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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1041 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-6+}-\frac{(28x^{2} - 84x + 7x^{3})}{(720x - 30x^{2} - 10x^{3} + 3240)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-6+}-\frac{(28x^{2} - 84x + 7x^{3})}{(720x - 30x^{2} - 10x^{3} + 3240)}\end{align*}$

Possible Answers:

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

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$\frac{ 28}{59}$ $\displaystyle \frac{ 28}{59}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-6\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(28x^{2} - 84x + 7x^{3})}{(720x - 30x^{2} - 10x^{3} + 3240)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(7x\cdot (x + 6)\cdot (x - 2))}{(10\cdot (x + 6)^{2}\cdot (x - 9))}\\&\frac{(7x\cdot (x - 2))}{(10\cdot (x + 6)\cdot (x - 9))}\\&lim_{x\rightarrow-6+}-\frac{(28x^{2} - 84x + 7x^{3})}{(720x - 30x^{2} - 10x^{3} + 3240)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-6+}\frac{(56x + 21x^{2} - 84)}{(60x + 30x^{2} - 720)}=-\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-6\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(28x^{2} - 84x + 7x^{3})}{(720x - 30x^{2} - 10x^{3} + 3240)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(7x\cdot (x + 6)\cdot (x - 2))}{(10\cdot (x + 6)^{2}\cdot (x - 9))}\\&\frac{(7x\cdot (x - 2))}{(10\cdot (x + 6)\cdot (x - 9))}\\&lim_{x\rightarrow-6+}-\frac{(28x^{2} - 84x + 7x^{3})}{(720x - 30x^{2} - 10x^{3} + 3240)}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-6+}\frac{(56x + 21x^{2} - 84)}{(60x + 30x^{2} - 720)}=-\infty\end{align*}$

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Example Question #1051 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7-}\frac{(102x + 17x^{2} - 119)}{(4x + 28)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7-}\frac{(102x + 17x^{2} - 119)}{(4x + 28)}\end{align*}$

Possible Answers:

$\displaystyle 0$

$\infty$ $\displaystyle \infty$

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

$\displaystyle -34$

Correct answer:

$\displaystyle -34$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(102x + 17x^{2} - 119)}{(4x + 28)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(17\cdot (x + 7)\cdot (x - 1))}{(4\cdot (x + 7))}\\&\frac{(17\cdot (x - 1))}{4}\\&lim_{x\rightarrow-7-}\frac{(102x + 17x^{2} - 119)}{(4x + 28)}=-34\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7-}\frac{(34x + 102)}{(4)}=-34\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(102x + 17x^{2} - 119)}{(4x + 28)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(17\cdot (x + 7)\cdot (x - 1))}{(4\cdot (x + 7))}\\&\frac{(17\cdot (x - 1))}{4}\\&lim_{x\rightarrow-7-}\frac{(102x + 17x^{2} - 119)}{(4x + 28)}=-34\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7-}\frac{(34x + 102)}{(4)}=-34\end{align*}$

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Example Question #1052 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}-\frac{(816x + 34x^{2} - 17x^{3})}{(10x)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}-\frac{(816x + 34x^{2} - 17x^{3})}{(10x)}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$-\frac{408}{5}$ $\displaystyle -\frac{408}{5}$

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

Correct answer:

$-\frac{408}{5}$ $\displaystyle -\frac{408}{5}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(816x + 34x^{2} - 17x^{3})}{(10x)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(17x\cdot (x + 6)\cdot (x - 8))}{(10x)}\\&\frac{(17\cdot (x + 6)\cdot (x - 8))}{10}\\&lim_{x\rightarrow0+}-\frac{(816x + 34x^{2} - 17x^{3})}{(10x)}=-\frac{408}{5}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0+}\frac{(51x^{2} - 68x - 816)}{(10)}=-\frac{408}{5}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(816x + 34x^{2} - 17x^{3})}{(10x)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(17x\cdot (x + 6)\cdot (x - 8))}{(10x)}\\&\frac{(17\cdot (x + 6)\cdot (x - 8))}{10}\\&lim_{x\rightarrow0+}-\frac{(816x + 34x^{2} - 17x^{3})}{(10x)}=-\frac{408}{5}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0+}\frac{(51x^{2} - 68x - 816)}{(10)}=-\frac{408}{5}\end{align*}$

