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Calculus 3 : Partial Derivatives

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 #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:

$\infty$

$-\frac{47}{83}$

$-\infty$

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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Example Question #62 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow5-}\frac{(9x - 45)}{(17x - 85)}\end{align*}$

Possible Answers:

$\frac{9}{17}$

$-\infty$

$\infty$

Correct answer:

$\frac{9}{17}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=5\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(9x - 45)}{(17x - 85)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(9\cdot (x - 5))}{(17\cdot (x - 5))}\\&\frac{9}{17}\\&lim_{x\rightarrow5-}\frac{(9x - 45)}{(17x - 85)}=\frac{9}{17}\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim_{x\rightarrow5-}\frac{(9)}{(17)}=\frac{9}{17}\end{align*}$

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Example Question #63 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7+}-\frac{(782x - 51x^{2} - 17x^{3} + 2856)}{(252x - 162x^{2} + 18x^{3})}\end{align*}$

Possible Answers:

$\frac{2431}{630}$

$\infty$

$-\infty$

Correct answer:

$\frac{2431}{630}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(782x - 51x^{2} - 17x^{3} + 2856)}{(252x - 162x^{2} + 18x^{3})}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(17\cdot (x - 7)\cdot (x + 4)\cdot (x + 6))}{(18x\cdot (x - 7)\cdot (x - 2))}\\&\frac{(17\cdot (x + 4)\cdot (x + 6))}{(18x\cdot (x - 2))}\\&lim_{x\rightarrow7+}-\frac{(782x - 51x^{2} - 17x^{3} + 2856)}{(252x - 162x^{2} + 18x^{3})}=\frac{2431}{630}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim_{x\rightarrow7+}\frac{(102x + 51x^{2} - 782)}{(54x^{2} - 324x + 252)}=\frac{2431}{630}\end{align*}$

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Example Question #64 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow6+}\frac{(6x^{2} - 216)}{(14x - 84)}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$\frac{36}{7}$

Correct answer:

$\frac{36}{7}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=6\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(6x^{2} - 216)}{(14x - 84)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(6\cdot (x - 6)\cdot (x + 6))}{(14\cdot (x - 6))}\\&\frac{(3\cdot (x + 6))}{7}\\&lim_{x\rightarrow6+}\frac{(6x^{2} - 216)}{(14x - 84)}=\frac{36}{7}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim_{x\rightarrow6+}\frac{(12x)}{(14)}=\frac{36}{7}\end{align*}$

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Example Question #65 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}\frac{(17x)}{(77x + 11x^{2})}\end{align*}$

Possible Answers:

$\frac{17}{77}$

$-\infty$

$\infty$

Correct answer:

$\frac{17}{77}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=0\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(17x)}{(77x + 11x^{2})}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(17x)}{(11x\cdot (x + 7))}\\&\frac{17}{(11\cdot (x + 7))}\\&lim_{x\rightarrow0+}\frac{(17x)}{(77x + 11x^{2})}=\frac{17}{77}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim_{x\rightarrow0+}\frac{(17)}{(22x + 77)}=\frac{17}{77}\end{align*}$

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Example Question #66 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-8+}\frac{(6x + 48)}{(72x + 4x^{2} + 320)}\end{align*}$

Possible Answers:

$\frac{3}{4}$

$-\infty$

$\infty$

$-\frac{3}{4}$

Correct answer:

$\frac{3}{4}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(6x + 48)}{(72x + 4x^{2} + 320)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(6\cdot (x + 8))}{(4\cdot (x + 8)\cdot (x + 10))}\\&\frac{3}{(2\cdot (x + 10))}\\&lim_{x\rightarrow-8+}\frac{(6x + 48)}{(72x + 4x^{2} + 320)}=\frac{3}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-8+}\frac{(6)}{(8x + 72)}=\frac{3}{4}\end{align*}$

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Example Question #67 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow6+}\frac{(8x - 48)}{(15x - 90)}\end{align*}$

Possible Answers:

$\frac{8}{15}$

$-\frac{8}{15}$

$\infty$

$-\infty$

Correct answer:

$\frac{8}{15}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=6\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(8x - 48)}{(15x - 90)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(8\cdot (x - 6))}{(15\cdot (x - 6))}\\&\frac{8}{15}\\&lim_{x\rightarrow6+}\frac{(8x - 48)}{(15x - 90)}=\frac{8}{15}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow6+}\frac{(8)}{(15)}=\frac{8}{15}\end{align*}$

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Example Question #68 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow10+}-\frac{(105x - 15x^{2} + 450)}{(2x - 20)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{195}{2}$

Correct answer:

$\frac{195}{2}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=10\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(105x - 15x^{2} + 450)}{(2x - 20)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(15\cdot (x - 10)\cdot (x + 3))}{(2\cdot (x - 10))}\\&\frac{(15\cdot (x + 3))}{2}\\&lim_{x\rightarrow10+}-\frac{(105x - 15x^{2} + 450)}{(2x - 20)}=\frac{195}{2}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow10+}\frac{(30x - 105)}{(2)}=\frac{195}{2}\end{align*}$

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Example Question #69 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4-}-\frac{(240x - 12x^{2} - 4x^{3} + 896)}{(70x + 10x^{2} + 120)}\end{align*}$

Possible Answers:

$\frac{72}{5}$

$\infty$

$-\infty$

Correct answer:

$\frac{72}{5}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-4\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(240x - 12x^{2} - 4x^{3} + 896)}{(70x + 10x^{2} + 120)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(4\cdot (x + 4)\cdot (x + 7)\cdot (x - 8))}{(10\cdot (x + 4)\cdot (x + 3))}\\&\frac{(2\cdot (x + 7)\cdot (x - 8))}{(5\cdot (x + 3))}\\&lim_{x\rightarrow-4-}-\frac{(240x - 12x^{2} - 4x^{3} + 896)}{(70x + 10x^{2} + 120)}=\frac{72}{5}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-4-}\frac{(24x + 12x^{2} - 240)}{(20x + 70)}=\frac{72}{5}\end{align*}$

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Example Question #70 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}-\frac{(3x - 3x^{2})}{(5x)}\end{align*}$

Possible Answers:

$\infty$

$\frac{3}{5}$

$-\infty$

$-\frac{3}{5}$

Correct answer:

$-\frac{3}{5}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=0\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(3x - 3x^{2})}{(5x)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(3x\cdot (x - 1))}{(5x)}\\&\frac{(3\cdot (x - 1))}{5}\\&lim_{x\rightarrow0+}-\frac{(3x - 3x^{2})}{(5x)}=-\frac{3}{5}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0+}\frac{(6x - 3)}{(5)}=-\frac{3}{5}\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #61 : Limits

Example Question #62 : Limits

Example Question #63 : Limits

Example Question #64 : Limits

Example Question #65 : Limits

Example Question #66 : Limits

Example Question #67 : Limits

Example Question #68 : Limits

Example Question #69 : Limits

Example Question #70 : Limits

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