Question

${lim}_{x\to \pi }\frac{1-sin\frac{x}{2}}{cos\frac{x}{2}(cosx\frac{x}{4}-sin\frac{x}{4})}$

Answer 1

Given, ${lim}_{x\to \pi }\frac{1-sin\frac{x}{2}}{cos\frac{x}{2}(cosx\frac{x}{4}-sin\frac{x}{4})}$

Given, ${lim}_{x\to \pi }\frac{co{s}^2\frac{x}{4}+si{n}^2-2.sin\frac{x}{4}.cos\frac{x}{4}}{cos\frac{x}{2}.(cos\frac{x}{4}-sin\frac{x}{4})}$ $[\because si{n}^2\theta +co{s}^2\theta =1si{n}^2\theta =2sin\theta cos\theta ]$

$={lim}_{x\to \pi }\frac{{(cos\frac{x}{4}-sin\frac{x}{4})}^2}{(co{s}^2\frac{x}{4}-si{n}^2\frac{x}{4})(cos\frac{x}{4}-sin\frac{x}{4})}$ $[\because co{s}^{2}2\theta =co{s}^2\theta -si{n}^2\theta ]$

$={lim}_{x\to \pi }\frac{(cos\frac{x}{4}-sin\frac{x}{4})}{(cos\frac{x}{4}+sin\frac{x}{4})(cos\frac{x}{4}-sin\frac{x}{4})}$ $[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$={lim}_{x\to \pi }\frac{1}{cos\frac{x}{4}+sin\frac{x}{4}}=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}$

$=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$