# 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.
$\underset{\left(x,y\right)\to \left(0,\pi /2\right)}{lim}\frac{1-\mathrm{cos}xy}{4{x}^{2}{y}^{3}}$
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nitruraviX
We have to solve the limit:
$\underset{\left(x,y\right)\to \left(0,\pi /2\right)}{lim}\frac{1-\mathrm{cos}xy}{4{x}^{2}{y}^{3}}$
We know the formula,
${\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=\mathrm{cos}2x$
$1-\mathrm{cos}2x=2{\mathrm{sin}}^{2}x$
for numerator,
$1-\mathrm{cos}xy=2{\mathrm{sin}}^{2}\left(\frac{xy}{2}\right)$
Therefore,
$\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}\frac{1-\mathrm{cos}xy}{4{x}^{2}{y}^{3}}=\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}\frac{2{\mathrm{sin}}^{2}\left(\frac{xy}{2}\right)}{4{x}^{2}{y}^{3}}$
We know the formula
$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$
$\underset{x\to 0}{lim}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}=1$
Multiplying and dividing by $\left(\frac{xy}{2}{\right)}^{2}$
$\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}\frac{2{\mathrm{sin}}^{2}\left(\frac{xy}{2}\right)}{4{x}^{2}{y}^{3}}=\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}\frac{2{\mathrm{sin}}^{2}\left(\frac{xy}{2}\right)}{\left(\frac{xy}{2}{\right)}^{2}}×\frac{\left(\frac{xy}{2}{\right)}^{2}}{4{x}^{2}{y}^{3}}$
$=\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}1×\frac{\left(\frac{xy}{2}{\right)}^{2}}{4{x}^{2}{y}^{3}}$
$=\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}\frac{\frac{{x}^{2}{y}^{2}}{4}}{4{x}^{2}{y}^{3}}$
$=\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}\frac{{x}^{2}{y}^{2}}{4×4{x}^{2}{y}^{3}}$
$=\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}\frac{{y}^{2}}{16{y}^{3}}$
$=\underset{\left(x,y\right)\to \left(0,\frac{\pi }{2}\right)}{lim}\frac{1}{16y}$
$=\frac{1}{16×\frac{\pi }{2}}$
$=\frac{2}{16\pi }$
$=\frac{1}{8\pi }$
Hence, value of limit is $\frac{1}{8\pi }$
Jeffrey Jordon