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Power rule

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{9 x^{56}}{-10 e^{x}-10 x^{97}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{-10 x^{13}}{-2 x^{2}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{-3 x^{23}+4^{x}}{3 \log _{4} x-5 x^{8}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{2 x^{16}-3 x^{24}}{-2 x^{24}+\log _{5} x}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{x^{71}+\log _{3} x}{3 x^{71}+x^{29}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{x^{37}}{\log _{4} x+4^{x}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{\ln x+10 x^{20}}{x^{88}+e^{x}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{9 x^{61}}{-9 x^{61}+x^{60}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{x^{68}}{6 x^{68}+x^{62}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{\ln x}{x^{46}+10 x^{91}}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Which statement best describes the limit shown below?

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\lim _{x \rightarrow \infty} \frac{-3(2)^{x}}{5 \log _{2} x}

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The limit equals zero

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The limit does not exist

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The limit exists and does not equal zero

Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-41 x^{5}+x^{6}}}{5 x+5 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt{-2+49 x^{4}}}{3 x+10 x^{3}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{64 x^{10}+2 x^{6}+41 x^{4}}}{3 x^{4}+3 x^{3}+8 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt{64 x^{9}+39 x+39 x^{8}}}{9 x+2 x^{4}+4 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt{-24 x^{2}+49 x^{6}}}{9+7 x^{3}+2 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt{64 x^{6}+12 x^{5}+42 x^{4}}}{7+6 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-8 x^{14}+16 x+x^{8}}}{9 x^{4}+9 x}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-6 x^{2}-27 x^{6}}}{1+9 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-33 x-27 x^{5}-33}}{5+9 x+5 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt{9 x^{6}+20 x^{4}}}{10 x^{3}+6 x}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt{x^{4}-8 x}}{2+x+3 x^{3}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{27 x^{6}-14 x^{3}}}{7+7 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{48 x^{5}+x^{6}-39 x^{2}}}{10 x^{2}+7 x^{3}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt{64 x^{8}-23 x^{6}}}{5 x^{2}+8 x^{4}+10}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-19 x^{3}-27 x^{6}}}{5 x+9 x^{2}+9}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{31-27 x^{6}}}{7+8 x+3 x^{2}}

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Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt{-11 x^{2}+22 x+4 x^{4}}}{9 x+4 x^{3}}

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m=\frac{k T^{2}}{2 \pi}

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An experimental physicist uses the given equation to model the relationship between an

m

kilogram mass connected to a spring with spring constant,

k

, and a period of oscillation,

T

. If the physicist changes the mass and the period of oscillation drops to

\frac{1}{4}

of its original value, which of the following describes the change in mass?

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1

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(A) The new mass is

\frac{1}{2}

of the original mass.

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(B) The new mass is

\frac{1}{4}

of the original mass.

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(C) The new mass is

\frac{1}{16}

of the original mass.

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(D) The new mass is quadrupled.

An architect has been commissioned to build a public landmark. She wants to design an enormous parabolic arch. The equation that models the height,

h

m

meters away from the center is given by:

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h(m)=-\frac{1}{50} m^{2}+200

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Which of the following equivalent expressions displays, as a constant or coefficient, the distance from the center to where the arch meets the ground?

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Note: Negative values of

m

are to the left of the center of the arch and positive values are to the right.

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1

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-\frac{1}{50}(m+100)(m-100)

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-\frac{1}{50}\left(m^{2}-10,000\right)

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-\frac{1}{50}(m+40)(m-40)+168

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-\frac{1}{50}(m+50)(m-50)+150

\left(-1-5 n+a n^{2}\right)-\left(7 n-4 n^{2}+9\right)=12 n^{2}-12 n-10

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If the given equation is true for all values of

n

a

is a constant, what is the value of

a

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1

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8

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12

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16

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19

Consider the curve given by the equation

x^{3}+x y=-2

. It can be shown that

\frac{d y}{d x}=\frac{-3 x^{2}-y}{x}

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Find the point on the curve where the line tangent to the curve is horizontal.

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(\square

\square)

g

be a twice differentiable function, and let

g(-5)=-6

g^{\prime}(-5)=0

g^{\prime \prime}(-5)=0

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What occurs in the graph of

g

(-5,-6)

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1

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(-5,-6)

is a minimum point.

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(-5,-6)

is a maximum point.

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(C) There's not enough information to tell.

h

be a twice differentiable function, and let

h(-4)=-3

h^{\prime}(-4)=0

h^{\prime \prime}(-4)=0

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What occurs in the graph of

h

(-4,-3)

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1

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(-4,-3)

is a minimum point.

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(-4,-3)

is a maximum point.

