Question

${lim}_{x\to \frac{\pi }{4}}\frac{1-sin2x}{1+cos4x}$

Answer 1

${lim}_{x\to \frac{\pi }{4}}\frac{1-sin2x}{1+cos4x}$

$\Rightarrow x\to \frac{\pi }{4},x-\frac{\pi }{4}\to 0.$ let $x-\frac{\pi }{4}=y$

$\Rightarrow y\to 0$

$\Rightarrow {lim}_{x\to \frac{\pi }{4}}\frac{1-sin2x}{1+cos4x}={lim}_{y\to 0}\frac{(1-sin2(y+\frac{\pi }{4}))}{1+cos4(y+\frac{\pi }{4})}$

$={lim}_{y\to 0}\left(\frac{1-sin(\frac{\pi }{2}+2y)}{1+cos(\pi +4y)}\right)$

$={lim}_{x\to \frac{\pi }{4}\to 0}\frac{1-cos2y}{1-cos4y}$

$={lim}_{y\to 0}\frac{2si{n}^{2}y}{2si{n}^{2}2y}$

$=\frac{{lim}_{y\to 0}si{n}^{2}y}{{lim}_{y\to 0}si{n}^{2}2y}$

$=\frac{{\left({lim}_{y\to 0}\frac{siny}{y}\right)}^2\times y^2}{{\left({lim}_{y\to 0}\frac{sin2y}{2y}\right)}^2}\times 4y^2$

$=\frac{1}{1\times 4}$ $[\because {lim}_{0\to 0}\frac{sin\theta }{\theta }=1]$

$=\frac{1}{4}$