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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #31 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-10}-\frac{(x + 10)}{(28x - 12x^{2} - x^{3} + 480)}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{ 1}{32}$

$\frac{ 1}{32}$

$\infty$

Correct answer:

$\frac{ 1}{32}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-10\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&-\frac{(x + 10)}{(28x - 12x^{2} - x^{3} + 480)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x + 10)}{((x - 6)\cdot (x + 8)\cdot (x + 10))}\\&\frac{1}{((x - 6)\cdot (x + 8))}\\&lim_{x\rightarrow-10}-\frac{(x + 10)}{(28x - 12x^{2} - x^{3} + 480)}=\frac{ 1}{32}\end{align*}$

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Example Question #31 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow9}-\frac{(65x + 8x^{2} - x^{3} - 504)}{(15x - 11x^{2} + x^{3} + 27)}\end{align*}$

Possible Answers:

$\infty$

$\frac{ 17}{30}$

$-\infty$

Correct answer:

$\frac{ 17}{30}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=9\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&-\frac{(65x + 8x^{2} - x^{3} - 504)}{(15x - 11x^{2} + x^{3} + 27)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x - 7)\cdot (x + 8)\cdot (x - 9))}{((x + 1)\cdot (x - 3)\cdot (x - 9))}\\&\frac{((x - 7)\cdot (x + 8))}{((x + 1)\cdot (x - 3))}\\&lim_{x\rightarrow9}-\frac{(65x + 8x^{2} - x^{3} - 504)}{(15x - 11x^{2} + x^{3} + 27)}=\frac{ 17}{30}\end{align*}$

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Example Question #31 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3}\frac{(x - 3)}{(101x - 18x^{2} + x^{3} - 168)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{ 1}{20}$

Correct answer:

$\frac{ 1}{20}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=3\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(x - 3)}{(101x - 18x^{2} + x^{3} - 168)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x - 3)}{((x - 3)\cdot (x - 7)\cdot (x - 8))}\\&\frac{1}{((x - 7)\cdot (x - 8))}\\&lim_{x\rightarrow3}\frac{(x - 3)}{(101x - 18x^{2} + x^{3} - 168)}=\frac{ 1}{20}\end{align*}$

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Example Question #33 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-1}-\frac{(76x + 17x^{2} + x^{3} + 60)}{(x - 6x^{2} - x^{3} + 6)}\end{align*}$

Possible Answers:

$-\frac{9}{2}$

$\infty$

$-\infty$

Correct answer:

$-\frac{9}{2}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-1\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&-\frac{(76x + 17x^{2} + x^{3} + 60)}{(x - 6x^{2} - x^{3} + 6)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x + 1)\cdot (x + 6)\cdot (x + 10))}{((x - 1)\cdot (x + 1)\cdot (x + 6))}\\&\frac{(x + 10)}{(x - 1)}\\&lim_{x\rightarrow-1}-\frac{(76x + 17x^{2} + x^{3} + 60)}{(x - 6x^{2} - x^{3} + 6)}= -\frac{9}{2}\end{align*}$

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Example Question #32 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow1}\frac{(9x + x^{2} - 10)}{(x + x^{2} - 2)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{ 11}{3}$

$-\frac{11}{3}$

Correct answer:

$\frac{ 11}{3}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=1\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(9x + x^{2} - 10)}{(x + x^{2} - 2)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x - 1)\cdot (x + 10))}{((x - 1)\cdot (x + 2))}\\&\frac{(x + 10)}{(x + 2)}\\&lim_{x\rightarrow1}\frac{(9x + x^{2} - 10)}{(x + x^{2} - 2)}=\frac{ 11}{3}\end{align*}$

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Example Question #33 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8}\frac{(x - 8)}{(2x + x^{2} - 80)}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$\frac{ 1}{18}$

$-\frac{1}{18}$

Correct answer:

$\frac{ 1}{18}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=8\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(x - 8)}{(2x + x^{2} - 80)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{(x - 8)}{((x - 8)\cdot (x + 10))}\\&\frac{1}{(x + 10)}\\&lim_{x\rightarrow8}\frac{(x - 8)}{(2x + x^{2} - 80)}=\frac{ 1}{18}\end{align*}$

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Example Question #36 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow6}\frac{(36x - 9x^{2} - x^{3} + 324)}{(32x - 10x^{2} - x^{3} + 384)}\end{align*}$

Possible Answers:

$\frac{ 45}{49}$

$-\infty$

$\infty$

$-\frac{45}{49}$

Correct answer:

$\frac{ 45}{49}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=6\text{, as that would}\\&\text{create a zero value in the denominator. However, note that the expression}\\&\frac{(36x - 9x^{2} - x^{3} + 324)}{(32x - 10x^{2} - x^{3} + 384)}\\&\text{can be factored. This will allow us to find the limit:}\\&\frac{((x - 6)\cdot (x + 6)\cdot (x + 9))}{((x - 6)\cdot (x + 8)^{2})}\\&\frac{((x + 6)\cdot (x + 9))}{(x + 8)^{2}}\\&lim_{x\rightarrow6}\frac{(36x - 9x^{2} - x^{3} + 324)}{(32x - 10x^{2} - x^{3} + 384)}=\frac{ 45}{49}\end{align*}$

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Example Question #34 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-3-}\frac{(246x + 72x^{2} + 6x^{3} + 252)}{(12x + 36)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-3\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is. However, note that the expression}\\&\frac{(246x + 72x^{2} + 6x^{3} + 252)}{(12x + 36)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(6\cdot (x + 3)\cdot (x + 2)\cdot (x + 7))}{(12\cdot (x + 3))}\\&\frac{((x + 2)\cdot (x + 7))}{2}\\&lim_{x\rightarrow-3-}\frac{(246x + 72x^{2} + 6x^{3} + 252)}{(12x + 36)}=-2\end{align*}$

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Example Question #34 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow5+}\frac{(13x - 65)}{(12x^{2} - 132x + 360)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{13}{12}$

$-\frac{13}{12}$

Correct answer:

$-\frac{13}{12}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=5\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is. However, note that the expression}\\&\frac{(13x - 65)}{(12x^{2} - 132x + 360)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(13\cdot (x - 5))}{(12\cdot (x - 5)\cdot (x - 6))}\\&\frac{13}{(12\cdot (x - 6))}\\&lim_{x\rightarrow5+}\frac{(13x - 65)}{(12x^{2} - 132x + 360)}=-\frac{13}{12}\end{align*}$

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Example Question #34 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow9+}\frac{(30x - 110x^{2} + 10x^{3} + 1350)}{(4x^{2} - 68x + 288)}\end{align*}$

Possible Answers:

120

$-\infty$

$\infty$

Correct answer:

120

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=9\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(30x - 110x^{2} + 10x^{3} + 1350)}{(4x^{2} - 68x + 288)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(10\cdot (x - 9)\cdot (x + 3)\cdot (x - 5))}{(4\cdot (x - 9)\cdot (x - 8))}\\&\frac{(5\cdot (x + 3)\cdot (x - 5))}{(2\cdot (x - 8))}\\&lim_{x\rightarrow9+}\frac{(30x - 110x^{2} + 10x^{3} + 1350)}{(4x^{2} - 68x + 288)}=120\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #31 : Limits

Example Question #31 : Partial Derivatives

Example Question #31 : Partial Derivatives

Example Question #33 : Limits

Example Question #32 : Partial Derivatives

Example Question #33 : Partial Derivatives

Example Question #36 : Limits

Example Question #34 : Partial Derivatives

Example Question #34 : Partial Derivatives

Example Question #34 : Partial Derivatives

All Calculus 3 Resources

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