is negative and has no horizontal asymptote).

Finding Horizontal Asymptotes is the same process as finding limits at positive and negative infinity.

A car is traveling along a straight road. It’s position in miles at any time $t$ in hours is given by $\sage {p12f1}$ . Calculate the car’s speed at time $t=\sage {p12target}$ . $\answer {\sage {p12ans}}$ mph.