Question

# 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}

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

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 $$\frac{1}{8\pi}$$