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(C) There's not enough information to tell.

h(x)=\log _{2}(x)

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Can we use the mean value theorem to say the equation

h^{\prime}(x)=1

has a solution where

1<x<2

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1

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(A) No, since the function is not differentiable on that interval.

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(B) No, since the average rate of change of

h

over the interval

1 \leq x \leq 2

1

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(C) Yes, both conditions for using the mean value theorem have been met.

f(x)=\frac{1}{x-2}

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Can we use the mean value theorem to say the equation

f^{\prime}(x)=\frac{1}{2}

has a solution where

1<x<4

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1

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(A) No, since the function is not differentiable on that interval.

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(B) No, since the average rate of change of

f

over the interval

1 \leq x \leq 4

\frac{1}{2}

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(C) Yes, both conditions for using the mean value theorem have been met.

f(x)=\left\{\begin{array}{ll}\frac{1}{x+1} & \text { for }-5<x \leq-4 \\ 2^{x} & \text { for } x>-4\end{array}\right.

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\lim _{x \rightarrow-4^{+}} f(x)

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1

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-4

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-\frac{1}{3}

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\frac{1}{16}

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(D) The limit doesn't exist.

h(x)=\left\{\begin{array}{ll} 2 x+1 & \text { for } x<-2 \\ \frac{1}{x-1} & \text { for }-2 \leq x<1 \end{array}\right.

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\lim _{x \rightarrow-3} h(x)

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1

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-5

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-3

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-\frac{1}{4}

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(D) The limit doesn't exist.

\lim _{x \rightarrow-4} \frac{x^{2}-15}{x^{2}-16}

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1

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-\frac{1}{8}

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\frac{1}{8}

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\frac{19}{20}

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(D) The limit doesn't exist

\lim _{x \rightarrow-5} \frac{(x-3)(x+4)}{x+5}

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1

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-\frac{4}{5}

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0

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\frac{9}{5}

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(D) The limit doesn't exist

g(x)=\left\{\begin{array}{ll} e^{x} & \text { for }-5<x<-1 \\ \frac{x}{e} & \text { for }-1 \leq x<0 \end{array}\right.

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\lim _{x \rightarrow-1^{-}} g(x)

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1

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-\frac{1}{e}

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0

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\frac{1}{e}

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(D) The limit doesn't exist.

h(x)=\left\{\begin{array}{ll} \frac{1}{2 x} & \text { for } x \leq-2 \\ 2^{x} & \text { for }-2<x \leq 0 \end{array}\right.

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\lim _{x \rightarrow-2} h(x)

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1

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-2

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-\frac{1}{4}

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\frac{1}{4}

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(D) The limit doesn't exist.

g(x)=\left\{\begin{array}{ll}\frac{x}{3}-2 & \text { for } 0<x<6 \\ \cos (x \cdot \pi) & \text { for } 6 \leq x \leq 10\end{array}\right.

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\lim _{x \rightarrow 6} g(x)

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1

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-1

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0

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1

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(D) The limit doesn't exist.

f(x)=\left\{\begin{array}{ll}\sqrt{7+x} & \text { for }-7 \leq x \leq-3 \\ x^{2}-5 & \text { for } x>-3\end{array}\right.

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\lim _{x \rightarrow-3} f(x)

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1

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-3

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2

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4

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(D) The limit doesn't exist.

\lim _{x \rightarrow-\frac{\pi}{2}} \tan (x)=?

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1

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-1

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-\frac{1}{2}

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0

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(D) The limit doesn't exist.

f(x)=\left\{\begin{array}{ll} 2^{x} & \text { for } 0<x \leq 4 \\ 8 \sqrt{x} & \text { for } x>4 \end{array}\right.

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\lim _{x \rightarrow 4} f(x)

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1

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4

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8

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16

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(D) The limit doesn't exist.

g(x)=\frac{5 x}{5 x^{4}-4}

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\lim _{x \rightarrow \infty} g(x)

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1

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-\frac{5}{4}

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1

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0

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(D) The limit is unbounded

h(x)=\frac{x^{2}+6}{x^{2}+3}

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\lim _{x \rightarrow \infty} h(x)

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1

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0

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2

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1

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(D) The limit is unbounded

\lim _{x \rightarrow \infty} \frac{5 x^{4}+x^{2}}{2 x^{4}-x^{3}-4}

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1

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\frac{5}{2}

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0

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-\frac{1}{4}

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(D) The limit is unbounded

g(x)=\frac{-x^{3}+4 x}{4 x^{3}-2 x^{2}+4}

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\lim _{x \rightarrow \infty} g(x)

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1

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-\frac{1}{4}

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1

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0

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(D) The limit is unbounded