Asymptotic behavior in terms of limits involving infinity - AP Calculus AB

Card 0 of 242

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-0.3\&\text{We find a zero denominator for the function: }\&-\frac{(x - 11)}{(4x + 1)}\end{align}$

Compare your answer with the correct one above

Question

Evaluate

$\lim_{x\rightarrow \infty}\frac{4x^3+2x^2-5x+1}{x^3-x^2+3x-5}$

Answer

The equation $\frac{4x^3+2x^2-5x+1}{x^3-x^2+3x-5}$ will have a horizontal asymptote y=4.

We can find the horizontal asymptote by looking at the terms with the highest power.

The terms with the highest power here are in the numerator and in the denominator. These terms will "take over" the function as x approaches infinity. That means the limit will reach the ratio of the two terms.

The ratio is $\frac{4x^3}{x^3}=\frac{4}{1}=4$

Compare your answer with the correct one above

Question

$\lim_{x\rightarrow -\infty}\frac{4x^5-3x^2+2}{2x^3-x^2+1}=?$

Answer

For this infinity limit, we need to consider the leading terms of both the numerator and the denominator. In our problem, the leading term of the numerator is larger than the leading term of the denominator. Therefore, it will be growing at a faster rate.

$\lim_{x\rightarrow -\infty}\frac{4x^5-3x^2+2}{2x^3-x^2+1}=\frac{4x^5}{2x^3}=2x^2.$

Now, simply input the limit value, and interpret the results.

$\lim_{x\rightarrow -\infty} 2x^2=2(-\infty)^2=2*\infty=\infty.$

Compare your answer with the correct one above

Question

$\lim_{x\rightarrow \infty}\frac{2x^2+3x}{10x^2+x}$

Answer

For infinity limits, we need only consider the leading term in both the numerator and the denominator. Here, we have the case that the exponents are equal in the leading terms. Therefore, the limit at infinity is simply the ratio of the coefficients of the leading terms.

$\lim_{x\rightarrow \infty}\frac{2x^2+3x}{10x^2+x}=\frac{2x^2}{10x^2}=\frac{2}{10}=\frac{1}{5}.$

Compare your answer with the correct one above

Question

$\lim_{x\rightarrow \infty}\frac{3x^7+4x^4-3x^2}{5x^5+2x^4}$

Answer

Infinity limits can be found by only considering the leading term in both the numerator and the denominator. In this problem, the numerator has a higher exponent than the denominator. Therefore, it will keep increasing and increasing at a much faster rate. These limits always tend to infinity.

$\lim_{x\rightarrow \infty}\frac{3x^7+4x^4-3x^2}{5x^5+2x^4}=\frac{3x^7}{5x^5}=\frac{3}{5}x^2.$

$\lim_{x\rightarrow \infty}\frac{3}{5}x^2=\infty.$

Compare your answer with the correct one above

Question

$\lim_{x\rightarrow -\infty} \frac{2x^3-3x^2+4x}{4x^5-3x^3}$

Answer

For infinity limits, we only consider the leading term in both the numerator and the denominator. Then, we need to consider the exponents of the leading terms. Here, the denominator has a higher degree than the numerator. Therefore, we have a bottom heavy fraction. Even though we are evaluating the limit at negative infinity, this will still tend to zero because the denominator is growing at a faster rate. You can convince yourself of this by plugging in larger and larger negative values. You will just get a longer and smaller decimal.

$\lim_{x\rightarrow -\infty} \frac{2x^3-3x^2+4x}{4x^5-3x^3}=\frac{2x^3}{4x^5}=\frac{1}{2x^2}$

