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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #381 : Ap Calculus Ab

$\begin{align*}&\text{Find the limit of }f(x)=\frac{x + 23}{112x + 5x^{2} - 69}\\&\text{At the location }x=-23\end{align*}$

Possible Answers:

$-\frac{1}{118}$

$\frac{1}{118}$

118

Correct answer:

$-\frac{1}{118}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=-23\text{ into the function}\\&\frac{x + 23}{112x + 5x^{2} - 69}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{x + 23}{(5x - 3)\cdot (x + 23)}\\&\text{Which reduces to:}\\&\frac{1}{(5x - 3)}\\&\text{At the value of }x=-23\text{ the function is}\\&-\frac{1}{118}\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&\text{For equations of the form: }ax^2+bx+c\\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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Example Question #141 : Functions, Graphs, And Limits

$\begin{align*}&\text{Find the limit of }f(x)=\frac{5x^{2} - 41x - 36}{x^{2} - 21x + 108}\\&\text{At the location }x=9\end{align*}$

Possible Answers:

$\frac{3}{49}$

$-\frac{49}{3}$

$\frac{49}{3}$

$-\frac{3}{49}$

Correct answer:

$-\frac{49}{3}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=9\text{ into the function}\\&\frac{5x^{2} - 41x - 36}{x^{2} - 21x + 108}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{(5x + 4)\cdot (x - 9)}{(x - 9)\cdot (x - 12)}\\&\text{Which reduces to:}\\&\frac{(5x + 4)}{(x - 12)}\\&\text{At the value of }x=9\text{ the function is}\\&-\frac{49}{3}\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&\text{For equations of the form: }ax^2+bx+c\\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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Example Question #381 : Ap Calculus Ab

$\begin{align*}&\text{Determine the limit if it exists of }\frac{73x + 3x^{2} + 342}{52x + 4x^{2} - 360}\\&\text{At }x=-18\end{align*}$

Possible Answers:

$\frac{92}{35}$

$\frac{35}{92}$

$-\frac{92}{35}$

$-\frac{35}{92}$

Correct answer:

$\frac{35}{92}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=-18\text{ into the function}\\&\frac{73x + 3x^{2} + 342}{52x + 4x^{2} - 360}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{(3x + 19)\cdot (x + 18)}{4\cdot (x - 5)\cdot (x + 18)}\\&\text{Which reduces to:}\\&\frac{(3x + 19)}{(4x - 20)}\\&\text{At the value of }x=-18\text{ the function is}\\&\frac{35}{92}\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&\text{For equations of the form: }ax^2+bx+c\\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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Example Question #382 : Ap Calculus Ab

$\begin{align*}&\text{Evaluate the limit of }\frac{3x^{2} - 24x + 36}{x - 6}\\&\text{At the point }x=6\end{align*}$

Possible Answers:

$-\frac{1}{12}$

$\frac{1}{12}$

Correct answer:

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=6\text{ into the function}\\&\frac{3x^{2} - 24x + 36}{x - 6}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{3\cdot (x - 2)\cdot (x - 6)}{x - 6}\\&\text{Which reduces to:}\\&3x - 6\\&\text{At the value of }x=6\text{ the function is}\\&12\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&\text{For equations of the form: }ax^2+bx+c\\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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Example Question #383 : Ap Calculus Ab

$\begin{align*}&\text{Evaluate the limit of }\frac{2x^{2} - 38x + 68}{5x^{2} - 110x + 425}\\&\text{At the point }x=17\end{align*}$

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=17\text{ into the function}\\&\frac{2x^{2} - 38x + 68}{5x^{2} - 110x + 425}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{2\cdot (x - 2)\cdot (x - 17)}{5\cdot (x - 5)\cdot (x - 17)}\\&\text{Which reduces to:}\\&\frac{(2x - 4)}{(5x - 25)}\\&\text{At the value of }x=17\text{ the function is}\\&\frac{1}{2}\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&\text{For equations of the form: }ax^2+bx+c\\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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Example Question #382 : Ap Calculus Ab

$\begin{align*}&\text{Find the limit of }f(x)=\frac{4x^{2} - 68x - 152}{2x^{2} - 62x + 456}\\&\text{At the location }x=19\end{align*}$

Possible Answers:

$\frac{1}{6}$

$-\frac{1}{6}$

Correct answer:

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=19\text{ into the function}\\&\frac{4x^{2} - 68x - 152}{2x^{2} - 62x + 456}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{4\cdot (x + 2)\cdot (x - 19)}{2\cdot (x - 12)\cdot (x - 19)}\\&\text{Which reduces to:}\\&\frac{(4x + 8)}{(2x - 24)}\\&\text{At the value of }x=19\text{ the function is}\\&6\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&\text{For equations of the form: }ax^2+bx+c\\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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Example Question #21 : Limits Of Functions (Including One Sided Limits)

$\begin{align*}&\text{Evaluate the limit of }\frac{85x + 5x^{2} - 90}{76x + 5x^{2} - 252}\\&\text{At the point }x=-18\end{align*}$

Possible Answers:

$-\frac{104}{95}$

$\frac{104}{95}$

$\frac{95}{104}$

$-\frac{95}{104}$

Correct answer:

