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Calculus 3 : Limits

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Example Questions

Example Question #1 : Limits

Evaluate:

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

Possible Answers:

Does not exist

$-\infty$

$\infty$

Correct answer:

Explanation:

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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Example Question #2 : Limits

Calculate the following limit.

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

Possible Answers:

$\small 2$

$\small 1$

$\small \frac{1}{2}$

$\small 0$

Correct answer:

$\small 1$

Explanation:

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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Example Question #3 : Limits

Calculate the following limit.

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

Possible Answers:

$\small -1$

$\small 1$

$\small \frac{1}{2}$

$\small 2$

Correct answer:

$\small -1$

Explanation:

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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Example Question #961 : Calculus 3

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

Possible Answers:

Does not exist

None of the other answers

Correct answer:

Does not exist

Explanation:

If we try evaluating the limit along two different paths, y=x, y=2x , 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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Example Question #1 : Limits

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

Possible Answers:

-1/14

$\infty$

$-\infty$

1/14

Correct answer:

1/14

Explanation:

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

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Example Question #6 : Limits

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

Possible Answers:

$-\frac{31}{897}$

$\frac{ 31}{897}$

$\infty$

$-\infty$

Correct answer:

$-\frac{31}{897}$

Explanation:

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

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Example Question #7 : Limits

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

Possible Answers:

$-\frac{9}{88}$

$-\infty$

$\infty$

$\frac{ 9}{88}$

Correct answer:

$-\frac{9}{88}$

Explanation:

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

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Example Question #1 : Partial Derivatives

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

Possible Answers:

$-\infty$

$\frac{ 17}{28}$

$\infty$

Correct answer:

$\frac{ 17}{28}$

Explanation:

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

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Example Question #9 : Limits

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

Possible Answers:

$-\frac{1}{3315}$

$-\infty$

$\infty$

Correct answer:

$-\frac{1}{3315}$

Explanation:

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

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Example Question #1 : Limits

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

Possible Answers:

$\infty$

$-\infty$

$-\frac{37}{8}$

$\frac{ 37}{8}$

Correct answer:

$-\frac{37}{8}$

Explanation:

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

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Calculus 3 : Limits

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Limits

Example Question #2 : Limits

Example Question #3 : Limits

Example Question #961 : Calculus 3

Example Question #1 : Limits

Example Question #6 : Limits

Example Question #7 : Limits

Example Question #1 : Partial Derivatives

Example Question #9 : Limits

Example Question #1 : Limits

All Calculus 3 Resources

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