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Calculus 3 : Limits

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Example Questions

Example Question #11 : Limits

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

Possible Answers:

$-\frac{23}{2088}$

$\infty$

$-\infty$

Correct answer:

$-\frac{23}{2088}$

Explanation:

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

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Example Question #971 : Calculus 3

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

Possible Answers:

$-\frac{9}{1024}$

$-\infty$

$\infty$

Correct answer:

$-\frac{9}{1024}$

Explanation:

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

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Example Question #11 : Limits

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

Possible Answers:

$-\infty$

$\frac{ 5}{9}$

$\infty$

$-\frac{5}{9}$

Correct answer:

$-\frac{5}{9}$

Explanation:

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

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Example Question #12 : Partial Derivatives

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

Possible Answers:

$\frac{ 1}{1100}$

$-\frac{1}{1100}$

$-\infty$

$\infty$

Correct answer:

$-\frac{1}{1100}$

Explanation:

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

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Example Question #12 : Limits

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

Possible Answers:

$-\frac{1}{63}$

$-\infty$

$\frac{ 1}{63}$

$\infty$

Correct answer:

$-\frac{1}{63}$

Explanation:

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

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Example Question #13 : Limits

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

Possible Answers:

$-\infty$

$-\frac{9}{2}$

$\infty$

Correct answer:

$-\frac{9}{2}$

Explanation:

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

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Example Question #16 : Limits

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

Possible Answers:

$-\frac{108}{31}$

$\infty$

$-\infty$

Correct answer:

$-\frac{108}{31}$

Explanation:

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

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Example Question #975 : Calculus 3

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

Possible Answers:

$-\infty$

$\infty$

$-\frac{1}{32}$

Correct answer:

$-\frac{1}{32}$

Explanation:

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

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Example Question #14 : Limits

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

Possible Answers:

$-\infty$

$\frac{ 1}{14}$

$-\frac{1}{14}$

$\infty$

Correct answer:

$-\frac{1}{14}$

Explanation:

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

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Example Question #14 : Partial Derivatives

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

Possible Answers:

$\infty$

$-\infty$

$\frac{ 99}{7}$

Correct answer:

$\frac{ 99}{7}$

Explanation:

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

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Calculus 3 : Limits

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Limits

Example Question #971 : Calculus 3

Example Question #11 : Limits

Example Question #12 : Partial Derivatives

Example Question #12 : Limits

Example Question #13 : Limits

Example Question #16 : Limits

Example Question #975 : Calculus 3

Example Question #14 : Limits

Example Question #14 : Partial Derivatives

All Calculus 3 Resources

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