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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1051 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow2+}-\frac{(42x^{2} - 4018x + 154x^{3} - 14x^{4} + 6860)}{(504x - 168x^{2} + 14x^{3} - 448)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow2+}-\frac{(42x^{2} - 4018x + 154x^{3} - 14x^{4} + 6860)}{(504x - 168x^{2} + 14x^{3} - 448)}\end{align*}$

Possible Answers:

$-\frac{1}{90}$ $\displaystyle -\frac{1}{90}$

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

$\frac{1}{90}$ $\displaystyle \frac{1}{90}$

$\displaystyle 0$

Correct answer:

$-\infty$ $\displaystyle -\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(42x^{2} - 4018x + 154x^{3} - 14x^{4} + 6860)}{(504x - 168x^{2} + 14x^{3} - 448)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(14\cdot (x - 2)\cdot (x + 5)\cdot (x - 7)^{2})}{(14\cdot (x - 2)^{2}\cdot (x - 8))}\\&\frac{((x + 5)\cdot (x - 7)^{2})}{((x - 2)\cdot (x - 8))}\\&lim_{x\rightarrow2+}-\frac{(42x^{2} - 4018x + 154x^{3} - 14x^{4} + 6860)}{(504x - 168x^{2} + 14x^{3} - 448)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow2+}\frac{(56x^{3} - 462x^{2} - 84x + 4018)}{(42x^{2} - 336x + 504)}=-\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(42x^{2} - 4018x + 154x^{3} - 14x^{4} + 6860)}{(504x - 168x^{2} + 14x^{3} - 448)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(14\cdot (x - 2)\cdot (x + 5)\cdot (x - 7)^{2})}{(14\cdot (x - 2)^{2}\cdot (x - 8))}\\&\frac{((x + 5)\cdot (x - 7)^{2})}{((x - 2)\cdot (x - 8))}\\&lim_{x\rightarrow2+}-\frac{(42x^{2} - 4018x + 154x^{3} - 14x^{4} + 6860)}{(504x - 168x^{2} + 14x^{3} - 448)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow2+}\frac{(56x^{3} - 462x^{2} - 84x + 4018)}{(42x^{2} - 336x + 504)}=-\infty\end{align*}$

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Example Question #1061 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7-}-\frac{(3824x + 48x^{2} - 60x^{3} - 4x^{4} + 13440)}{(64x^{3} - 240x^{2} - 1792x + 8x^{4} + 1960)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7-}-\frac{(3824x + 48x^{2} - 60x^{3} - 4x^{4} + 13440)}{(64x^{3} - 240x^{2} - 1792x + 8x^{4} + 1960)}\end{align*}$

Possible Answers:

$-\frac{ 16}{7}$ $\displaystyle -\frac{ 16}{7}$

$-\infty$ $\displaystyle -\infty$

$\displaystyle 0$

$\infty$ $\displaystyle \infty$

$\frac{ 16}{7}$ $\displaystyle \frac{ 16}{7}$

Correct answer:

$-\infty$ $\displaystyle -\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(3824x + 48x^{2} - 60x^{3} - 4x^{4} + 13440)}{(64x^{3} - 240x^{2} - 1792x + 8x^{4} + 1960)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(4\cdot (x + 7)\cdot (x + 6)\cdot (x - 8)\cdot (x + 10))}{(8\cdot (x + 7)^{2}\cdot (x - 1)\cdot (x - 5))}\\&\frac{((x + 6)\cdot (x - 8)\cdot (x + 10))}{(2\cdot (x - 1)\cdot (x - 5)\cdot (x + 7))}\\&lim_{x\rightarrow-7-}-\frac{(3824x + 48x^{2} - 60x^{3} - 4x^{4} + 13440)}{(64x^{3} - 240x^{2} - 1792x + 8x^{4} + 1960)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7-}\frac{(180x^{2} - 96x + 16x^{3} - 3824)}{(192x^{2} - 480x + 32x^{3} - 1792)}=-\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(3824x + 48x^{2} - 60x^{3} - 4x^{4} + 13440)}{(64x^{3} - 240x^{2} - 1792x + 8x^{4} + 1960)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(4\cdot (x + 7)\cdot (x + 6)\cdot (x - 8)\cdot (x + 10))}{(8\cdot (x + 7)^{2}\cdot (x - 1)\cdot (x - 5))}\\&\frac{((x + 6)\cdot (x - 8)\cdot (x + 10))}{(2\cdot (x - 1)\cdot (x - 5)\cdot (x + 7))}\\&lim_{x\rightarrow-7-}-\frac{(3824x + 48x^{2} - 60x^{3} - 4x^{4} + 13440)}{(64x^{3} - 240x^{2} - 1792x + 8x^{4} + 1960)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7-}\frac{(180x^{2} - 96x + 16x^{3} - 3824)}{(192x^{2} - 480x + 32x^{3} - 1792)}=-\infty\end{align*}$