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Example Question #1053 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}-\frac{(10x - 10)}{(14x + 84x^{2} - 14x^{3} - 84)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1+}-\frac{(10x - 10)}{(14x + 84x^{2} - 14x^{3} - 84)}\end{align*}$

Possible Answers:

$-\frac{1}{14}$ $\displaystyle -\frac{1}{14}$

$\frac{1}{14}$ $\displaystyle \frac{1}{14}$

$\displaystyle 0$

$\infty$ $\displaystyle \infty$

Correct answer:

$-\frac{1}{14}$ $\displaystyle -\frac{1}{14}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=1\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(10x - 10)}{(14x + 84x^{2} - 14x^{3} - 84)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(10\cdot (x - 1))}{(14\cdot (x - 1)\cdot (x + 1)\cdot (x - 6))}\\&\frac{5}{(7\cdot (x + 1)\cdot (x - 6))}\\&lim_{x\rightarrow1+}-\frac{(10x - 10)}{(14x + 84x^{2} - 14x^{3} - 84)}=-\frac{1}{14}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow1+}\frac{(10)}{(42x^{2} - 168x - 14)}=-\frac{1}{14}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=1\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(10x - 10)}{(14x + 84x^{2} - 14x^{3} - 84)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(10\cdot (x - 1))}{(14\cdot (x - 1)\cdot (x + 1)\cdot (x - 6))}\\&\frac{5}{(7\cdot (x + 1)\cdot (x - 6))}\\&lim_{x\rightarrow1+}-\frac{(10x - 10)}{(14x + 84x^{2} - 14x^{3} - 84)}=-\frac{1}{14}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow1+}\frac{(10)}{(42x^{2} - 168x - 14)}=-\frac{1}{14}\end{align*}$

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Example Question #1054 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow5+}-\frac{(x - 5)}{(225x + 15x^{2} - 5x^{3} - 875)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow5+}-\frac{(x - 5)}{(225x + 15x^{2} - 5x^{3} - 875)}\end{align*}$

Possible Answers:

$\frac{83}{93}$ $\displaystyle \frac{83}{93}$

$-\frac{83}{93}$ $\displaystyle -\frac{83}{93}$

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

Correct answer:

$\infty$ $\displaystyle \infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=5\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(x - 5)}{(225x + 15x^{2} - 5x^{3} - 875)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(x - 5)}{(5\cdot (x - 5)^{2}\cdot (x + 7))}\\&\frac{1}{(5\cdot (x - 5)\cdot (x + 7))}\\&lim_{x\rightarrow5+}-\frac{(x - 5)}{(225x + 15x^{2} - 5x^{3} - 875)}=\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow5+}\frac{(1)}{(15x^{2} - 30x - 225)}=\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=5\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(x - 5)}{(225x + 15x^{2} - 5x^{3} - 875)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(x - 5)}{(5\cdot (x - 5)^{2}\cdot (x + 7))}\\&\frac{1}{(5\cdot (x - 5)\cdot (x + 7))}\\&lim_{x\rightarrow5+}-\frac{(x - 5)}{(225x + 15x^{2} - 5x^{3} - 875)}=\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow5+}\frac{(1)}{(15x^{2} - 30x - 225)}=\infty\end{align*}$

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Example Question #1055 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8-}\frac{(19x - 152)}{(x - 8)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8-}\frac{(19x - 152)}{(x - 8)}\end{align*}$

Possible Answers:

$\displaystyle 19$

$\infty$ $\displaystyle \infty$

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

$\displaystyle 0$

Correct answer:

