Partial Derivatives - AP Calculus BC

Card 0 of 7240

Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow0-}\frac{x^{2}}{sin(13\cdot \pi x)^{2}}\end{align}$

Answer

$\begin{align}&\text{Examining this problem, we find that both}\&\text{numerator and denominator have zero values at the limit specified.}\&\text{Noting this, L'Hopital's method may be of use:}\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\&\frac{x^{2}}{sin(13\cdot \pi x)^{2}}\rightarrow\frac{0}{0}\&\frac{2x}{26\cdot \pi cos(13\cdot \pi x)sin(13\cdot \pi x)}\rightarrow\frac{0}{0}\&\frac{2}{338\cdot \pi ^{2}cos(13\cdot \pi x)^{2} - 338\cdot \pi ^{2}sin(13\cdot \pi x)^{2}}\rightarrow\frac{2}{338\cdot \pi ^{2}}\&lim_{x\rightarrow0-}\frac{x^{2}}{sin(13\cdot \pi x)^{2}}=\frac{1}{(169\cdot \pi ^{2})}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(12x^{4}y^{2} - 11xy^{5})}{(20x^{4}y^{2} + 9xy^{5} + 3x^{6} + 36y^{6})}\&\text{along the x-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\&\frac{(12(0)^{4}y^{2} - 11(0)y^{5})}{(20(0)^{4}y^{2} + 9(0)y^{5} + 3(0)^{6} + 36y^{6})}=\frac{0}{36y^{6}}\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{36y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(12x^{4}(x)^{2} - 11x(x)^{5})}{(20x^{4}(x)^{2} + 9x(x)^{5} + 3x^{6} + 36(x)^{6})}=\frac{1}{68}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0+,0+)}\frac{(10xy^{2} + 2y^{3})}{(19x^{2}y + 10x^{3} - 5y^{3})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&\frac{(10(0)y^{2} + 2y^{3})}{(19(0)^{2}y + 10(0)^{3} - 5y^{3})}=\frac{-2y^{3}}{5y^{3}}\&lim_{(0,y)\rightarrow(0+,0+)}\frac{-2y^{3}}{5y^{3}}=-\frac{2}{5}\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(10x(x)^{2} + 2(x)^{3})}{(19x^{2}(x) + 10x^{3} - 5(x)^{3})}=\frac{1}{2}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0+,0-)}-\frac{(6x^{2}y^{3} - 6xy^{4})}{(4x^{5} + 5y^{5})}\&\text{along the y-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(6(0)^{2}y^{3} - 6(0)y^{4})}{(4(0)^{5} + 5y^{5})}=\frac{0}{5y^{5}}\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{5y^{5}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(6x^{2}(x)^{3} - 6x(x)^{4})}{(4x^{5} + 5(x)^{5})}=0\&\text{Since the two limits are equal, it's possible the limit exists.}\&\text{Testing additional lines could prove or disprove this.}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0-,0+)}\frac{(11x^{4}y^{2} - 17x^{6})}{(2x^{4}y^{2} - 12x^{3}y^{3} - 19x^{5}y + 17x^{6} + y^{6})}\&\text{along the x-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\&\frac{(11(0)^{4}y^{2} - 17(0)^{6})}{(2(0)^{4}y^{2} - 12(0)^{3}y^{3} - 19(0)^{5}y + 17(0)^{6} + y^{6})}=\frac{0}{y^{6}}\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(11x^{4}(x)^{2} - 17x^{6})}{(2x^{4}(x)^{2} - 12x^{3}(x)^{3} - 19x^{5}(x) + 17x^{6} + (x)^{6})}=\frac{6}{11}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(9x^{4}y^{2} + 9xy^{5} - 18x^{6})}{(3x^{6} + 2y^{6})}\&\text{along the x-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\&-\frac{(9(0)^{4}y^{2} + 9(0)y^{5} - 18(0)^{6})}{(3(0)^{6} + 2y^{6})}=\frac{0}{2y^{6}}\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{2y^{6}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(9x^{4}(x)^{2} + 9x(x)^{5} - 18x^{6})}{(3x^{6} + 2(x)^{6})}=0\&\text{Since the two limits are equal, it's possible the limit exists.}\&\text{Testing additional lines could prove or disprove this.}\end{align}$

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Question

Evaluate:

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}$

Answer

Since we won't have any divided by zero problems, we can just evaluate at the point.

