mathop {lim }limits_{alpha to pi /4} frac{{si | vedclass

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Question

(A) Let $L = \mathop {\lim }\limits_{\alpha 	o \pi /4} \frac{{\sin \alpha - \cos \alpha }}{{\alpha - \pi /4}}$.We can rewrite the numerator as:$\sin

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) Let $L = \mathop {\lim }\limits_{\alpha 	o \pi /4} \frac{{\sin \alpha - \cos \alpha }}{{\alpha - \pi /4}}$.We can rewrite the numerator as:$\sin \alpha - \cos \alpha = \sqrt{2} \left( \sin \alpha \cdot \frac{1}{\sqrt{2}} - \cos \alpha \cdot \frac{1}{\sqrt{2}} ight) = \sqrt{2} \left( \sin \alpha \cos \frac{\pi}{4} - \cos \alpha \sin \frac{\pi}{4} ight) = \sqrt{2} \sin \left( \alpha - \frac{\pi}{4} ight)$.Substituting this into the limit:$L = \mathop {\lim }\limits_{\alpha 	o \pi /4} \frac{\sqrt{2} \sin \left( \alpha - \frac{\pi}{4} ight)}{\alpha - \frac{\pi}{4}}$.Using the standard limit $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$,where $x = \alpha - \frac{\pi}{4}$:$L = \sqrt{2} 	imes 1 = \sqrt{2}$.Alternatively,using $L$-Hospital's rule:$L = \mathop {\lim }\limits_{\alpha 	o \pi /4} \frac{\frac{d}{d\alpha}(\sin \alpha - \cos \alpha)}{\frac{d}{d\alpha}(\alpha - \pi/4)} = \mathop {\lim }\limits_{\alpha 	o \pi /4} \frac{\cos \alpha + \sin \alpha}{1} = \cos \frac{\pi}{4} + \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

$\mathop {\lim }\limits_{\alpha \to \pi /4} \frac{{\sin \alpha - \cos \alpha }}{{\alpha - \frac{\pi }{4}}} = $

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