Use the method of your choice to evaluate the following limits. lim_{(x,y)rightarrow(0,pi/2)}frac{1-cos xy}{4x^2y^3}

Use the method of your choice to evaluate the following limits. lim_{(x,y)rightarrow(0,pi/2)}frac{1-cos xy}{4x^2y^3}

Question
Limits and continuity
asked 2020-11-27
Use the method of your choice to evaluate the following limits.
\(\lim_{(x,y)\rightarrow(0,\pi/2)}\frac{1-\cos xy}{4x^2y^3}\)

Answers (1)

2020-11-28
We have to solve the limit:
\(\lim_{(x,y)\rightarrow(0,\pi/2)}\frac{1-\cos xy}{4x^2y^3}\)
We know the formula,
\(\cos^2x-\sin^2x=\cos2x\)
\(1-\cos2x=2\sin^2x\)
for numerator,
\(1-\cos xy=2\sin^2(\frac{xy}{2})\)
Therefore,
\(\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}\frac{1-\cos xy}{4x^2y^3}=\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}\frac{2\sin^2(\frac{xy}{2})}{4x^2y^3}\)
We know the formula
\(\lim_{x\rightarrow0}\frac{\sin x}{x}=1\)
\(\lim_{x\rightarrow0}\frac{\sin^2x}{x^2}=1\)
Multiplying and dividing by \((\frac{xy}{2})^2\)
\(\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}\frac{2\sin^2(\frac{xy}{2})}{4x^2y^3}=\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}\frac{2\sin^2(\frac{xy}{2})}{(\frac{xy}{2})^2}\times\frac{(\frac{xy}{2})^2}{4x^2y^3}\)
\(=\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}1\times\frac{(\frac{xy}{2})^2}{4x^2y^3}\)
\(=\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}\frac{\frac{x^2y^2}{4}}{4x^2y^3}\)
\(=\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}\frac{{x^2y^2}}{4\times4x^2y^3}\)
\(=\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}\frac{{y^2}}{16y^3}\)
\(=\lim_{(x,y)\rightarrow(0,\frac{\pi}{2})}\frac{{1}}{16y}\)
\(=\frac{{1}}{16\times\frac{\pi}{2}}\)
\(=\frac{{2}}{16\pi}\)
\(=\frac{1}{8\pi}\)
Hence, value of limit is \(\frac{1}{8\pi}\)
0

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