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}=\frac{0^2+0\cdot1}{1}=0$

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Question

Calculate the following limit.

$\small \lim_{(x,y)\to (1,0)} x^2+2xy+y^2$

Answer

We can calculate the limit

$\small \lim_{(x,y)\to (1,0)} x^2+2xy+y^2$

by simply plugging in $\small (x,y)=(1,0)$ because the value is defined at $\small (1,0)$ and around that point.

So we get

$\small \small \lim_{(x,y)\to (1,0)} x^2+2xy+y^2=1^2+2\cdot 0\cdot 1+0^2=1$ .

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Question

Calculate the following limit.

$\small \small \lim_{(x,y)\to (2,1)} 2x-5y$

Answer

We can calculate the limit

$\small \small \lim_{(x,y)\to (2,1)} 2x-5y$

by simply plugging in $\small \small (x,y)=(2,1)$ because the value is defined at $\small \small (2,1)$ and at every value in $\small \mathbb{R}^2$ .

So we get

$\small \small \small \lim_{(x,y)\to (2,1)} 2x-5y=2\cdot 2-5\cdot1=4-5=-1$ .

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Question

Evaluate $\lim_{(x,y)\to(0,0)}\frac{6x}{x^2+y}$

Answer

If we try evaluating the limit along two different paths, , we have

$\lim_{(x,x)\to(0,0)}\frac{6x}{x^2+x}= \lim_{(x,x)\to(0,0)}\frac{6}{x+1} = 6$

And

$\lim_{(x,2x)\to(0,0)}\frac{6x}{x^2+(2x)} = \lim_{(x,2x)\to(0,0)}\frac{6}{x+2} = 3$

Since evaluating along these two paths yielded diffferent values, the limit does not exist.

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow13}-\frac{(x - 13)}{(12x - x^{2} + 13)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=13\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(x - 13)}{(12x - x^{2} + 13)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{(x - 13)}{((x + 1)\cdot (x - 13))}\&\frac{1}{(x + 1)}\&lim_{x\rightarrow13}-\frac{(x - 13)}{(12x - x^{2} + 13)}= 1/14\end{align}$

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Question

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

Answer

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

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow12}\frac{(7x + x^{2} - 228)}{(1863x - 133x^{2} - 15x^{3} + x^{4} + 1980)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=12\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(7x + x^{2} - 228)}{(1863x - 133x^{2} - 15x^{3} + x^{4} + 1980)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 12)\cdot (x + 19))}{((x + 1)\cdot (x + 11)\cdot (x - 12)\cdot (x - 15))}\&\frac{(x + 19)}{((x + 1)\cdot (x + 11)\cdot (x - 15))}\&lim_{x\rightarrow12}\frac{(7x + x^{2} - 228)}{(1863x - 133x^{2} - 15x^{3} + x^{4} + 1980)}= -\frac{31}{897}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow2}-\frac{(14x + x^{2} - 32)}{(144x + 11x^{2} - x^{3} - 324)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=2\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(14x + x^{2} - 32)}{(144x + 11x^{2} - x^{3} - 324)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 2)\cdot (x + 16))}{((x - 2)\cdot (x + 9)\cdot (x - 18))}\&\frac{(x + 16)}{((x + 9)\cdot (x - 18))}\&lim_{x\rightarrow2}-\frac{(14x + x^{2} - 32)}{(144x + 11x^{2} - x^{3} - 324)}= -\frac{9}{88}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow13}-\frac{(9x - x^{2} + 52)}{(2x + x^{2} - 195)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=13\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(9x - x^{2} + 52)}{(2x + x^{2} - 195)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 4)\cdot (x - 13))}{((x - 13)\cdot (x + 15))}\&\frac{(x + 4)}{(x + 15)}\&lim_{x\rightarrow13}-\frac{(9x - x^{2} + 52)}{(2x + x^{2} - 195)}=\frac{ 17}{28}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-1}-\frac{(x + 1)}{(3576x + 98x^{2} - 23x^{3} - x^{4} + 3456)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-1\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(x + 1)}{(3576x + 98x^{2} - 23x^{3} - x^{4} + 3456)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{(x + 1)}{((x + 1)\cdot (x - 12)\cdot (x + 16)\cdot (x + 18))}\&\frac{1}{((x - 12)\cdot (x + 16)\cdot (x + 18))}\&lim_{x\rightarrow-1}-\frac{(x + 1)}{(3576x + 98x^{2} - 23x^{3} - x^{4} + 3456)}= -\frac{1}{3315}\end{align}$