$\displaystyle 19$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=8\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(19x - 152)}{(x - 8)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(19\cdot (x - 8))}{(x - 8)}\\&19\\&lim_{x\rightarrow8-}\frac{(19x - 152)}{(x - 8)}=19\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8-}\frac{(19)}{(1)}=19\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=8\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(19x - 152)}{(x - 8)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(19\cdot (x - 8))}{(x - 8)}\\&19\\&lim_{x\rightarrow8-}\frac{(19x - 152)}{(x - 8)}=19\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8-}\frac{(19)}{(1)}=19\end{align*}$

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Example Question #1056 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2-}-\frac{(84x - 162x^{2} + 6x^{4} + 720)}{(8320x + 624x^{2} - 130x^{3} - 13x^{4} + 13312)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2-}-\frac{(84x - 162x^{2} + 6x^{4} + 720)}{(8320x + 624x^{2} - 130x^{3} - 13x^{4} + 13312)}\end{align*}$

Possible Answers:

$\frac{3}{26}$ $\displaystyle \frac{3}{26}$

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

$\infty$ $\displaystyle \infty$

$-\frac{3}{26}$ $\displaystyle -\frac{3}{26}$

Correct answer:

$-\frac{3}{26}$ $\displaystyle -\frac{3}{26}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(84x - 162x^{2} + 6x^{4} + 720)}{(8320x + 624x^{2} - 130x^{3} - 13x^{4} + 13312)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(6\cdot (x + 2)\cdot (x - 3)\cdot (x - 4)\cdot (x + 5))}{(13\cdot (x + 2)\cdot (x - 8)\cdot (x + 8)^{2})}\\&\frac{(6\cdot (x - 3)\cdot (x - 4)\cdot (x + 5))}{(13\cdot (x - 8)\cdot (x + 8)^{2})}\\&lim_{x\rightarrow-2-}-\frac{(84x - 162x^{2} + 6x^{4} + 720)}{(8320x + 624x^{2} - 130x^{3} - 13x^{4} + 13312)}=-\frac{3}{26}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-2-}\frac{(24x^{3} - 324x + 84)}{(390x^{2} - 1248x + 52x^{3} - 8320)}=-\frac{3}{26}\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(84x - 162x^{2} + 6x^{4} + 720)}{(8320x + 624x^{2} - 130x^{3} - 13x^{4} + 13312)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(6\cdot (x + 2)\cdot (x - 3)\cdot (x - 4)\cdot (x + 5))}{(13\cdot (x + 2)\cdot (x - 8)\cdot (x + 8)^{2})}\\&\frac{(6\cdot (x - 3)\cdot (x - 4)\cdot (x + 5))}{(13\cdot (x - 8)\cdot (x + 8)^{2})}\\&lim_{x\rightarrow-2-}-\frac{(84x - 162x^{2} + 6x^{4} + 720)}{(8320x + 624x^{2} - 130x^{3} - 13x^{4} + 13312)}=-\frac{3}{26}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-2-}\frac{(24x^{3} - 324x + 84)}{(390x^{2} - 1248x + 52x^{3} - 8320)}=-\frac{3}{26}\end{align*}$

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Example Question #1057 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4+}-\frac{(1260x + 240x^{2} + 15x^{3} + 2160)}{(384x + 432x^{2} - 12x^{4} - 2304)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4+}-\frac{(1260x + 240x^{2} + 15x^{3} + 2160)}{(384x + 432x^{2} - 12x^{4} - 2304)}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$\frac{ 89}{67}$ $\displaystyle \frac{ 89}{67}$

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

$\displaystyle 0$

Correct answer:

