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Question Number 175160 by thean last updated on 21/Aug/22

Commented by Alisajadrajaee last updated on 21/Aug/22

$$\underset{x\to \frac{\pi }{4}}{\lim }\left(\frac{\sqrt{2}cosx-1}{tanx-1}\right)=\frac{0}{0}$$$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{{(\sqrt{2}cosx-1)}^{{}^{\prime }}}{{(tanx-1)}^{{}^{\prime }}}=\frac{\sqrt{2}(-sinx)}{{sec}^{2}x}$$$$\underset{x\to \frac{\pi }{4}}{\lim }-\frac{\sqrt{2}\left(sin\frac{\pi }{4}\right)}{{sec}^2\frac{\pi }{4}}=-\frac{\sqrt{2}\left(\frac{\sqrt{2}}{2}\right)}{{\left(\sqrt{2}\right)}^2}=-\frac{1}{2}$$$$$$

Answered by cortano1 last updated on 21/Aug/22

$$L=\underset{x\to \frac{\pi }{4}}{\lim }\left(\frac{\sqrt{2}\cos x-1}{\tan x-1}\right)$$$$L=\underset{x\to \frac{\pi }{4}}{\lim }\left(\frac{\sqrt{2}(\cos x-\cos \frac{\pi }{4})}{\left(\frac{\sin (x-\frac{\pi }{4})}{\cos \frac{\pi }{4}\cos x}\right)}\right)$$$$L=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}.\frac{1}{\sqrt{2}}.\cos x(-2\sin \left(\frac{x+\frac{\pi }{4}}{2}\right)\sin \left(\frac{x-\frac{\pi }{4}}{2}\right))}{\sin (x-\frac{\pi }{4})}$$$$L=-\frac{\sqrt{2}}{2}.\frac{\sqrt{2}}{2}.\underset{x\to \frac{\pi }{4}}{\lim }\frac{2\sin \frac{1}{2}(x-\frac{\pi }{4})}{\sin (x-\frac{\pi }{4})}$$$$L=-\frac{1}{4}.2=\frac{-1}{2}$$

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