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Example Question #1062 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7+}-\frac{(4x - 28)}{(81x + 7x^{2} - x^{3} - 567)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7+}-\frac{(4x - 28)}{(81x + 7x^{2} - x^{3} - 567)}\end{align*}$

Possible Answers:

$-\frac{1}{8}$ $\displaystyle -\frac{1}{8}$

$\frac{1}{8}$ $\displaystyle \frac{1}{8}$

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

$\displaystyle 0$

Correct answer:

$-\frac{1}{8}$ $\displaystyle -\frac{1}{8}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(4x - 28)}{(81x + 7x^{2} - x^{3} - 567)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(4\cdot (x - 7))}{((x - 7)\cdot (x - 9)\cdot (x + 9))}\\&\frac{4}{((x - 9)\cdot (x + 9))}\\&lim_{x\rightarrow7+}-\frac{(4x - 28)}{(81x + 7x^{2} - x^{3} - 567)}=-\frac{1}{8}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow7+}\frac{(4)}{(3x^{2} - 14x - 81)}=-\frac{1}{8}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(4x - 28)}{(81x + 7x^{2} - x^{3} - 567)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(4\cdot (x - 7))}{((x - 7)\cdot (x - 9)\cdot (x + 9))}\\&\frac{4}{((x - 9)\cdot (x + 9))}\\&lim_{x\rightarrow7+}-\frac{(4x - 28)}{(81x + 7x^{2} - x^{3} - 567)}=-\frac{1}{8}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow7+}\frac{(4)}{(3x^{2} - 14x - 81)}=-\frac{1}{8}\end{align*}$

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Example Question #1063 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3-}-\frac{(2x^{2} - 24x + 54)}{(72x + 8x^{2} - 8x^{3} - 72)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3-}-\frac{(2x^{2} - 24x + 54)}{(72x + 8x^{2} - 8x^{3} - 72)}\end{align*}$

Possible Answers:

$\displaystyle 0$

$\infty$ $\displaystyle \infty$

$-\frac{1}{8}$ $\displaystyle -\frac{1}{8}$

$\frac{1}{8}$ $\displaystyle \frac{1}{8}$

$-\infty$ $\displaystyle -\infty$

Correct answer:

$-\frac{1}{8}$ $\displaystyle -\frac{1}{8}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=3\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(2x^{2} - 24x + 54)}{(72x + 8x^{2} - 8x^{3} - 72)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(2\cdot (x - 3)\cdot (x - 9))}{(8\cdot (x - 3)\cdot (x - 1)\cdot (x + 3))}\\&\frac{(x - 9)}{(4\cdot (x - 1)\cdot (x + 3))}\\&lim_{x\rightarrow3-}-\frac{(2x^{2} - 24x + 54)}{(72x + 8x^{2} - 8x^{3} - 72)}=-\frac{1}{8}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow3-}\frac{(4x - 24)}{(24x^{2} - 16x - 72)}=-\frac{1}{8}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=3\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(2x^{2} - 24x + 54)}{(72x + 8x^{2} - 8x^{3} - 72)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(2\cdot (x - 3)\cdot (x - 9))}{(8\cdot (x - 3)\cdot (x - 1)\cdot (x + 3))}\\&\frac{(x - 9)}{(4\cdot (x - 1)\cdot (x + 3))}\\&lim_{x\rightarrow3-}-\frac{(2x^{2} - 24x + 54)}{(72x + 8x^{2} - 8x^{3} - 72)}=-\frac{1}{8}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow3-}\frac{(4x - 24)}{(24x^{2} - 16x - 72)}=-\frac{1}{8}\end{align*}$

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Example Question #1064 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow9-}-\frac{(x - 9)}{(900x + 15x^{2} - 15x^{3} + 1620)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow9-}-\frac{(x - 9)}{(900x + 15x^{2} - 15x^{3} + 1620)}\end{align*}$

Possible Answers:

$\frac{1}{2475}$ $\displaystyle \frac{1}{2475}$

$\infty$ $\displaystyle \infty$

$-\frac{1}{2475}$ $\displaystyle -\frac{1}{2475}$

$-\infty$ $\displaystyle -\infty$

$\displaystyle 0$

Correct answer:

