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Calculus 3 : Calculus 3

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Example Questions

Example Question #1071 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3+}\frac{(84x - 28x^{2} - 21x^{3} + 7x^{4})}{(12x - 36)}\end{align*}$

Possible Answers:

$\infty$

$-\frac{35}{4}$

$\frac{35}{4}$

$-\infty$

Correct answer:

$\frac{35}{4}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=3\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(84x - 28x^{2} - 21x^{3} + 7x^{4})}{(12x - 36)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(7x\cdot (x - 3)\cdot (x - 2)\cdot (x + 2))}{(12\cdot (x - 3))}\\&\frac{(7x\cdot (x - 2)\cdot (x + 2))}{12}\\&lim_{x\rightarrow3+}\frac{(84x - 28x^{2} - 21x^{3} + 7x^{4})}{(12x - 36)}=\frac{35}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow3+}\frac{(28x^{3} - 63x^{2} - 56x + 84)}{(12)}=\frac{35}{4}\end{align*}$

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Example Question #1072 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-3+}\frac{(2x + 6)}{(21x + 3x^{2} + 36)}\end{align*}$

Possible Answers:

$-\frac{2}{3}$

$\frac{2}{3}$

$-\infty$

$\infty$

Correct answer:

$\frac{2}{3}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-3\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(2x + 6)}{(21x + 3x^{2} + 36)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(2\cdot (x + 3))}{(3\cdot (x + 3)\cdot (x + 4))}\\&\frac{2}{(3\cdot (x + 4))}\\&lim_{x\rightarrow-3+}\frac{(2x + 6)}{(21x + 3x^{2} + 36)}=\frac{2}{3}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-3+}\frac{(2)}{(6x + 21)}=\frac{2}{3}\end{align*}$

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Example Question #1073 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-1-}\frac{(720x + 390x^{2} + 80x^{3} + 5x^{4} + 405)}{(40x + 48x^{2} + 8x^{3})}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-1\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(720x + 390x^{2} + 80x^{3} + 5x^{4} + 405)}{(40x + 48x^{2} + 8x^{3})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(5\cdot (x + 1)\cdot (x + 3)^{2}\cdot (x + 9))}{(8x\cdot (x + 1)\cdot (x + 5))}\\&\frac{(5\cdot (x + 3)^{2}\cdot (x + 9))}{(8x\cdot (x + 5))}\\&lim_{x\rightarrow-1-}\frac{(720x + 390x^{2} + 80x^{3} + 5x^{4} + 405)}{(40x + 48x^{2} + 8x^{3})}=-5\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-1-}\frac{(780x + 240x^{2} + 20x^{3} + 720)}{(96x + 24x^{2} + 40)}=-5\end{align*}$

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Example Question #1074 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-5+}-\frac{(800x + 720x^{2} + 192x^{3} + 16x^{4})}{(268x - 24x^{2} - 4x^{3} + 1440)}\end{align*}$

Possible Answers:

$\frac{ 32}{31}$

$-\infty$

$\infty$

$-\frac{ 32}{31}$

Correct answer:

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-5\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(800x + 720x^{2} + 192x^{3} + 16x^{4})}{(268x - 24x^{2} - 4x^{3} + 1440)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(16x\cdot (x + 5)^{2}\cdot (x + 2))}{(4\cdot (x + 5)\cdot (x - 8)\cdot (x + 9))}\\&\frac{(4x\cdot (x + 2)\cdot (x + 5))}{((x - 8)\cdot (x + 9))}\\&lim_{x\rightarrow-5+}-\frac{(800x + 720x^{2} + 192x^{3} + 16x^{4})}{(268x - 24x^{2} - 4x^{3} + 1440)}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-5+}\frac{(1440x + 576x^{2} + 64x^{3} + 800)}{(48x + 12x^{2} - 268)}=0\end{align*}$

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Example Question #1075 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1-}-\frac{(13x - 13x^{3})}{(36x^{2} - 378x + 18x^{3} + 324)}\end{align*}$

Possible Answers:

$\frac{13}{126}$

$-\infty$ 0

$-\frac{13}{126}$

$\infty$

Correct answer:

$-\frac{13}{126}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=1\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(13x - 13x^{3})}{(36x^{2} - 378x + 18x^{3} + 324)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(13x\cdot (x - 1)\cdot (x + 1))}{(18\cdot (x - 1)\cdot (x - 3)\cdot (x + 6))}\\&\frac{(13x\cdot (x + 1))}{(18\cdot (x - 3)\cdot (x + 6))}\\&lim_{x\rightarrow1-}-\frac{(13x - 13x^{3})}{(36x^{2} - 378x + 18x^{3} + 324)}=-\frac{13}{126}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow1-}\frac{(39x^{2} - 13)}{(72x + 54x^{2} - 378)}=-\frac{13}{126}\end{align*}$

