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

lim_(x→−(π/4)) (((π/( (√8))) −(√2) x. tan x)/(sin x+cos x)) =?

$$\:\underset{{x}\rightarrow−\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\frac{\pi}{\:\sqrt{\mathrm{8}}}\:−\sqrt{\mathrm{2}}\:{x}.\:\mathrm{tan}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:=? \\ $$

Commented by cortano last updated on 14/Nov/21

lim_(x→−(π/4)) (((π cos x−4x sin x)/( (√8)))/( (√2) sin (x+(π/4))cos x)) = (1/( 2(√2))) lim_(x→−(π/4)) ((πcos x−4x sin x)/(sin (x+(π/4)))) = (1/( 2(√2))) lim_(x→−(π/4)) ((−πsin x−4(sin x+xcos x))/(cos (x+(π/4)))) =(1/(2(√2)))((((π/( (√2)))+2(√2)+(π/( (√2))))/1)) =(1/(2(√2)))(2(√2)+π(√2))=((2+π)/2)

$$\:\underset{{x}\rightarrow−\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\frac{\pi\:\mathrm{cos}\:{x}−\mathrm{4}{x}\:\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{8}}}}{\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}\:{x}} \\ $$$$=\:\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}\:\underset{{x}\rightarrow−\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\pi\mathrm{cos}\:{x}−\mathrm{4}{x}\:\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}\:\underset{{x}\rightarrow−\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{−\pi\mathrm{sin}\:{x}−\mathrm{4}\left(\mathrm{sin}\:{x}+{x}\mathrm{cos}\:{x}\right)}{\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\frac{\frac{\pi}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\sqrt{\mathrm{2}}+\frac{\pi}{\:\sqrt{\mathrm{2}}}}{\mathrm{1}}\right) \\ $$$$\underline{=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{2}\sqrt{\mathrm{2}}+\pi\sqrt{\mathrm{2}}\right)=\frac{\mathrm{2}+\pi}{\mathrm{2}}\:} \\ $$

Answered by FongXD last updated on 13/Nov/21

L=lim_(x→−(π/4)) (((π/( (√8)))(tanx+1)−((π/( (√8)))+(√2)x)tanx)/( (√2)(sin(π/4)sinx+cos(π/4)cosx))) ⇔ L=(1/4)lim_(x→−(π/4)) ((((πsin(x+(π/4)))/(cosxcos(π/4)))−(π+4x)tanx)/(cos(x−(π/4)))) ⇔ L=(1/4)lim_(x→−(π/4)) ((πsin(x+(π/4))−(π+4x)tanxcosxcos(π/4))/(sin(x+(π/4))))×(1/(cosxcos(π/4))) ⇔ L=(π/2)−(1/2)lim_(x→−(π/4)) ((π+4x)/(sin(x+(π/4))))×tanxcosxcos(π/4) ⇔ L=(π/2)+(1/4)lim_(x→−(π/4)) ((π+4x)/(sin(x+(π/4)))) put t=x+(π/4), if x→−(π/4), ⇒ t→0 we get: L=(π/2)+(1/4)lim_(t→0) ((4t)/(sint))=(π/2)+(1/4)(4×1)=(π/2)+1 so. determinant (((L=lim_(x→−(π/4)) (((π/( (√8)))−(√2)xtanx)/(sinx+cosx))=(π/2)+1)))

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

Answered by tounghoungko last updated on 14/Nov/21

= (1/2) lim_(x→−(π/4)) ((π cos x−4x sin x)/( (√2) sin (x+(π/4)))) =(1/(2(√2))) lim_(x→0) ((π cos (x−(π/4))−(4x−π)sin (x−(π/4)))/(sin x)) =(1/(2(√2))) lim_(x→0) (((π(√2)−2(√2)x)sin x+2(√2)x cos x)/(sin x)) = (1/(2(√2))) lim_(x→0) (π(√2)−2(√2)x+2(√2) cos x) = ((π(√2) +2(√2))/(2(√2))) = (π/2)+1

$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{x}\rightarrow−\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\pi\:\mathrm{cos}\:{x}−\mathrm{4}{x}\:\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\pi\:\mathrm{cos}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)−\left(\mathrm{4}{x}−\pi\right)\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{sin}\:{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\pi\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{2}}{x}\right)\mathrm{sin}\:{x}+\mathrm{2}\sqrt{\mathrm{2}}{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\pi\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{2}}{x}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cos}\:{x}\right) \\ $$$$=\:\frac{\pi\sqrt{\mathrm{2}}\:+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{1}\: \\ $$

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