$\frac{1}{2475}$ $\displaystyle \frac{1}{2475}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=9\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(x - 9)}{(900x + 15x^{2} - 15x^{3} + 1620)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(x - 9)}{(15\cdot (x - 9)\cdot (x + 2)\cdot (x + 6))}\\&\frac{1}{(15\cdot (x + 2)\cdot (x + 6))}\\&lim_{x\rightarrow9-}-\frac{(x - 9)}{(900x + 15x^{2} - 15x^{3} + 1620)}=\frac{1}{2475}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow9-}\frac{(1)}{(45x^{2} - 30x - 900)}=\frac{1}{2475}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=9\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(x - 9)}{(900x + 15x^{2} - 15x^{3} + 1620)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(x - 9)}{(15\cdot (x - 9)\cdot (x + 2)\cdot (x + 6))}\\&\frac{1}{(15\cdot (x + 2)\cdot (x + 6))}\\&lim_{x\rightarrow9-}-\frac{(x - 9)}{(900x + 15x^{2} - 15x^{3} + 1620)}=\frac{1}{2475}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow9-}\frac{(1)}{(45x^{2} - 30x - 900)}=\frac{1}{2475}\end{align*}$

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Example Question #1065 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}-\frac{(225x + 30x^{2} - 5x^{3} - 250)}{(7x^{2} - 14x + 7)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}-\frac{(225x + 30x^{2} - 5x^{3} - 250)}{(7x^{2} - 14x + 7)}\end{align*}$

Possible Answers:

$-\frac{ 11}{26}$ $\displaystyle -\frac{ 11}{26}$

$\displaystyle 0$

$\frac{ 11}{26}$ $\displaystyle \frac{ 11}{26}$

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

Correct answer:

$-\infty$ $\displaystyle -\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=1\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(225x + 30x^{2} - 5x^{3} - 250)}{(7x^{2} - 14x + 7)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(5\cdot (x - 1)\cdot (x + 5)\cdot (x - 10))}{(7\cdot (x - 1)^{2})}\\&\frac{(5\cdot (x + 5)\cdot (x - 10))}{(7\cdot (x - 1))}\\&lim_{x\rightarrow1+}-\frac{(225x + 30x^{2} - 5x^{3} - 250)}{(7x^{2} - 14x + 7)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow1+}\frac{(15x^{2} - 60x - 225)}{(14x - 14)}=-\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=1\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(225x + 30x^{2} - 5x^{3} - 250)}{(7x^{2} - 14x + 7)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(5\cdot (x - 1)\cdot (x + 5)\cdot (x - 10))}{(7\cdot (x - 1)^{2})}\\&\frac{(5\cdot (x + 5)\cdot (x - 10))}{(7\cdot (x - 1))}\\&lim_{x\rightarrow1+}-\frac{(225x + 30x^{2} - 5x^{3} - 250)}{(7x^{2} - 14x + 7)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow1+}\frac{(15x^{2} - 60x - 225)}{(14x - 14)}=-\infty\end{align*}$

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Example Question #1066 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3+}-\frac{(12x^{2} - 72x + 108)}{(306x + 153x^{2} - 34x^{3} - 17x^{4})}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3+}-\frac{(12x^{2} - 72x + 108)}{(306x + 153x^{2} - 34x^{3} - 17x^{4})}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

$\frac{ 3}{35}$ $\displaystyle \frac{ 3}{35}$

$\displaystyle 0$

$-\frac{ 3}{35}$ $\displaystyle -\frac{ 3}{35}$

Correct answer:

$\displaystyle 0$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=3\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(12x^{2} - 72x + 108)}{(306x + 153x^{2} - 34x^{3} - 17x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(12\cdot (x - 3)^{2})}{(17x\cdot (x - 3)\cdot (x + 2)\cdot (x + 3))}\\&\frac{(12\cdot (x - 3))}{(17x\cdot (x + 2)\cdot (x + 3))}\\&lim_{x\rightarrow3+}-\frac{(12x^{2} - 72x + 108)}{(306x + 153x^{2} - 34x^{3} - 17x^{4})}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow3+}\frac{(24x - 72)}{(102x^{2} - 306x + 68x^{3} - 306)}=0\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=3\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(12x^{2} - 72x + 108)}{(306x + 153x^{2} - 34x^{3} - 17x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(12\cdot (x - 3)^{2})}{(17x\cdot (x - 3)\cdot (x + 2)\cdot (x + 3))}\\&\frac{(12\cdot (x - 3))}{(17x\cdot (x + 2)\cdot (x + 3))}\\&lim_{x\rightarrow3+}-\frac{(12x^{2} - 72x + 108)}{(306x + 153x^{2} - 34x^{3} - 17x^{4})}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow3+}\frac{(24x - 72)}{(102x^{2} - 306x + 68x^{3} - 306)}=0\end{align*}$

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Example Question #1067 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow4+}\frac{(19x - 76)}{(2x - 8)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow4+}\frac{(19x - 76)}{(2x - 8)}\end{align*}$