$\frac{95}{104}$

Explanation:

$\begin{align*}&\text{Simply plugging in the value of }x=-18\text{ into the function}\\&\frac{85x + 5x^{2} - 90}{76x + 5x^{2} - 252}\\&\text{will not tell us much, because it simply gives us a zero in}\\&\text{the numerator and denominator: }\frac{0}{0}\\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\\&\text{We can do this because we're not actually evaluating the function at these zero points,}\\&\text{just close to them. Factoring our function we find:}\\&\frac{5\cdot (x - 1)\cdot (x + 18)}{(5x - 14)\cdot (x + 18)}\\&\text{Which reduces to:}\\&\frac{(5x - 5)}{(5x - 14)}\\&\text{At the value of }x=-18\text{ the function is}\\&\frac{95}{104}\\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\\&\text{For equations of the form: }ax^2+bx+c\\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

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Example Question #381 : Ap Calculus Ab

Find the limit of the following function as x approaches infinity.

$\frac{1}{x}$

Possible Answers:

Infinity

$\frac{1}{4}$

Correct answer:

Explanation:

As x becomes infinitely large, $\frac{1}{x}$ approaches .

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Example Question #22 : Calculating Limits Using Algebra

$f(x) = \left\{\begin{matrix} \frac{2}{x+2}&x<0 \\ 0 &x= 0 \\ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

Possible Answers:

$\lim_{x\rightarrow 0 }f(x)$ does not exist, because either or both of $\lim_{x\rightarrow 0 - }f(x)$ and $\lim_{x\rightarrow 0 + }f(x)$ is unequal to f(0) .

$\lim_{x\rightarrow 0 }f(x)$ does not exist, because $\lim_{x\rightarrow 0 - }f(x) \ne \lim_{x\rightarrow 0 + }f(x)$ .

$\frac{1}{2}$

Correct answer:

Explanation:

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

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Example Question #151 : Functions, Graphs, And Limits

Define $f (x) = \begin{Bmatrix} x^{2}+2x+ 3& x< 0 \\ a \cos\theta x & x \ge 0 \end{matrix}$ for some real $a,\theta$ .

Evaluate and $\theta$ so that is both continuous and differentiable at x = 0 .

Possible Answers:

No such values exist.

$a = 3, \theta = 2$

$a = 3, \theta = 1$

$a = 2, \theta = 3$

$a = 1, \theta = 3$

Correct answer:

No such values exist.

Explanation:

For to be continuous, it must hold that

$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x)$ .

To find $\lim_{x \rightarrow 0^{-}} f(x)$ , we can use the definition of f(x) for all negative values of :

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

It must hold that $\lim_{x \rightarrow 0^{+}} f(x) =3$ as well; using the definition of f(x) for all positive values of :

$\lim_{x \rightarrow 0^{+}} f(x) =\lim_{x \rightarrow 0^{+}} a \cos \theta x = a \cos 0 = a \cdot 1 = a$

$\lim_{x \rightarrow 0^{+}} f(x) =3$ , so a = 3 .

Now examine f'(x) . For to be differentiable, it must hold that

$\lim_{x \rightarrow 0^{-}} f'(x) = \lim_{x \rightarrow 0^{+}} f'(x)$ .

To find $\lim_{x \rightarrow 0^{-}} f'(x)$ , we can differentiate the expression for f(x) for all negative values of :

$f(x)= x^{2}+ 2x + 3$

f'(x) = 2x + 2

$\lim_{x \rightarrow 0^{-}} f'(x) = \lim_{x \rightarrow 0^{-}} (2x +2)= 2(0)+2 = 2$

To find $\lim_{x \rightarrow 0^{+}} f'(x)$ , we can differentiate the expression for f(x) for all positive values of :

$f(x) = a \cos \theta x$

$f'(x) = - a \theta \sin \theta x$

$\lim_{x \rightarrow 0^{+}} f'(x) =\lim_{x \rightarrow 0^{+}} ( - a \theta \sin \theta x)$

We know that a = 3 , so

$\lim_{x \rightarrow 0^{+}} f'(x) =\lim_{x \rightarrow 0^{+}} ( - 3 \theta \sin \theta x) = -3 \theta \sin 0 = -3 \theta \cdot 0 = 0$

Since $\lim_{x \rightarrow 0^{-}} f'(x) = 2$ and $\lim_{x \rightarrow 0^{+}} f'(x)=0$ ,

$\lim_{x \rightarrow 0 } f'(x)$ cannot exist regardless of the values of and $\theta$ .

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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 #381 : Ap Calculus Ab

Example Question #141 : Functions, Graphs, And Limits

Example Question #381 : Ap Calculus Ab

Example Question #382 : Ap Calculus Ab

Example Question #383 : Ap Calculus Ab

Example Question #382 : Ap Calculus Ab

Example Question #21 : Limits Of Functions (Including One Sided Limits)

Example Question #381 : Ap Calculus Ab

Example Question #22 : Calculating Limits Using Algebra

Example Question #151 : Functions, Graphs, And Limits

All AP Calculus AB Resources

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