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Question Number 159142 by tounghoungko last updated on 13/Nov/21

$$\underset{x\to -\frac{\pi }{4}}{\lim }\frac{\frac{\pi }{\sqrt{8}}-\sqrt{2}x.\tan x}{\sin x+\cos x}=?$$

Commented by cortano last updated on 14/Nov/21

$$\underset{x\to -\frac{\pi }{4}}{\lim }\frac{\frac{\pi \cos x-4x\sin x}{\sqrt{8}}}{\sqrt{2}\sin (x+\frac{\pi }{4})\cos x}$$$$=\frac{1}{2\sqrt{2}}\underset{x\to -\frac{\pi }{4}}{\lim }\frac{\pi \cos x-4x\sin x}{\sin (x+\frac{\pi }{4})}$$$$=\frac{1}{2\sqrt{2}}\underset{x\to -\frac{\pi }{4}}{\lim }\frac{-\pi \sin x-4(\sin x+x\cos x)}{\cos (x+\frac{\pi }{4})}$$$$=\frac{1}{2\sqrt{2}}\left(\frac{\frac{\pi }{\sqrt{2}}+2\sqrt{2}+\frac{\pi }{\sqrt{2}}}{1}\right)$$$$\underset{―}{=\frac{1}{2\sqrt{2}}(2\sqrt{2}+\pi \sqrt{2})=\frac{2+\pi }{2}}$$

Answered by FongXD last updated on 13/Nov/21

$$\mathrm{L}=\underset{\mathrm{x}\to -\frac{\pi }{4}}{\lim }\frac{\frac{\pi }{\sqrt{8}}(\mathrm{tanx}+1)-(\frac{\pi }{\sqrt{8}}+\sqrt{2}\mathrm{x})\mathrm{tanx}}{\sqrt{2}(\sin \frac{\pi }{4}\mathrm{sinx}+\cos \frac{\pi }{4}\mathrm{cosx})}$$$$\iff \mathrm{L}=\frac{1}{4}\underset{\mathrm{x}\to -\frac{\pi }{4}}{\lim }\frac{\frac{\pi \sin (\mathrm{x}+\frac{\pi }{4})}{\mathrm{cosxcos}\frac{\pi }{4}}-(\pi +4\mathrm{x})\mathrm{tanx}}{\cos (\mathrm{x}-\frac{\pi }{4})}$$$$\iff \mathrm{L}=\frac{1}{4}\underset{\mathrm{x}\to -\frac{\pi }{4}}{\lim }\frac{\pi \sin (\mathrm{x}+\frac{\pi }{4})-(\pi +4\mathrm{x})\mathrm{tanxcosxcos}\frac{\pi }{4}}{\sin (\mathrm{x}+\frac{\pi }{4})}\times \frac{1}{\mathrm{cosxcos}\frac{\pi }{4}}$$$$\iff \mathrm{L}=\frac{\pi }{2}-\frac{1}{2}\underset{\mathrm{x}\to -\frac{\pi }{4}}{\lim }\frac{\pi +4\mathrm{x}}{\sin (\mathrm{x}+\frac{\pi }{4})}\times \mathrm{tanxcosxcos}\frac{\pi }{4}$$$$\iff \mathrm{L}=\frac{\pi }{2}+\frac{1}{4}\underset{\mathrm{x}\to -\frac{\pi }{4}}{\lim }\frac{\pi +4\mathrm{x}}{\sin (\mathrm{x}+\frac{\pi }{4})}$$$$\mathrm{put}\mathrm{t}=\mathrm{x}+\frac{\pi }{4},\mathrm{if}\mathrm{x}\to -\frac{\pi }{4},\Rightarrow \mathrm{t}\to 0$$$$\mathrm{we}\mathrm{get}:\mathrm{L}=\frac{\pi }{2}+\frac{1}{4}\underset{\mathrm{t}\to 0}{\lim }\frac{4\mathrm{t}}{\mathrm{sint}}=\frac{\pi }{2}+\frac{1}{4}\left(4\times 1\right)=\frac{\pi }{2}+1$$$$\mathrm{so}.\overline{)\begin{array}{c}\mathrm{L}=\underset{\mathrm{x}\to -\frac{\pi }{4}}{\lim }\frac{\frac{\pi }{\sqrt{8}}-\sqrt{2}\mathrm{xtanx}}{\mathrm{sinx}+\mathrm{cosx}}=\frac{\pi }{2}+1\end{array}}$$

Answered by tounghoungko last updated on 14/Nov/21

$$=\frac{1}{2}\underset{x\to -\frac{\pi }{4}}{\lim }\frac{\pi \cos x-4x\sin x}{\sqrt{2}\sin (x+\frac{\pi }{4})}$$$$=\frac{1}{2\sqrt{2}}\underset{x\to 0}{\lim }\frac{\pi \cos (x-\frac{\pi }{4})-(4x-\pi )\sin (x-\frac{\pi }{4})}{\sin x}$$$$=\frac{1}{2\sqrt{2}}\underset{x\to 0}{\lim }\frac{(\pi \sqrt{2}-2\sqrt{2}x)\sin x+2\sqrt{2}x\cos x}{\sin x}$$$$=\frac{1}{2\sqrt{2}}\underset{x\to 0}{\lim }(\pi \sqrt{2}-2\sqrt{2}x+2\sqrt{2}\cos x)$$$$=\frac{\pi \sqrt{2}+2\sqrt{2}}{2\sqrt{2}}=\frac{\pi }{2}+1$$

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