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Calculus 3 : Calculus 3

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Example Questions

Example Question #1231 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(9x^{3}y^{3} + x^{5}y)}{(19xy^{5} + 3x^{6} + 18y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{9}{2}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{1}{2}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{1}{4}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{1}{24}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{1}{4}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(9(0)^{3}y^{3} + (0)^{5}y)}{(19(0)y^{5} + 3(0)^{6} + 18y^{6})}=\frac{0}{-18y^{6}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{-18y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(9x^{3}(x)^{3} + x^{5}(x))}{(19x(x)^{5} + 3x^{6} + 18(x)^{6})}=-\frac{1}{4}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1232 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(10x^{3} - 13xy^{2} + 3y^{3})}{(5x^{3} + 5y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{3}{5}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{54}{5}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{1}{10}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{21}{5}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{3}{5}\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(10(0)^{3} - 13(0)y^{2} + 3y^{3})}{(5(0)^{3} + 5y^{3})}=\frac{-3y^{3}}{5y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{-3y^{3}}{5y^{3}}=-\frac{3}{5}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(10x^{3} - 13x(x)^{2} + 3(x)^{3})}{(5x^{3} + 5(x)^{3})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1233 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(6x^{3} - 19xy^{2} + 13y^{3})}{(2x^{3} + 4y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{13}{4}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{13}{60}\end{align*}$

$\begin{align*}&\text{y-axis: }13\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{91}{2}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{13}{4}\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(6(0)^{3} - 19(0)y^{2} + 13y^{3})}{(2(0)^{3} + 4y^{3})}=\frac{-13y^{3}}{4y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{-13y^{3}}{4y^{3}}=-\frac{13}{4}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(6x^{3} - 19x(x)^{2} + 13(x)^{3})}{(2x^{3} + 4(x)^{3})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1234 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}-\frac{(xy^{4} - x^{4}y)}{(x^{5} + 4y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{20}{19}\\&y=x:\frac{41}{17}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{27}{16}\\&y=x:-\frac{15}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{45}{7}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{((0)y^{4} - (0)^{4}y)}{((0)^{5} + 4y^{5})}=\frac{0}{4y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{4y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(x(x)^{4} - x^{4}(x))}{(x^{5} + 4(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #1235 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}\frac{(3xy^{2} + 17x^{2}y)}{(3x^{3} - 4y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-380\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }220\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-20\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{5}{4}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-20\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(3(0)y^{2} + 17(0)^{2}y)}{(3(0)^{3} - 4y^{3})}=\frac{0}{-4y^{3}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{-4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(3x(x)^{2} + 17x^{2}(x))}{(3x^{3} - 4(x)^{3})}=-20\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1236 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}\frac{(14x^{3})}{(2xy^{2} - 4x^{3} + 4y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:7\end{align*}$

$\begin{align*}&\text{y-axis: }70\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-14\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{7}{13}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:7\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(14(0)^{3})}{(2(0)y^{2} - 4(0)^{3} + 4y^{3})}=\frac{0}{4y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(14x^{3})}{(2x(x)^{2} - 4x^{3} + 4(x)^{3})}=7\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1237 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(17x^{3}y^{2} + 3x^{4}y - 20y^{5})}{(3x^{5} + 4y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-90\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:45\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{5}{7}\end{align*}$

$\begin{align*}&\text{y-axis: }5\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }5\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(17(0)^{3}y^{2} + 3(0)^{4}y - 20y^{5})}{(3(0)^{5} + 4y^{5})}=\frac{20y^{5}}{4y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{20y^{5}}{4y^{5}}=5\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(17x^{3}(x)^{2} + 3x^{4}(x) - 20(x)^{5})}{(3x^{5} + 4(x)^{5})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1238 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0-)}-\frac{(9xy^{5} - 2x^{6})}{(x^{6} - x^{5}y + 8y^{6})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-7\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{7}{128}\end{align*}$

$\begin{align*}&\text{y-axis: }7\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{7}{8}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{7}{8}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(9(0)y^{5} - 2(0)^{6})}{((0)^{6} - (0)^{5}y + 8y^{6})}=\frac{0}{8y^{6}}\\&\lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{8y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(9x(x)^{5} - 2x^{6})}{(x^{6} - x^{5}(x) + 8(x)^{6})}=-\frac{7}{8}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1239 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(3xy^{3} - 13x^{3}y + 10x^{4})}{(5x^{4} + 4y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{45}{13}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{33}{5}\\&y=x:-\frac{11}{4}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{46}{9}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(3(0)y^{3} - 13(0)^{3}y + 10(0)^{4})}{(5(0)^{4} + 4y^{4})}=\frac{0}{4y^{4}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{4y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(3x(x)^{3} - 13x^{3}(x) + 10x^{4})}{(5x^{4} + 4(x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #1231 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(2xy^{2} + 2x^{2}y - 4x^{3})}{(3x^{3} + 5y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }\frac{50}{19}\\&y=x:-\frac{24}{19}\end{align*}$

$\begin{align*}&\text{y-axis: }-18\\&y=x:-\frac{29}{18}\end{align*}$

$\begin{align*}&\text{y-axis: }-45\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)y^{2} + 2(0)^{2}y - 4(0)^{3})}{(3(0)^{3} + 5y^{3})}=\frac{0}{5y^{3}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{5y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(2x(x)^{2} + 2x^{2}(x) - 4x^{3})}{(3x^{3} + 5(x)^{3})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1231 : Calculus 3

Example Question #1232 : Calculus 3

Example Question #1233 : Calculus 3

Example Question #1234 : Calculus 3

Example Question #1235 : Calculus 3

Example Question #1236 : Calculus 3

Example Question #1237 : Calculus 3

Example Question #1238 : Calculus 3

Example Question #1239 : Calculus 3

Example Question #1231 : Calculus 3

All Calculus 3 Resources

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