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Calculus 3 : Limits

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Example Questions

Example Question #1161 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0-)}\frac{(9x^{2}y^{4} + 12xy^{5})}{(16x^{4}y^{2} + 19xy^{5} - 17x^{6} - 5y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{21}{13}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{336}{13}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{21}{260}\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{252}{13}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{21}{13}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(9(0)^{2}y^{4} + 12(0)y^{5})}{(16(0)^{4}y^{2} + 19(0)y^{5} - 17(0)^{6} - 5y^{6})}=\frac{0}{5y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{5y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(9x^{2}(x)^{4} + 12x(x)^{5})}{(16x^{4}(x)^{2} + 19x(x)^{5} - 17x^{6} - 5(x)^{6})}=\frac{21}{13}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #212 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(20x^{4}y^{2})}{(4x^{2}y^{4} + 14x^{3}y^{3} + 10xy^{5} - x^{5}y + x^{6} + y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{180}{29}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{400}{29}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{20}{29}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{10}{261}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{20}{29}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(20(0)^{4}y^{2})}{(4(0)^{2}y^{4} + 14(0)^{3}y^{3} + 10(0)y^{5} - (0)^{5}y + (0)^{6} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(20x^{4}(x)^{2})}{(4x^{2}(x)^{4} + 14x^{3}(x)^{3} + 10x(x)^{5} - x^{5}(x) + x^{6} + (x)^{6})}=-\frac{20}{29}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #213 : Limits

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

Possible Answers:

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

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{10}{9}\end{align*}$

$\begin{align*}&\text{x-axis: }-39\\&y=x:-5\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{34}{3}\\&y=x:0\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{((0)^{2}y^{3} + 11(0)^{3}y^{2} - 12(0)^{5})}{(4(0)^{5} + 3y^{5})}=\frac{0}{3y^{5}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(x^{2}(x)^{3} + 11x^{3}(x)^{2} - 12x^{5})}{(4x^{5} + 3(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #1171 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0+)}-\frac{(5x^{4}y^{2} - 20x^{6})}{(19x^{4}y^{2} - 5x^{2}y^{4} + 7x^{5}y + x^{6} + y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{15}{23}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{210}{23}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }\frac{75}{23}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{15}{299}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{15}{23}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(5(0)^{4}y^{2} - 20(0)^{6})}{(19(0)^{4}y^{2} - 5(0)^{2}y^{4} + 7(0)^{5}y + (0)^{6} + y^{6})}=\frac{0}{y^{6}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(5x^{4}(x)^{2} - 20x^{6})}{(19x^{4}(x)^{2} - 5x^{2}(x)^{4} + 7x^{5}(x) + x^{6} + (x)^{6})}=\frac{15}{23}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #215 : Limits

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

Possible Answers:

$\begin{align*}&\text{x-axis: }-\frac{17}{6}\\&y=x:3\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{49}{10}\end{align*}$

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

$\begin{align*}&\text{x-axis: }-\frac{43}{15}\\&y=x:-15\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(13(0)^{2}y^{2} - 13(0)y^{3})}{(13(0)^{4} - 19(0)^{3}y + y^{4})}=\frac{0}{y^{4}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(13x^{2}(x)^{2} - 13x(x)^{3})}{(13x^{4} - 19x^{3}(x) + (x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Example Question #1172 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0+,0+)}\frac{(28xy^{4})}{(22x^{2}y^{3} + 2xy^{4} - 5x^{4}y + x^{5} + 5y^{5})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{308}{25}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{28}{25}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{224}{25}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{28}{425}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{28}{25}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(28(0)y^{4})}{(22(0)^{2}y^{3} + 2(0)y^{4} - 5(0)^{4}y + (0)^{5} + 5y^{5})}=\frac{0}{5y^{5}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{5y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(28x(x)^{4})}{(22x^{2}(x)^{3} + 2x(x)^{4} - 5x^{4}(x) + x^{5} + 5(x)^{5})}=\frac{28}{25}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #217 : Limits

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

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{36}{7}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{27}{7}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{9}{14}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{3}{14}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{9}{14}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(11(0)^{3}y^{3} - 20(0)^{5}y)}{(4(0)^{6} - 20(0)^{4}y^{2} + 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}-\frac{(11x^{3}(x)^{3} - 20x^{5}(x))}{(4x^{6} - 20x^{4}(x)^{2} + 2(x)^{6})}=-\frac{9}{14}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #218 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}\frac{(15x^{4}y - 7x^{3}y^{2} + 18x^{5})}{(34x^{4}y + 3x^{5} + 5y^{5})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{26}{21}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{13}{21}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{169}{21}\\&y=x:0\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{13}{105}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:\frac{13}{21}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{4}y - 7(0)^{3}y^{2} + 18(0)^{5})}{(34(0)^{4}y + 3(0)^{5} + 5y^{5})}=\frac{0}{5y^{5}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{5y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(15x^{4}(x) - 7x^{3}(x)^{2} + 18x^{5})}{(34x^{4}(x) + 3x^{5} + 5(x)^{5})}=\frac{13}{21}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #1173 : Calculus 3

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0+)}-\frac{(8x^{2}y^{4})}{(19x^{2}y^{4} + 13xy^{5} + 4x^{6} - 2y^{6})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{32}{17}\\&y=x:0\end{align*}$

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

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{4}{17}\end{align*}$

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{2}{51}\end{align*}$

Correct answer:

$\begin{align*}&\text{x-axis: }0\\&y=x:-\frac{4}{17}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(8(0)^{2}y^{4})}{(19(0)^{2}y^{4} + 13(0)y^{5} + 4(0)^{6} - 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(8x^{2}(x)^{4})}{(19x^{2}(x)^{4} + 13x(x)^{5} + 4x^{6} - 2(x)^{6})}=-\frac{4}{17}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #220 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&lim_{(x,y)\rightarrow(0-,0-)}-\frac{(5xy^{3} - 17x^{3}y + 12x^{4})}{(2x^{4} + 3y^{4})}\\&\text{along the x-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{x-axis: }\frac{10}{3}\\&y=x:-\frac{41}{15}\end{align*}$

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

$\begin{align*}&\text{x-axis: }-\frac{37}{17}\\&y=x:\frac{7}{3}\end{align*}$

$\begin{align*}&\text{x-axis: }-\frac{43}{8}\\&y=x:-\frac{33}{19}\end{align*}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(5(0)y^{3} - 17(0)^{3}y + 12(0)^{4})}{(2(0)^{4} + 3y^{4})}=\frac{0}{3y^{4}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{3y^{4}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(5x(x)^{3} - 17x^{3}(x) + 12x^{4})}{(2x^{4} + 3(x)^{4})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

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Calculus 3 : Limits

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1161 : Calculus 3

Example Question #212 : Limits

Example Question #213 : Limits

Example Question #1171 : Calculus 3

Example Question #215 : Limits

Example Question #1172 : Calculus 3

Example Question #217 : Limits

Example Question #218 : Limits

Example Question #1173 : Calculus 3

Example Question #220 : Limits

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