lim _{x rightarrow pi} frac{1-sin (x/2)}{left | vedclass

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Question

(C) Let $x = \pi + h$. As $x \rightarrow \pi$,$h \rightarrow 0$. The expression becomes $\lim _{h \rightarrow 0} \frac{1-\sin (\frac{\pi+h}{2})}{\cos

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $x = \pi + h$. As $x \rightarrow \pi$,$h \rightarrow 0$. The expression becomes $\lim _{h \rightarrow 0} \frac{1-\sin (\frac{\pi+h}{2})}{\cos (\frac{\pi+h}{2}) (\cos (\frac{\pi+h}{4})-\sin (\frac{\pi+h}{4}))}$. Using trigonometric identities,$\sin (\frac{\pi}{2} + \frac{h}{2}) = \cos (\frac{h}{2})$ and $\cos (\frac{\pi}{2} + \frac{h}{2}) = -\sin (\frac{h}{2})$. The expression simplifies to $\lim _{h \rightarrow 0} \frac{1-\cos (h/2)}{-\sin (h/2) (\cos (\frac{\pi}{4} + \frac{h}{4}) - \sin (\frac{\pi}{4} + \frac{h}{4}))}$. Since $\cos (\frac{\pi}{4} + \frac{h}{4}) - \sin (\frac{\pi}{4} + \frac{h}{4}) = \sqrt{2} \sin (\frac{\pi}{4} - (\frac{\pi}{4} + \frac{h}{4})) = \sqrt{2} \sin (-h/4) = -\sqrt{2} \sin (h/4)$. Substituting this back: $\lim _{h \rightarrow 0} \frac{2 \sin^2 (h/4)}{-\sin (h/2) \cdot (-\sqrt{2} \sin (h/4))} = \lim _{h \rightarrow 0} \frac{2 \sin (h/4)}{\sqrt{2} \sin (h/2)}$. Using $\sin (h/2) = 2 \sin (h/4) \cos (h/4)$,we get $\lim _{h \rightarrow 0} \frac{2 \sin (h/4)}{\sqrt{2} \cdot 2 \sin (h/4) \cos (h/4)} = \lim _{h \rightarrow 0} \frac{1}{\sqrt{2} \cos (h/4)} = \frac{1}{\sqrt{2} \cdot 1} = \frac{1}{\sqrt{2}}$.

$\lim _{x \rightarrow \pi} \frac{1-\sin (x/2)}{\left(\cos \frac{x}{2}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)} =$

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