$\infty$ $\displaystyle \infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-4\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(1260x + 240x^{2} + 15x^{3} + 2160)}{(384x + 432x^{2} - 12x^{4} - 2304)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(15\cdot (x + 4)\cdot (x + 6)^{2})}{(12\cdot (x + 4)^{2}\cdot (x - 2)\cdot (x - 6))}\\&\frac{(5\cdot (x + 6)^{2})}{(4\cdot (x - 2)\cdot (x + 4)\cdot (x - 6))}\\&lim_{x\rightarrow-4+}-\frac{(1260x + 240x^{2} + 15x^{3} + 2160)}{(384x + 432x^{2} - 12x^{4} - 2304)}=\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-4+}\frac{(480x + 45x^{2} + 1260)}{(48x^{3} - 864x - 384)}=\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-4\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(1260x + 240x^{2} + 15x^{3} + 2160)}{(384x + 432x^{2} - 12x^{4} - 2304)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(15\cdot (x + 4)\cdot (x + 6)^{2})}{(12\cdot (x + 4)^{2}\cdot (x - 2)\cdot (x - 6))}\\&\frac{(5\cdot (x + 6)^{2})}{(4\cdot (x - 2)\cdot (x + 4)\cdot (x - 6))}\\&lim_{x\rightarrow-4+}-\frac{(1260x + 240x^{2} + 15x^{3} + 2160)}{(384x + 432x^{2} - 12x^{4} - 2304)}=\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-4+}\frac{(480x + 45x^{2} + 1260)}{(48x^{3} - 864x - 384)}=\infty\end{align*}$

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Example Question #1058 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{(13x^{2})}{(65x + 78x^{2} + 13x^{3})}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{(13x^{2})}{(65x + 78x^{2} + 13x^{3})}\end{align*}$

Possible Answers:

$\frac{ 9}{23}$ $\displaystyle \frac{ 9}{23}$

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$-\frac{ 9}{23}$ $\displaystyle -\frac{ 9}{23}$

Correct answer:

$\displaystyle 0$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(13x^{2})}{(65x + 78x^{2} + 13x^{3})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(13x^{2})}{(13x\cdot (x + 1)\cdot (x + 5))}\\&\frac{x}{((x + 1)\cdot (x + 5))}\\&lim_{x\rightarrow0+}\frac{(13x^{2})}{(65x + 78x^{2} + 13x^{3})}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0+}\frac{(26x)}{(156x + 39x^{2} + 65)}=0\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(13x^{2})}{(65x + 78x^{2} + 13x^{3})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(13x^{2})}{(13x\cdot (x + 1)\cdot (x + 5))}\\&\frac{x}{((x + 1)\cdot (x + 5))}\\&lim_{x\rightarrow0+}\frac{(13x^{2})}{(65x + 78x^{2} + 13x^{3})}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0+}\frac{(26x)}{(156x + 39x^{2} + 65)}=0\end{align*}$

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Example Question #1059 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7+}-\frac{(14x + 98)}{(1330x + 60x^{2} - 50x^{3} - 5x^{4} + 1225)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7+}-\frac{(14x + 98)}{(1330x + 60x^{2} - 50x^{3} - 5x^{4} + 1225)}\end{align*}$

Possible Answers:

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

$\infty$ $\displaystyle \infty$

$\frac{32}{19}$ $\displaystyle \frac{32}{19}$

$-\frac{32}{19}$ $\displaystyle -\frac{32}{19}$

Correct answer:

$\infty$ $\displaystyle \infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(14x + 98)}{(1330x + 60x^{2} - 50x^{3} - 5x^{4} + 1225)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(14\cdot (x + 7))}{(5\cdot (x + 7)^{2}\cdot (x + 1)\cdot (x - 5))}\\&\frac{14}{(5\cdot (x + 1)\cdot (x - 5)\cdot (x + 7))}\\&lim_{x\rightarrow-7+}-\frac{(14x + 98)}{(1330x + 60x^{2} - 50x^{3} - 5x^{4} + 1225)}=\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7+}\frac{(14)}{(150x^{2} - 120x + 20x^{3} - 1330)}=\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(14x + 98)}{(1330x + 60x^{2} - 50x^{3} - 5x^{4} + 1225)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(14\cdot (x + 7))}{(5\cdot (x + 7)^{2}\cdot (x + 1)\cdot (x - 5))}\\&\frac{14}{(5\cdot (x + 1)\cdot (x - 5)\cdot (x + 7))}\\&lim_{x\rightarrow-7+}-\frac{(14x + 98)}{(1330x + 60x^{2} - 50x^{3} - 5x^{4} + 1225)}=\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7+}\frac{(14)}{(150x^{2} - 120x + 20x^{3} - 1330)}=\infty\end{align*}$

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Calculus 3 : Partial Derivatives

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