$\lim_{x\rightarrow -\infty}\frac{1}{2x^2}=0.$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Of the following functions, which most likely represents the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-1.8\&\text{We find a zero denominator for the function: }\&-\frac{(10x + 11)}{(11x + 20)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=0.75\&\text{This matches the ratio of coefficients for the function: }\&\frac{(6x + 1)}{(8x - 17)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Which of the four following functions is depicted in the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-1.1\&\text{We find a zero denominator for the function: }\&\frac{(8x - 2)}{(10x + 11)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-2.67\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(8x - 18)}{(3x + 3)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Which of the four following functions is depicted in the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-0.8\&\text{We find a zero denominator for the function: }\&\frac{(10x - 1)}{(6x + 5)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Of the following functions, which most likely represents the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=2\&\text{This matches the ratio of coefficients for the function: }\&\frac{(16x - 12)}{(8x - 18)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-0.73\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(11x - 18)}{(15x - 17)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Of the following functions, which most likely represents the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-0.2\&\text{We find a zero denominator for the function: }\&\frac{(11x - 18)}{(17x + 4)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-1.5\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(18x + 17)}{(12x - 11)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Which of the four following functions is most likely depicted in the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-1.3\&\text{We find a zero denominator for the function: }\&-\frac{(x + 1)}{(10x + 13)}\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{Which of the four following functions is probably depicted in the sample data above?}\end{align}$

Answer

$\begin{align}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-3.8\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(19x - 15)}{(5x - 15)}\end{align}$

Compare your answer with the correct one above

Question

Compute $\lim_{g\rightarrow \infty } \left ( \frac{g^4 -2g^5}{g^5 + g^3} \right )$

Answer

Firstly, recall the rules for evaluating the limits of rational expressions as they go to infinity:

If the degree on the top is greater than the degree on the bottom, the limit does not exist.

If the degree on the top is less than the degree on the bottom, the limit is 0.

If the degrees in both the numerator and denominator are equal, the limit is the ratio of the leading coefficients.

These rules follow from how the function grows as the inputs get larger, ignoring everything but the leading terms.

Then, note that, while the numerator is not written in the correct order for you, that the highest power in the numerator is 5. Likewise, the highest power in the denominator is 5. Thus, the limit will be the ratio of the coefficients on the terms, which is $\frac{-2}{1} = -2$ .

Compare your answer with the correct one above

Question

Find $\lim_{x\rightarrow \infty}\frac{ln(x)}{x}$ .

Answer

First step for finding limits: evaluate the function at the limit.

$\frac{ln(\infty )}{\infty }=\frac{\infty }{\infty }$ which is an Indeterminate Form.

This got us nowhere. However, since we have an indeterminate form, we can use L'Hopital's Rule (take the derivative of the top and bottom and the limit's value won't change).

$\lim_{x\rightarrow \infty}\frac{ln(x)}{x}=\lim_{x\rightarrow \infty }\frac{1/x}{1}=\lim_{x\rightarrow \infty}\frac{1}{x}$

This is something we can evaluate:

$\lim_{x\to\infty}\frac{1}{x}\rightarrow \frac{1}{\infty}\rightarrow 0$

The value of this limit is .

Compare your answer with the correct one above

Question

Assume that a population of bunnies grows at a rate of $\frac{\mathrm{d} P}{\mathrm{d} t} = 5P\left ( \frac{350 - P}{350} \right )$ where is the number of bunnies in the population at time . If the population begins with 14 bunnies, given unlimited time to grow, how many bunnies do you expect there to be in the population?

Note: No integration is required for this problem.

Answer

The equation given is a model of logistic growth. Note that one of the other common forms this equation might be given in is $\frac{\mathrm{d} P}{\mathrm{d} t} = 5P\left ( 1-\frac{P}{350} \right )$ . What we want to know is what the population will be given unlimited time, or what $\lim_{t\rightarrow \infty} P(t)$ is.

Looking at the form of the derivative above, note that if we start with 14 bunnies, we get a positive derivative. This means that the population is increasing. As our function is continuous, the population will keep growing until the derivative hits 0. When does this occur?

In the above form, it should be more clear that the derivative is only 0 at and . Thus, the population will keep growing, but never go above 350, because if it were to hit 350, the derivative would be 0 and growth would stop. (Alternatively, if the population were above 350, you can see the derivative would be negative and that the population would shrink back down to 350. Indeed, it wasn't necessary to tell you we started with 14 bunnies -- the limit will be the same for any positive starting value.)

Compare your answer with the correct one above

Tap the card to reveal the answer