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AP Calculus AB : Functions, Graphs, and Limits

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Example Questions

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 #22 : Limits Of Functions (Including One Sided Limits)

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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Example Question #23 : Limits Of Functions (Including One Sided Limits)

Find the following limit as x approaches infinity.

$\ln (x)$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

$\infty$

Explanation:

As x becomes infinitely large, $\ln (x)$ approaches infinity.

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Example Question #392 : Ap Calculus Ab

Find the limit:

$\lim_{x\rightarrow \propto }\frac{4x^3+3x^2+10}{10x^3+x^2+8x}$

Possible Answers:

$\propto$

$\frac{4}{10}$

DNE (Does not exist)

Correct answer:

$\frac{4}{10}$

Explanation:

So for limits involving infinity, there is one important concept regarding fractions that is important to understand.

So if you have a fraction with a number on the bottom that is getting larger and larger, the whole fraction becomes smaller.

For example $\frac{1}{1000}>\frac{1}{10000}$

So the way to solve for limits involving infinity is to divide each term on the top, and each term on the bottom by the largest power of the variable. In the case of this problem, that is x^3 .

So if you do this you get:

$\lim_{x\rightarrow \propto }\frac{\frac{4x^3}{x^3}+\frac{3x^2}{x^3}+\frac{10}{x^3}}{\frac{10x^3}{x^3}+\frac{x^2}{x^3}+\frac{8x}{x^3}}$

So anything divided by itself is 1, which means that the first two terms cancel to one. Then the rest of the terms with a higher power of x on the bottom than on the top will have some power of x left on the bottom. This will look like:

$\lim_{x\rightarrow \propto }\frac{4+\frac{3}{x}+\frac{10}{x^3}}{10+\frac{1}{x}+\frac{8}{x^2}}$

Then, if you let the limit go to infinity, the terms with an x left on the bottom will go to zero.

This leaves you with the answer of $\frac{4}{10}$

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AP Calculus AB : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

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 #22 : Limits Of Functions (Including One Sided Limits)

Example Question #22 : Calculating Limits Using Algebra

Example Question #151 : Functions, Graphs, And Limits

Example Question #23 : Limits Of Functions (Including One Sided Limits)

Example Question #392 : Ap Calculus Ab

All AP Calculus AB Resources

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