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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #4 : Limits Of Functions (Including One Sided Limits)

Evaluate the following limit:

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

Possible Answers:

$\frac{1}{5}$

$\infty$

Correct answer:

$\infty$

Explanation:

As x approaches infinity, the x values that are raised to the highest power dominate the function and all other values become negligible. In this limit, x³has the highest order in the numerator and 5x²has the highest power of all the values in the denominator. We can simplify the limit to the following:

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

After factoring out x² it is easy to see that the limit diverges to infinity.

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Example Question #1 : Calculating Limits Using Algebra

$\lim_{x\rightarrow 0}\frac{x^2-2x+3}{x-2}$

Possible Answers:

$\frac{-3}{2}$

Undefined.

$\frac{3}{2}$

Correct answer:

$\frac{-3}{2}$

Explanation:

The first step to evaluating limits is always attempt to direct substitute the value into the function. In this case, that is all you need to do.

$\lim_{x\rightarrow 0}\frac{x^2-2x+3}{x-2}=\frac{0^2-2(0)+3}{0-2}=\frac{-3}{2}.$

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Example Question #2 : Calculating Limits Using Algebra

$\lim_{x\rightarrow 2}\frac{x^2+5x-14}{x-2}=?$

Possible Answers:

$-\infty$

$\infty$

Undefined

Correct answer:

Explanation:

To start this problem, we need to factor the quadratic equation. This will result in a hole at x=2 , which will allow us to directly substitute the limit value into the resulting function to determine the limit.

$\frac{x^2+5x-14}{x-2}=\frac{(x-2)(x+7)}{x-2}=x+7.$

$\lim_{x\rightarrow 2}{x+7}=2+7=9.$

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Example Question #3 : Calculating Limits Using Algebra

$\lim_{x\rightarrow -3}\frac{2x+4}{x-3}=?$

Possible Answers:

Undefined.

$\frac{-1}{3}$

$\frac{1}{6}$

$\frac{-1}{6}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

To evaluate this limit, we need to direct substitute the value in question into our function. Note, this value is in the domain of the function, so direct substitution is the only method we need here.

$\lim_{x\rightarrow -3}\frac{2x+4}{x-3}=\frac{2(-3)+4}{(-3-3)}=\frac{-2}{-6}=\frac{1}{3}.$

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Example Question #1 : Limits Of Functions (Including One Sided Limits)

Evaluate $\lim_{x \to 3} \frac{x^2-9}{x-3}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

This limit can't be evaluated by a simple substitution; f(x) is not defined at x=3 , so some simplification is in order first.

$\lim_{x \to 3} \frac{x^2-9}{x-3}$ . (Start)

$=\lim_{x \to 3} \frac{(x+3)(x-3)}{x-3}$ . (Factor the numerator)

$=\lim_{x \to 3} x+3$ . (Cancel the x-3 terms)

=3+3 . (Substitute in the )

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Example Question #1 : Calculating Limits Using Algebra

$\begin{align*}&\text{Determine the limit if it exists of }\frac{x - 8}{x^{2} - 21x + 104}\\&\text{At }x=8\end{align*}$

Possible Answers:

$-\frac{1}{5}$

$\frac{1}{5}$

Correct answer:

$-\frac{1}{5}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=8\text{ into the function}\\&\frac{x - 8}{x^{2} - 21x + 104}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points, just close to them.}\\&\text{Factoring our function we find:}\\&\frac{x - 8}{(x - 8)\cdot (x - 13)}\\&\text{Which reduces to:}\\&\frac{1}{(x - 13)}\\&\text{At the value of }x=8\text{ the function is}\\&-\frac{1}{5}\end{align*}$

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Example Question #371 : Ap Calculus Ab

$\begin{align*}&\text{Evaluate the limit of }\frac{4x + x^{2} - 60}{2x + x^{2} - 80}\\&\text{At the point }x=-10\end{align*}$

Possible Answers:

$-\frac{8}{9}$

$-\frac{9}{8}$

$\frac{8}{9}$

$\frac{9}{8}$

Correct answer:

$\frac{8}{9}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=-10\text{ into the function}\\&\frac{4x + x^{2} - 60}{2x + x^{2} - 80}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{(x - 6)\cdot (x + 10)}{(x - 8)\cdot (x + 10)}\\&\text{Which reduces to:}\\&\frac{(x - 6)}{(x - 8)}\\&\text{At the value of }x=-10\text{ the function is}\\&\frac{8}{9}\end{align*}$

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Example Question #11 : Calculating Limits Using Algebra

$\begin{align*}&\text{Find the limit of }f(x)=\frac{42x + 3x^{2} + 144}{43x + 5x^{2} + 24}\\&\text{At the location }x=-8\end{align*}$

Possible Answers:

$-\frac{6}{37}$

$-\frac{37}{6}$

$\frac{37}{6}$

$\frac{6}{37}$

Correct answer:

$\frac{6}{37}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=-8\text{ into the function}\\&\frac{42x + 3x^{2} + 144}{43x + 5x^{2} + 24}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{3\cdot (x + 6)\cdot (x + 8)}{(5x + 3)\cdot (x + 8)}\\&\text{Which reduces to:}\\&\frac{(3x + 18)}{(5x + 3)}\\&\text{At the value of }x=-8\text{ the function is}\\&\frac{6}{37}\end{align*}$

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Example Question #141 : Functions, Graphs, And Limits

$\begin{align*}&\text{Find the limit of }f(x)=\frac{x^{2} - 32x + 252}{x^{2} - 20x + 84}\\&\text{At the location }x=14\end{align*}$

Possible Answers:

$-\frac{1}{2}$

$\frac{1}{2}$

Correct answer:

$-\frac{1}{2}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=14\text{ into the function}\\&\frac{x^{2} - 32x + 252}{x^{2} - 20x + 84}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{(x - 14)\cdot (x - 18)}{(x - 6)\cdot (x - 14)}\\&\text{Which reduces to:}\\&\frac{(x - 18)}{(x - 6)}\\&\text{At the value of }x=14\text{ the function is}\\&-\frac{1}{2}\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&ax^2+bx+c\\&x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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Example Question #371 : Ap Calculus Ab

$\begin{align*}&\text{Determine the limit if it exists of }\frac{9x + 2x^{2} - 35}{59x + 5x^{2} + 168}\\&\text{At }x=-7\end{align*}$

Possible Answers:

$\frac{19}{11}$

$-\frac{11}{19}$

$\frac{11}{19}$

$-\frac{19}{11}$

Correct answer:

$\frac{19}{11}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=-7\text{ into the function}\\&\frac{9x + 2x^{2} - 35}{59x + 5x^{2} + 168}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{(2x - 5)\cdot (x + 7)}{(5x + 24)\cdot (x + 7)}\\&\text{Which reduces to:}\\&\frac{(2x - 5)}{(5x + 24)}\\&\text{At the value of }x=-7\text{ the function is}\\&\frac{19}{11}\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&\text{For equations of the form: }ax^2+bx+c\\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #4 : Limits Of Functions (Including One Sided Limits)

Example Question #1 : Calculating Limits Using Algebra

Example Question #2 : Calculating Limits Using Algebra

Example Question #3 : Calculating Limits Using Algebra

Example Question #1 : Limits Of Functions (Including One Sided Limits)

Example Question #1 : Calculating Limits Using Algebra

Example Question #371 : Ap Calculus Ab

Example Question #11 : Calculating Limits Using Algebra

Example Question #141 : Functions, Graphs, And Limits

Example Question #371 : Ap Calculus Ab

All AP Calculus AB Resources

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