Partial Derivatives - AP Calculus BC

Card 0 of 7240

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

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

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

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

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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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow7}\frac{(x^{2} - 9x + 14)}{(x^{2} - 23x + 112)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(x^{2} - 9x + 14)}{(x^{2} - 23x + 112)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 2)\cdot (x - 7))}{((x - 7)\cdot (x - 16))}\&\frac{(x - 2)}{(x - 16)}\&lim_{x\rightarrow7}\frac{(x^{2} - 9x + 14)}{(x^{2} - 23x + 112)}= -\frac{5}{9}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-13}-\frac{(25x + x^{2} + 156)}{(2930x + 103x^{2} - 20x^{3} - x^{4} + 5304)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-13\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(25x + x^{2} + 156)}{(2930x + 103x^{2} - 20x^{3} - x^{4} + 5304)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 12)\cdot (x + 13))}{((x + 2)\cdot (x - 12)\cdot (x + 13)\cdot (x + 17))}\&\frac{(x + 12)}{((x + 2)\cdot (x - 12)\cdot (x + 17))}\&lim_{x\rightarrow-13}-\frac{(25x + x^{2} + 156)}{(2930x + 103x^{2} - 20x^{3} - x^{4} + 5304)}= -\frac{1}{1100}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow4}\frac{(x - 4)}{(x - 14x^{2} + x^{3} + 156)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=4\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(x - 4)}{(x - 14x^{2} + x^{3} + 156)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{(x - 4)}{((x + 3)\cdot (x - 4)\cdot (x - 13))}\&\frac{1}{((x + 3)\cdot (x - 13))}\&lim_{x\rightarrow4}\frac{(x - 4)}{(x - 14x^{2} + x^{3} + 156)}= -\frac{1}{63}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-6}-\frac{(33x^{2} - 144x + 14x^{3} + x^{4} - 324)}{(12x - x^{2} + 108)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-6\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(33x^{2} - 144x + 14x^{3} + x^{4} - 324)}{(12x - x^{2} + 108)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 2)\cdot (x - 3)\cdot (x + 6)\cdot (x + 9))}{((x + 6)\cdot (x - 18))}\&\frac{((x + 2)\cdot (x - 3)\cdot (x + 9))}{(x - 18)}\&lim_{x\rightarrow-6}-\frac{(33x^{2} - 144x + 14x^{3} + x^{4} - 324)}{(12x - x^{2} + 108)}= -\frac{9}{2}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-16}-\frac{(12416x + 156x^{2} - 31x^{3} - x^{4} + 97280)}{(31x^{2} - 7200x + 32x^{3} + x^{4} - 57600)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-16\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&-\frac{(12416x + 156x^{2} - 31x^{3} - x^{4} + 97280)}{(31x^{2} - 7200x + 32x^{3} + x^{4} - 57600)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 16)^{2}\cdot (x + 19)\cdot (x - 20))}{((x - 15)\cdot (x + 15)\cdot (x + 16)^{2})}\&\frac{((x + 19)\cdot (x - 20))}{((x - 15)\cdot (x + 15))}\&lim_{x\rightarrow-16}-\frac{(12416x + 156x^{2} - 31x^{3} - x^{4} + 97280)}{(31x^{2} - 7200x + 32x^{3} + x^{4} - 57600)}= -\frac{108}{31}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-2}\frac{(x + x^{2} - 2)}{(28x - 14x^{2} + x^{3} + 120)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-2\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(x + x^{2} - 2)}{(28x - 14x^{2} + x^{3} + 120)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x - 1)\cdot (x + 2))}{((x + 2)\cdot (x - 6)\cdot (x - 10))}\&\frac{(x - 1)}{((x - 6)\cdot (x - 10))}\&lim_{x\rightarrow-2}\frac{(x + x^{2} - 2)}{(28x - 14x^{2} + x^{3} + 120)}= -\frac{1}{32}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-10}\frac{(x + 10)}{(6x + x^{2} - 40)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-10\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(x + 10)}{(6x + x^{2} - 40)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{(x + 10)}{((x - 4)\cdot (x + 10))}\&\frac{1}{(x - 4)}\&lim_{x\rightarrow-10}\frac{(x + 10)}{(6x + x^{2} - 40)}= -\frac{1}{14}\end{align}$

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Question

$\begin{align}&\text{Evaluate the following limit:}\&lim_{x\rightarrow-1}\frac{(62x - 17x^{2} + x^{3} + 80)}{(9x + x^{2} + 8)}\end{align}$

Answer

$\begin{align}&\text{We cannot simply plug in the value of }x=-1\text{, as that would}\&\text{create a zero value in the denominator. However, note that the expression}\&\frac{(62x - 17x^{2} + x^{3} + 80)}{(9x + x^{2} + 8)}\&\text{can be factored. This will allow us to find the limit:}\&\frac{((x + 1)\cdot (x - 8)\cdot (x - 10))}{((x + 1)\cdot (x + 8))}\&\frac{((x - 8)\cdot (x - 10))}{(x + 8)}\&lim_{x\rightarrow-1}\frac{(62x - 17x^{2} + x^{3} + 80)}{(9x + x^{2} + 8)}=\frac{ 99}{7}\end{align}$

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