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Example Question #1076 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-8-}\frac{(180x + 12x^{2} + 672)}{(240x + 15x^{2} + 960)}\end{align*}$

Possible Answers:

$\frac{ 2}{13}$

$-\infty$

$\infty$

$-\frac{ 2}{13}$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-8\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(180x + 12x^{2} + 672)}{(240x + 15x^{2} + 960)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(12\cdot (x + 8)\cdot (x + 7))}{(15\cdot (x + 8)^{2})}\\&\frac{(4\cdot (x + 7))}{(5\cdot (x + 8))}\\&lim_{x\rightarrow-8-}\frac{(180x + 12x^{2} + 672)}{(240x + 15x^{2} + 960)}=\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-8-}\frac{(24x + 180)}{(30x + 240)}=\infty\end{align*}$

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Example Question #1077 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow4+}-\frac{(2x^{2} - 16x + 32)}{(4000x - 1800x^{2} + 240x^{3} - 10x^{4})}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{We can't just plug in }x=4\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(2x^{2} - 16x + 32)}{(4000x - 1800x^{2} + 240x^{3} - 10x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(2\cdot (x - 4)^{2})}{(10x\cdot (x - 4)\cdot (x - 10)^{2})}\\&\frac{(x - 4)}{(5x\cdot (x - 10)^{2})}\\&lim_{x\rightarrow4+}-\frac{(2x^{2} - 16x + 32)}{(4000x - 1800x^{2} + 240x^{3} - 10x^{4})}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow4+}\frac{(4x - 16)}{(3600x - 720x^{2} + 40x^{3} - 4000)}=0\end{align*}$

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Example Question #1078 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-10+}-\frac{(1360x^{2} - 12800x + 128x^{3} - 16x^{4} + 24000)}{(75x^{2} + 5x^{3} - 2500)}\end{align*}$

Possible Answers:

$-\frac{ 9}{23}$

$-\infty$

$\infty$

$\frac{ 9}{23}$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-10\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(1360x^{2} - 12800x + 128x^{3} - 16x^{4} + 24000)}{(75x^{2} + 5x^{3} - 2500)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(16\cdot (x + 10)\cdot (x - 3)\cdot (x - 5)\cdot (x - 10))}{(5\cdot (x + 10)^{2}\cdot (x - 5))}\\&\frac{(16\cdot (x - 3)\cdot (x - 10))}{(5\cdot (x + 10))}\\&lim_{x\rightarrow-10+}-\frac{(1360x^{2} - 12800x + 128x^{3} - 16x^{4} + 24000)}{(75x^{2} + 5x^{3} - 2500)}=\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-10+}\frac{(64x^{3} - 384x^{2} - 2720x + 12800)}{(150x + 15x^{2})}=\infty\end{align*}$

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Example Question #1079 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-10-}\frac{(28x + 2x^{2} + 80)}{(2660x + 418x^{2} + 19x^{3} + 3800)}\end{align*}$

Possible Answers:

$\frac{ 71}{50}$

$-\frac{ 71}{50}$

$\infty$

$-\infty$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-10\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(28x + 2x^{2} + 80)}{(2660x + 418x^{2} + 19x^{3} + 3800)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(2\cdot (x + 10)\cdot (x + 4))}{(19\cdot (x + 10)^{2}\cdot (x + 2))}\\&\frac{(2\cdot (x + 4))}{(19\cdot (x + 2)\cdot (x + 10))}\\&lim_{x\rightarrow-10-}\frac{(28x + 2x^{2} + 80)}{(2660x + 418x^{2} + 19x^{3} + 3800)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-10-}\frac{(4x + 28)}{(836x + 57x^{2} + 2660)}=-\infty\end{align*}$

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Example Question #121 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow6+}-\frac{(51x - 17x^{2} + 306)}{(x^{2} - 36)}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$\frac{51}{4}$

$-\frac{51}{4}$

Correct answer:

$\frac{51}{4}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=6\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(51x - 17x^{2} + 306)}{(x^{2} - 36)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(17\cdot (x - 6)\cdot (x + 3))}{((x - 6)\cdot (x + 6))}\\&\frac{(17\cdot (x + 3))}{(x + 6)}\\&lim_{x\rightarrow6+}-\frac{(51x - 17x^{2} + 306)}{(x^{2} - 36)}=\frac{51}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow6+}\frac{(34x - 51)}{(2x)}=\frac{51}{4}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1071 : Calculus 3

Example Question #1072 : Calculus 3

Example Question #1073 : Calculus 3

Example Question #1074 : Calculus 3

Example Question #1075 : Calculus 3

Example Question #1076 : Calculus 3

Example Question #1077 : Calculus 3

Example Question #1078 : Calculus 3

Example Question #1079 : Calculus 3

Example Question #121 : Limits

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