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AP Calculus AB : Limits of Functions (including one-sided limits)

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Example Questions

Example Question #371 : Ap Calculus Ab

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

Possible Answers:

$-\frac{8}{9}$ $\displaystyle -\frac{8}{9}$

$-\frac{9}{8}$ $\displaystyle -\frac{9}{8}$

$\frac{8}{9}$ $\displaystyle \frac{8}{9}$

$\frac{9}{8}$ $\displaystyle \frac{9}{8}$

Correct answer:

$\frac{8}{9}$ $\displaystyle \frac{8}{9}$

Explanation:

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

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Example Question #11 : Calculating Limits Using Algebra

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

Possible Answers:

$-\frac{6}{37}$ $\displaystyle -\frac{6}{37}$

$-\frac{37}{6}$ $\displaystyle -\frac{37}{6}$

$\frac{37}{6}$ $\displaystyle \frac{37}{6}$

$\frac{6}{37}$ $\displaystyle \frac{6}{37}$

Correct answer:

$\frac{6}{37}$ $\displaystyle \frac{6}{37}$

Explanation:

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

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Example Question #11 : Calculating Limits Using Algebra

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

Possible Answers:

$-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\displaystyle 2$

$\displaystyle -2$

Correct answer:

$-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$

Explanation:

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

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Example Question #371 : Ap Calculus Ab

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

Possible Answers:

$\frac{19}{11}$ $\displaystyle \frac{19}{11}$

$-\frac{11}{19}$ $\displaystyle -\frac{11}{19}$

$\frac{11}{19}$ $\displaystyle \frac{11}{19}$

$-\frac{19}{11}$ $\displaystyle -\frac{19}{11}$

Correct answer:

$\frac{19}{11}$ $\displaystyle \frac{19}{11}$

Explanation:

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

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Example Question #381 : Ap Calculus Ab

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

Possible Answers:

$-\frac{1}{118}$ $\displaystyle -\frac{1}{118}$

$\displaystyle -118$

$\frac{1}{118}$ $\displaystyle \frac{1}{118}$

118 $\displaystyle 118$

Correct answer:

$-\frac{1}{118}$ $\displaystyle -\frac{1}{118}$

Explanation:

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

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Example Question #141 : Functions, Graphs, And Limits

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

Possible Answers:

$\frac{3}{49}$ $\displaystyle \frac{3}{49}$

$-\frac{49}{3}$ $\displaystyle -\frac{49}{3}$

$\frac{49}{3}$ $\displaystyle \frac{49}{3}$

$-\frac{3}{49}$ $\displaystyle -\frac{3}{49}$

Correct answer:

$-\frac{49}{3}$ $\displaystyle -\frac{49}{3}$

Explanation:

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

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Example Question #381 : Ap Calculus Ab

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

Possible Answers:

$\frac{92}{35}$ $\displaystyle \frac{92}{35}$

$\frac{35}{92}$ $\displaystyle \frac{35}{92}$

$-\frac{92}{35}$ $\displaystyle -\frac{92}{35}$

$-\frac{35}{92}$ $\displaystyle -\frac{35}{92}$

Correct answer:

$\frac{35}{92}$ $\displaystyle \frac{35}{92}$

Explanation:

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

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Example Question #382 : Ap Calculus Ab

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

Possible Answers:

$-\frac{1}{12}$ $\displaystyle -\frac{1}{12}$

$\displaystyle 12$

$\displaystyle -12$

$\frac{1}{12}$ $\displaystyle \frac{1}{12}$

Correct answer:

$\displaystyle 12$

Explanation:

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

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Example Question #383 : Ap Calculus Ab

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

Possible Answers:

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$

$\displaystyle -2$

$\displaystyle 2$

Correct answer:

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Explanation:

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

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Example Question #382 : Ap Calculus Ab

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

Possible Answers:

$\frac{1}{6}$ $\displaystyle \frac{1}{6}$

$\displaystyle 6$

$-\frac{1}{6}$ $\displaystyle -\frac{1}{6}$

$\displaystyle -6$

Correct answer:

$\displaystyle 6$

Explanation:

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

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AP Calculus AB : Limits of Functions (including one-sided limits)

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