Possible Answers:

$-\infty$ $\displaystyle -\infty$

$-\frac{19}{2}$ $\displaystyle -\frac{19}{2}$

$\displaystyle 0$

$\infty$ $\displaystyle \infty$

$\frac{19}{2}$ $\displaystyle \frac{19}{2}$

Correct answer:

$\frac{19}{2}$ $\displaystyle \frac{19}{2}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=4\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(19x - 76)}{(2x - 8)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(19\cdot (x - 4))}{(2\cdot (x - 4))}\\&\frac{19}{2}\\&lim_{x\rightarrow4+}\frac{(19x - 76)}{(2x - 8)}=\frac{19}{2}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow4+}\frac{(19)}{(2)}=\frac{19}{2}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=4\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(19x - 76)}{(2x - 8)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(19\cdot (x - 4))}{(2\cdot (x - 4))}\\&\frac{19}{2}\\&lim_{x\rightarrow4+}\frac{(19x - 76)}{(2x - 8)}=\frac{19}{2}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow4+}\frac{(19)}{(2)}=\frac{19}{2}\end{align*}$

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Example Question #1068 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}-\frac{(3x^{3} - x^{4})}{(560x + 752x^{2} + 208x^{3} + 16x^{4})}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0-}-\frac{(3x^{3} - x^{4})}{(560x + 752x^{2} + 208x^{3} + 16x^{4})}\end{align*}$

Possible Answers:

$-\frac{ 64}{75}$ $\displaystyle -\frac{ 64}{75}$

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

$\frac{ 64}{75}$ $\displaystyle \frac{ 64}{75}$

$\displaystyle 0$

Correct answer:

$\displaystyle 0$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(3x^{3} - x^{4})}{(560x + 752x^{2} + 208x^{3} + 16x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(x^{3}\cdot (x - 3))}{(16x\cdot (x + 1)\cdot (x + 5)\cdot (x + 7))}\\&\frac{(x^{2}\cdot (x - 3))}{(16\cdot (x + 1)\cdot (x + 5)\cdot (x + 7))}\\&lim_{x\rightarrow0-}-\frac{(3x^{3} - x^{4})}{(560x + 752x^{2} + 208x^{3} + 16x^{4})}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0-}\frac{(4x^{3} - 9x^{2})}{(1504x + 624x^{2} + 64x^{3} + 560)}=0\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(3x^{3} - x^{4})}{(560x + 752x^{2} + 208x^{3} + 16x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(x^{3}\cdot (x - 3))}{(16x\cdot (x + 1)\cdot (x + 5)\cdot (x + 7))}\\&\frac{(x^{2}\cdot (x - 3))}{(16\cdot (x + 1)\cdot (x + 5)\cdot (x + 7))}\\&lim_{x\rightarrow0-}-\frac{(3x^{3} - x^{4})}{(560x + 752x^{2} + 208x^{3} + 16x^{4})}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0-}\frac{(4x^{3} - 9x^{2})}{(1504x + 624x^{2} + 64x^{3} + 560)}=0\end{align*}$

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Example Question #1069 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow10+}\frac{(13x^{2} - 260x + 1300)}{(9x^{2} - 108x + 180)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow10+}\frac{(13x^{2} - 260x + 1300)}{(9x^{2} - 108x + 180)}\end{align*}$

Possible Answers:

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

$\frac{ 7}{18}$ $\displaystyle \frac{ 7}{18}$

$-\frac{ 7}{18}$ $\displaystyle -\frac{ 7}{18}$

$\infty$ $\displaystyle \infty$

Correct answer:

$\displaystyle 0$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=10\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(13x^{2} - 260x + 1300)}{(9x^{2} - 108x + 180)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(13\cdot (x - 10)^{2})}{(9\cdot (x - 10)\cdot (x - 2))}\\&\frac{(13\cdot (x - 10))}{(9\cdot (x - 2))}\\&lim_{x\rightarrow10+}\frac{(13x^{2} - 260x + 1300)}{(9x^{2} - 108x + 180)}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow10+}\frac{(26x - 260)}{(18x - 108)}=0\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=10\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(13x^{2} - 260x + 1300)}{(9x^{2} - 108x + 180)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(13\cdot (x - 10)^{2})}{(9\cdot (x - 10)\cdot (x - 2))}\\&\frac{(13\cdot (x - 10))}{(9\cdot (x - 2))}\\&lim_{x\rightarrow10+}\frac{(13x^{2} - 260x + 1300)}{(9x^{2} - 108x + 180)}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow10+}\frac{(26x - 260)}{(18x - 108)}=0\end{align*}$

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