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Calculus 3 : Limits

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Example Questions

Example Question #311 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}-\frac{(13x^{3}y - 6x^{2}y^{2} + 14y^{4})}{(13xy^{3} - x^{4} + 18y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{7}{99}\\&y=x:-\frac{7}{170}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{77}{10}\\&y=x:-\frac{70}{9}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{7}{9}\\&y=x:-\frac{7}{10}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{119}{10}\\&y=x:\frac{49}{9}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{7}{9}\\&y=x:-\frac{7}{10}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(13(0)^{3}y - 6(0)^{2}y^{2} + 14y^{4})}{(13(0)y^{3} - (0)^{4} + 18y^{4})}=\frac{14y^{4}}{-18y^{4}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{14y^{4}}{-18y^{4}}=-\frac{7}{9}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(13x^{3}(x) - 6x^{2}(x)^{2} + 14(x)^{4})}{(13x(x)^{3} - x^{4} + 18(x)^{4})}=-\frac{7}{10}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #312 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}-\frac{(4x^{3}y^{2} + 3xy^{4} - 7x^{5})}{(3x^{5} + 2y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{48}{19}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-\frac{38}{5}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-11\\&y=x:-\frac{7}{2}\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{3}y^{2} + 3(0)y^{4} - 7(0)^{5})}{(3(0)^{5} + 2y^{5})}=\frac{0}{2y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{2y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}-\frac{(4x^{3}(x)^{2} + 3x(x)^{4} - 7x^{5})}{(3x^{5} + 2(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #313 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}\frac{(17xy^{2})}{(5x^{2}y + 18x^{3} + 3y^{3})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{17}{26}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{17}{2}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{17}{234}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{119}{13}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{17}{26}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(17(0)y^{2})}{(5(0)^{2}y + 18(0)^{3} + 3y^{3})}=\frac{0}{3y^{3}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(17x(x)^{2})}{(5x^{2}(x) + 18x^{3} + 3(x)^{3})}=\frac{17}{26}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #314 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{1}{25}\end{align*}$

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

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{33}{5}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{48}{5}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

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

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Example Question #315 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{13}{15}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{26}{3}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{143}{15}\\&y=x:0\end{align*}$

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

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{13}{15}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(13(0)^{5})}{(5(0)y^{4} + 13(0)^{5} - 3y^{5})}=\frac{0}{3y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(13x^{5})}{(5x(x)^{4} + 13x^{5} - 3(x)^{5})}=\frac{13}{15}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #316 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }-\frac{9}{2}\\&y=x:-\frac{10}{3}\end{align*}$

$\begin{align*}&\text{y-axis: }2\\&y=x:-\frac{21}{8}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{29}{9}\\&y=x:0\end{align*}$

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(11(0)^{3}y^{2} - 11(0)^{4}y)}{(3(0)^{5} + y^{5})}=\frac{0}{y^{5}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(11x^{3}(x)^{2} - 11x^{4}(x))}{(3x^{5} + (x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #317 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}\frac{(4x^{3}y^{2} + 7x^{4}y + 13x^{5})}{(16x^{2}y^{3} + x^{4}y - 2x^{5} - 3y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

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

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{2}{13}\end{align*}$

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

$\begin{align*}&\text{y-axis: }-18\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(4(0)^{3}y^{2} + 7(0)^{4}y + 13(0)^{5})}{(16(0)^{2}y^{3} + (0)^{4}y - 2(0)^{5} - 3y^{5})}=\frac{0}{3y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(4x^{3}(x)^{2} + 7x^{4}(x) + 13x^{5})}{(16x^{2}(x)^{3} + x^{4}(x) - 2x^{5} - 3(x)^{5})}=2\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #318 : Limits

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

Possible Answers:

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

$\begin{align*}&\text{y-axis: }0\\&y=x:65\end{align*}$

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

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

Correct answer:

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

Explanation:

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

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Example Question #319 : Limits

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

Possible Answers:

$\begin{align*}&\text{y-axis: }12\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:228\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{12}{11}\end{align*}$

$\begin{align*}&\text{y-axis: }-240\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }12\\&y=x:0\end{align*}$

Explanation:

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

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Example Question #320 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0-)}\frac{(15x^{5})}{(18x^{3}y^{2} + 8x^{4}y + 4x^{5} + 3y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{5}{88}\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{45}{11}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{5}{11}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{45}{11}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{5}{11}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{5})}{(18(0)^{3}y^{2} + 8(0)^{4}y + 4(0)^{5} + 3y^{5})}=\frac{0}{3y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(15x^{5})}{(18x^{3}(x)^{2} + 8x^{4}(x) + 4x^{5} + 3(x)^{5})}=\frac{5}{11}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Calculus 3 : Limits

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #311 : Limits

Example Question #312 : Limits

Example Question #313 : Limits

Example Question #314 : Limits

Example Question #315 : Limits

Example Question #316 : Limits

Example Question #317 : Limits

Example Question #318 : Limits

Example Question #319 : Limits

Example Question #320 : Limits

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