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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #51 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{(21x + 14x^{2} - 7x^{3})}{(28x^{2} - 14x^{3})}\end{align*}$

Possible Answers:

$-\infty$

$\frac{ 9}{92}$

$\infty$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=0\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(21x + 14x^{2} - 7x^{3})}{(28x^{2} - 14x^{3})}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(7x\cdot (x + 1)\cdot (x - 3))}{(14x^{2}\cdot (x - 2))}\\&\frac{((x + 1)\cdot (x - 3))}{(2x\cdot (x - 2))}\\&lim_{x\rightarrow0+}\frac{(21x + 14x^{2} - 7x^{3})}{(28x^{2} - 14x^{3})}=\infty\end{align*}$

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Example Question #53 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-8+}\frac{(28x + 7x^{2} - 224)}{(17x + 136)}\end{align*}$

Possible Answers:

$\infty$

$-\frac{84}{17}$

$-\infty$

Correct answer:

$-\frac{84}{17}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(28x + 7x^{2} - 224)}{(17x + 136)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(7\cdot (x + 8)\cdot (x - 4))}{(17\cdot (x + 8))}\\&\frac{(7\cdot (x - 4))}{17}\\&lim_{x\rightarrow-8+}\frac{(28x + 7x^{2} - 224)}{(17x + 136)}=-\frac{84}{17}\end{align*}$

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Example Question #52 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-9+}\frac{(x + 9)}{(144x + 12x^{2} + 324)}\end{align*}$

Possible Answers:

$-\frac{1}{72}$

$\infty$

$-\infty$

Correct answer:

$-\frac{1}{72}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-9\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(x + 9)}{(144x + 12x^{2} + 324)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(x + 9)}{(12\cdot (x + 9)\cdot (x + 3))}\\&\frac{1}{(12\cdot (x + 3))}\\&lim_{x\rightarrow-9+}\frac{(x + 9)}{(144x + 12x^{2} + 324)}=-\frac{1}{72}\end{align*}$

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Example Question #51 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow2-}-\frac{(32x - 16x^{2})}{(5x - 10)}\end{align*}$

Possible Answers:

$\infty$

$\frac{32}{5}$

$-\infty$

Correct answer:

$\frac{32}{5}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=2\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(32x - 16x^{2})}{(5x - 10)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(16x\cdot (x - 2))}{(5\cdot (x - 2))}\\&\frac{(16x)}{5}\\&lim_{x\rightarrow2-}-\frac{(32x - 16x^{2})}{(5x - 10)}=\frac{32}{5}\end{align*}$

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Example Question #56 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4-}\frac{(20x + 80)}{(646x + 209x^{2} + 19x^{3} + 456)}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{10}{57}$

$\frac{10}{57}$

$\infty$

Correct answer:

$-\frac{10}{57}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-4\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(20x + 80)}{(646x + 209x^{2} + 19x^{3} + 456)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(20\cdot (x + 4))}{(19\cdot (x + 4)\cdot (x + 1)\cdot (x + 6))}\\&\frac{20}{(19\cdot (x + 1)\cdot (x + 6))}\\&lim_{x\rightarrow-4-}\frac{(20x + 80)}{(646x + 209x^{2} + 19x^{3} + 456)}=-\frac{10}{57}\end{align*}$

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Example Question #1011 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow6-}-\frac{(1444x - 323x^{2} + 19x^{3} - 1140)}{(620x + 50x^{2} - 10x^{3} - 3360)}\end{align*}$

Possible Answers:

$\frac{19}{7}$

$-\infty$

$\infty$

$-\frac{19}{7}$

Correct answer:

$\frac{19}{7}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=6\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(1444x - 323x^{2} + 19x^{3} - 1140)}{(620x + 50x^{2} - 10x^{3} - 3360)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(19\cdot (x - 6)\cdot (x - 1)\cdot (x - 10))}{(10\cdot (x - 6)\cdot (x - 7)\cdot (x + 8))}\\&\frac{(19\cdot (x - 1)\cdot (x - 10))}{(10\cdot (x - 7)\cdot (x + 8))}\\&lim_{x\rightarrow6-}-\frac{(1444x - 323x^{2} + 19x^{3} - 1140)}{(620x + 50x^{2} - 10x^{3} - 3360)}=\frac{19}{7}\end{align*}$

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Example Question #55 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-9-}\frac{(14x + 126)}{(16x^{2} - 960x + 16x^{3} + 1728)}\end{align*}$

Possible Answers:

$-\infty$

$\frac{7}{1320}$

$\infty$

Correct answer:

$\frac{7}{1320}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-9\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(14x + 126)}{(16x^{2} - 960x + 16x^{3} + 1728)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(14\cdot (x + 9))}{(16\cdot (x + 9)\cdot (x - 2)\cdot (x - 6))}\\&\frac{7}{(8\cdot (x - 2)\cdot (x - 6))}\\&lim_{x\rightarrow-9-}\frac{(14x + 126)}{(16x^{2} - 960x + 16x^{3} + 1728)}=\frac{7}{1320}\end{align*}$

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Example Question #59 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{(180x + 96x^{2} + 12x^{3})}{(19x)}\end{align*}$

Possible Answers:

$-\infty$

$\frac{180}{19}$

$\infty$

Correct answer:

$\frac{180}{19}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=0\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(180x + 96x^{2} + 12x^{3})}{(19x)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(12x\cdot (x + 3)\cdot (x + 5))}{(19x)}\\&\frac{(12\cdot (x + 3)\cdot (x + 5))}{19}\\&lim_{x\rightarrow0+}\frac{(180x + 96x^{2} + 12x^{3})}{(19x)}=\frac{180}{19}\end{align*}$

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Example Question #60 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4+}\frac{(4x + 16)}{(108x + 12x^{2} + 240)}\end{align*}$

Possible Answers:

$\infty$

$\frac{1}{3}$

$-\infty$

Correct answer:

$\frac{1}{3}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-4\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(4x + 16)}{(108x + 12x^{2} + 240)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(4\cdot (x + 4))}{(12\cdot (x + 4)\cdot (x + 5))}\\&\frac{1}{(3\cdot (x + 5))}\\&lim_{x\rightarrow-4+}\frac{(4x + 16)}{(108x + 12x^{2} + 240)}=\frac{1}{3}\end{align*}$

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Example Question #61 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2-}-\frac{(80x + 5x^{2} - 5x^{3} + 100)}{(5x + 10)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$-\frac{47}{83}$

Correct answer:

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-2\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(80x + 5x^{2} - 5x^{3} + 100)}{(5x + 10)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(5\cdot (x + 2)^{2}\cdot (x - 5))}{(5\cdot (x + 2))}\\&(x + 2)\cdot (x - 5)\\&lim_{x\rightarrow-2-}-\frac{(80x + 5x^{2} - 5x^{3} + 100)}{(5x + 10)}=0\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #51 : Partial Derivatives

Example Question #53 : Partial Derivatives

Example Question #52 : Limits

Example Question #51 : Partial Derivatives

Example Question #56 : Partial Derivatives

Example Question #1011 : Calculus 3

Example Question #55 : Limits

Example Question #59 : Partial Derivatives

Example Question #60 : Partial Derivatives

Example Question #61 : Limits

All Calculus 3 Resources

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