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AP Calculus BC : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Limits

Limitplot

$\begin{align*}&\text{Which of the values most likely represents the value of f(x) at}\\&x=\infty\end{align*}$

Possible Answers:

$\infty$

$\text{Nonexistent.}$

$\frac{19}{15}$

Correct answer:

$\text{Nonexistent.}$

Explanation:

$\begin{align*}&\text{Although actually observing a function at an infinite value}\\&\text{of x is impossible, we can often determine via observation if}\\&\text{the function approaches a certain limit. Looking at the values}\\&\text{the function approaches as the absolute value of x increases}\\&\text{gives a clue. Things to look out for are if the function continuously}\\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\\&\text{If the function continuously climbs/decreases, it likely has}\\&\text{a limit that approaches positive/negative infinity. If it flattens,}\\&\text{the limit is a real number. If it oscillates without control,}\\&\text{the limit does not exist.}\\&\text{The limit of }f(\infty)\text{ is most likely }\text{Nonexistent.}\end{align*}$

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Example Question #11 : Estimating Limits From Graphs And Tables

Limitplot

$\begin{align*}&\text{What is the likely value of}\\&f(-\infty)\\&\text{Based on the graph?}\end{align*}$

Possible Answers:

$\frac{2}{7}$

$\text{Nonexistent.}$

$-\infty$

Correct answer:

$\text{Nonexistent.}$

Explanation:

$\begin{align*}&\text{Although actually observing a function at an infinite value}\\&\text{of x is impossible, we can often determine via observation if}\\&\text{the function approaches a certain limit. Looking at the values}\\&\text{the function approaches as the absolute value of x increases}\\&\text{gives a clue. Things to look out for are if the function continuously}\\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\\&\text{If the function continuously climbs/decreases, it likely has}\\&\text{a limit that approaches positive/negative infinity. If it flattens,}\\&\text{the limit is a real number. If it oscillates without control,}\\&\text{the limit does not exist.}\\&\text{The limit of }f(-\infty)\text{ is most likely }\text{Nonexistent.}\end{align*}$

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Example Question #12 : Limits

Limitplot

$\begin{align*}&\text{What is the likely value of}\\&f(-\infty)\\&\text{Based on the graph?}\end{align*}$

Possible Answers:

$-\infty$

$\text{Nonexistent.}$

$\frac{17}{18}$

Correct answer:

Explanation:

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Example Question #12 : Limits

Limitplot

$\begin{align*}&\text{What is the likely value of}\\&f(-\infty)\\&\text{Based on the graph?}\end{align*}$

Possible Answers:

$\text{Nonexistent.}$

$\frac{1}{4}$

$-\infty$

Correct answer:

$\frac{1}{4}$

Explanation:

$\begin{align*}&\text{Although actually observing a function at an infinite value}\\&\text{of x is impossible, we can often determine via observation if}\\&\text{the function approaches a certain limit. Looking at the values}\\&\text{the function approaches as the absolute value of x increases}\\&\text{gives a clue. Things to look out for are if the function continuously}\\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\\&\text{If the function continuously climbs/decreases, it likely has}\\&\text{a limit that approaches positive/negative infinity. If it flattens,}\\&\text{the limit is a real number. If it oscillates without control,}\\&\text{the limit does not exist.}\\&\text{The limit of }f(-\infty)\text{ is most likely }\frac{1}{4}\end{align*}$

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Example Question #11 : Limits

Limitplot

$\begin{align*}&\text{Which of the values most likely represents the value of f(x) at}\\&x=\infty\end{align*}$

Possible Answers:

$\text{Nonexistent.}$

$\infty$

$-\frac{1331}{512}$

Correct answer:

$-\frac{1331}{512}$

Explanation:

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Example Question #11 : Limits

Limitplot

$\begin{align*}&\text{Based on the graph, decide which of the following functions values likely depicts}\\&f(\infty)\end{align*}$

Possible Answers:

$\infty$

$\frac{17}{12}$

$\text{Nonexistent.}$

Correct answer:

Explanation:

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Example Question #11 : Limits

Limitplot

$\begin{align*}&\text{Which of the values most likely represents the value of f(x) at}\\&x=\infty\end{align*}$

Possible Answers:

$\text{Nonexistent.}$

$-\infty$

$-\frac{15}{17}$

Correct answer:

$-\infty$

Explanation:

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Example Question #1 : Limits And Continuity

Rational_graph

The graph above is a sketch of the function . For what intervals is continuous?

Possible Answers:

$(-\infty, -2) \cup (-2, 2)$

$(-\infty, -2) \cup (-2, 2) \cup \(2, \infty)$

$(-\infty, \infty)$

$(-2, 2) \cup \(2, \infty)$

$(-\infty, -2) \cup \(2, \infty)$

Correct answer:

$(-\infty, -2) \cup (-2, 2) \cup \(2, \infty)$

Explanation:

For a function to be continuous at a point , f(a) must exist and $\lim_{x \rightarrow a}f(x)=f(a)$ .

This is true for all values of except x=-2 and x=2 .

Therefore, the interval of continuity is $(-\infty, -2) \cup (-2, 2) \cup \(2, \infty)$ .

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Example Question #2 : Limits And Continuity

Consider the piecewise function:

$y=\left\{\begin{matrix} 1;\: x<-5\\0;\:x>-5 \end{matrix}\right.$

What is $\lim_{x \to -5}$ ?

Possible Answers:

0<x<1

Limit does not exist.

$\frac{1}{2}$

Correct answer:

Limit does not exist.

Explanation:

The piecewise function

$\left\{\begin{matrix} 1;\: x<-5\\0;\:x>-5 \end{matrix}\right.$

indicates that is one when is less than five, and is zero if the variable is greater than five. At x=-5 , there is a hole at the end of the split.

The limit does not indicate whether we want to find the limit from the left or right, which means that it is necessary to check the limit from the left and right. From the left to right, the limit approaches 1 as approaches negative five. From the right, the limit approaches zero as approaches negative five.

Since the limits do not coincide, the limit does not exist for $\lim_{x \to -5}$ .

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Example Question #3 : Limits And Continuity

Consider the function $f(x)=\frac{x^2-7x+12}{x-3}$ .

Which of the following statements are true about this function?

I. $f(x)\ is\ continuous\ at\ x = 3.$

II. $\lim_{x\to3}f(x)=-1$

III. $\lim_{x\to3}f(x)\ does\ not\ exist$

Possible Answers:

I and II

I and III

III only

II only

Correct answer:

I and II

Explanation:

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

$f(3)=\frac{3^2-7(3)+12}{3-3}=\frac{(x-3)(x-4)}{(x-3)}=(x-4)=(3-4)=-1$ .

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

$f(x)=\frac{x^2-7x+12}{x-3}=\frac{(x-4)(x-3)}{(x-3)}=x-4$

$\lim_{x\to3}(x-4)=-1$

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

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AP Calculus BC : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Limits

Example Question #11 : Estimating Limits From Graphs And Tables

Example Question #12 : Limits

Example Question #12 : Limits

Example Question #11 : Limits

Example Question #11 : Limits

Example Question #11 : Limits

Example Question #1 : Limits And Continuity

Example Question #2 : Limits And Continuity

Example Question #3 : Limits And Continuity

All AP Calculus BC Resources

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