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Question

$\begin{align}&\text{Calculate the limits of the function}\&lim_{(x,y)\rightarrow(0+,0+)}\frac{(6xy^{2} - 13x^{2}y + 5x^{3})}{(xy^{2} - 20x^{2}y + 9x^{3} - 3y^{3})}\&\text{along the x-axis and the line, }y=x\end{align}$

Answer

$\begin{align}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\&\frac{(6(0)y^{2} - 13(0)^{2}y + 5(0)^{3})}{((0)y^{2} - 20(0)^{2}y + 9(0)^{3} - 3y^{3})}=\frac{0}{3y^{3}}\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{3}}=0\&\text{We can make a similar substitution for }y=x:\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(6x(x)^{2} - 13x^{2}(x) + 5x^{3})}{(x(x)^{2} - 20x^{2}(x) + 9x^{3} - 3(x)^{3})}=\frac{2}{13}\&\text{As these two limits are different, the limit does not exist.}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-18}-\frac{(x - x^{2} + 342)}{(764x + 48x^{2} + x^{3} + 4032)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-18\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(x - x^{2} + 342)}{(764x + 48x^{2} + x^{3} + 4032)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 18)\cdot (x - 19))}{((x + 14)\cdot (x + 16)\cdot (x + 18))}\&\frac{(x - 19)}{((x + 14)\cdot (x + 16))}\&lim_{x\rightarrow-18}-\frac{(x - x^{2} + 342)}{(764x + 48x^{2} + x^{3} + 4032)}= -\frac{37}{8}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-17}\frac{(11x + x^{2} - 102)}{(5x^{3} - 565x^{2} - 1805x + x^{4} + 73644)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-17\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(11x + x^{2} - 102)}{(5x^{3} - 565x^{2} - 1805x + x^{4} + 73644)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 6)\cdot (x + 17))}{((x - 12)\cdot (x + 17)\cdot (x - 19)\cdot (x + 19))}\&\frac{(x - 6)}{((x - 12)\cdot (x - 19)\cdot (x + 19))}\&lim_{x\rightarrow-17}\frac{(11x + x^{2} - 102)}{(5x^{3} - 565x^{2} - 1805x + x^{4} + 73644)}= -\frac{23}{2088}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow12}\frac{(3x + x^{2} - 180)}{(13x^{3} - 500x^{2} - 3600x + x^{4} + 72000)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=12\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(3x + x^{2} - 180)}{(13x^{3} - 500x^{2} - 3600x + x^{4} + 72000)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 12)\cdot (x + 15))}{((x - 12)\cdot (x - 15)\cdot (x + 20)^{2})}\&\frac{(x + 15)}{((x - 15)\cdot (x + 20)^{2})}\&lim_{x\rightarrow12}\frac{(3x + x^{2} - 180)}{(13x^{3} - 500x^{2} - 3600x + x^{4} + 72000)}= -\frac{9}{1024}\end{align}$

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