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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 #1191 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}\frac{(8x^{5} - 15y^{5})}{(8x^{2}y^{3} + 3xy^{4} - 4x^{5} + 9y^{5})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{5}{21}\\&y=x:-\frac{1}{16}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{5}{3}\\&y=x:-\frac{7}{16}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{63}{8}\\&y=x:-\frac{20}{3}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{21}{8}\\&y=x:\frac{40}{3}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }-\frac{5}{3}\\&y=x:-\frac{7}{16}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(8(0)^{5} - 15y^{5})}{(8(0)^{2}y^{3} + 3(0)y^{4} - 4(0)^{5} + 9y^{5})}=\frac{15y^{5}}{-9y^{5}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{15y^{5}}{-9y^{5}}=-\frac{5}{3}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(8x^{5} - 15(x)^{5})}{(8x^{2}(x)^{3} + 3x(x)^{4} - 4x^{5} + 9(x)^{5})}=-\frac{7}{16}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #232 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(14xy^{3} + 12x^{4} + 16y^{4})}{(8x^{2}y^{2} + 19xy^{3} + 5x^{4} + y^{4})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{182}{11}\\&y=x:144\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{16}{5}\\&y=x:-\frac{14}{99}\end{align*}$

$\begin{align*}&\text{x-axis: }-16\\&y=x:-\frac{14}{11}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{252}{11}\\&y=x:-240\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }-16\\&y=x:-\frac{14}{11}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(14(0)y^{3} + 12(0)^{4} + 16y^{4})}{(8(0)^{2}y^{2} + 19(0)y^{3} + 5(0)^{4} + y^{4})}=\frac{-16y^{4}}{y^{4}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{-16y^{4}}{y^{4}}=-16\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(14x(x)^{3} + 12x^{4} + 16(x)^{4})}{(8x^{2}(x)^{2} + 19x(x)^{3} + 5x^{4} + (x)^{4})}=-\frac{14}{11}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1192 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}\frac{(14x^{3})}{(4x^{2}y + 15x^{3} + 4y^{3})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{7}{184}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{84}{23}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{14}{23}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{266}{23}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{14}{23}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(14(0)^{3})}{(4(0)^{2}y + 15(0)^{3} + 4y^{3})}=\frac{0}{4y^{3}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{4y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(14x^{3})}{(4x^{2}(x) + 15x^{3} + 4(x)^{3})}=\frac{14}{23}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #234 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}-\frac{(13x^{2}y^{4} - 18x^{3}y^{3} + 5x^{4}y^{2})}{(4x^{6} + 4y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }-28\\&y=x:-\frac{33}{7}\end{align*}$

$\begin{align*}&\text{x-axis: }-2\\&y=x:\frac{29}{7}\end{align*}$

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(13(0)^{2}y^{4} - 18(0)^{3}y^{3} + 5(0)^{4}y^{2})}{(4(0)^{6} + 4y^{6})}=\frac{0}{4y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{4y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(13x^{2}(x)^{4} - 18x^{3}(x)^{3} + 5x^{4}(x)^{2})}{(4x^{6} + 4(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #1193 : Calculus 3

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

Possible Answers:

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

$\begin{align*}&\text{x-axis: }-\frac{22}{17}\\&y=x:-\frac{27}{17}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{27}{2}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{42}{17}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #1194 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(12x^{4}y^{2} - 11xy^{5})}{(20x^{4}y^{2} + 9xy^{5} + 3x^{6} + 36y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{1}{68}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{7}{34}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{7}{34}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{1}{680}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{1}{68}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(12(0)^{4}y^{2} - 11(0)y^{5})}{(20(0)^{4}y^{2} + 9(0)y^{5} + 3(0)^{6} + 36y^{6})}=\frac{0}{36y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{36y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(12x^{4}(x)^{2} - 11x(x)^{5})}{(20x^{4}(x)^{2} + 9x(x)^{5} + 3x^{6} + 36(x)^{6})}=\frac{1}{68}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #237 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}\frac{(9x^{3})}{(16x^{2}y - 4x^{3} + y^{3})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{162}{13}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{9}{13}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{54}{13}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{9}{130}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{9}{13}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(9(0)^{3})}{(16(0)^{2}y - 4(0)^{3} + y^{3})}=\frac{0}{-y^{3}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{-y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(9x^{3})}{(16x^{2}(x) - 4x^{3} + (x)^{3})}=\frac{9}{13}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1195 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(2xy^{5} - 15x^{3}y^{3} + 13y^{6})}{(x^{6} + 5y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:-52\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{13}{70}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{104}{5}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{13}{5}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }-\frac{13}{5}\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)y^{5} - 15(0)^{3}y^{3} + 13y^{6})}{((0)^{6} + 5y^{6})}=\frac{-13y^{6}}{5y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{-13y^{6}}{5y^{6}}=-\frac{13}{5}\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(2x(x)^{5} - 15x^{3}(x)^{3} + 13(x)^{6})}{(x^{6} + 5(x)^{6})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #239 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}\frac{(6x^{2}y^{3})}{(9x^{2}y^{3} + 4xy^{4} - 5x^{4}y + 3x^{5} + y^{5})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

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

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{1}{40}\end{align*}$

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

$\begin{align*}&\text{x-axis: }\frac{15}{2}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(6(0)^{2}y^{3})}{(9(0)^{2}y^{3} + 4(0)y^{4} - 5(0)^{4}y + 3(0)^{5} + y^{5})}=\frac{0}{y^{5}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(6x^{2}(x)^{3})}{(9x^{2}(x)^{3} + 4x(x)^{4} - 5x^{4}(x) + 3x^{5} + (x)^{5})}=\frac{1}{2}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1196 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(4x^{3}y^{3} + 5x^{4}y^{2} - 9x^{5}y)}{(x^{6} + 2y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{16}{5}\\&y=x:-\frac{37}{19}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{32}{9}\\&y=x:-5\end{align*}$

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

$\begin{align*}&\text{x-axis: }-\frac{45}{19}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{3}y^{3} + 5(0)^{4}y^{2} - 9(0)^{5}y)}{((0)^{6} + 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(4x^{3}(x)^{3} + 5x^{4}(x)^{2} - 9x^{5}(x))}{(x^{6} + 2(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1191 : Calculus 3

Example Question #232 : Limits

Example Question #1192 : Calculus 3

Example Question #234 : Limits

Example Question #1193 : Calculus 3

Example Question #1194 : Calculus 3

Example Question #237 : Limits

Example Question #1195 : Calculus 3

Example Question #239 : Limits

Example Question #1196 : Calculus 3

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