Question Number 190824 by cortano12 last updated on 12/Apr/23
$$\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2}\sqrt{\mathrm{tan}\:\mathrm{x}}}{\mathrm{2cos}\:\mathrm{x}−\mathrm{2sin}\:\mathrm{x}}\:=? \\ $$
Commented by 0670322918 last updated on 12/Apr/23
![(1/2)lim_(x→(π/4)) ((tan^2 (x)+1−2+2−2(√(tan(x))))/(cos(x)[1−tan(x)]))= cos((π/4))=((√2)/2) (1/( (√2)))lim_(x→(π/4)) [((tan^2 (x)−1)/(1−tan(x)))+2((1−(√(tan(x))))/(1−tan(x)))]= (1/( (√2)))[lim_(x→+(π/4)) [2((1−(√(tan(x))))/([1−(√(tan(x)))][1+(√(tan(x)))))−(([tan(x)−1][tan(x)+1])/(tan(x)−1))]= (1/( (√2)))(lim_(x→(π/4)) [(2/(1+(√(tan(x)))))−tan(x)−1])=(1/( (√2)))[1−1−1]=−((√2)/2) lim_(x→(π/4)) ((sec^2 (x)−2(√(tan(x))))/(2cos(x)−2sin(x)))=−((√2)/2)](https://www.tinkutara.com/question/Q190832.png)
$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{{tan}^{\mathrm{2}} \left({x}\right)+\mathrm{1}−\mathrm{2}+\mathrm{2}−\mathrm{2}\sqrt{{tan}\left({x}\right)}}{{cos}\left({x}\right)\left[\mathrm{1}−{tan}\left({x}\right)\right]}= \\ $$$${cos}\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\left[\frac{{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{1}}{\mathrm{1}−{tan}\left({x}\right)}+\mathrm{2}\frac{\mathrm{1}−\sqrt{{tan}\left({x}\right)}}{\mathrm{1}−{tan}\left({x}\right)}\right]= \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\underset{{x}\rightarrow+\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\left[\mathrm{2}\frac{\mathrm{1}−\sqrt{{tan}\left({x}\right)}}{\left[\mathrm{1}−\sqrt{{tan}\left({x}\right)}\right]\left[\mathrm{1}+\sqrt{{tan}\left({x}\right)}\right.}−\frac{\left[{tan}\left({x}\right)−\mathrm{1}\right]\left[{tan}\left({x}\right)+\mathrm{1}\right]}{{tan}\left({x}\right)−\mathrm{1}}\right]=\right. \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\left[\frac{\mathrm{2}}{\mathrm{1}+\sqrt{{tan}\left({x}\right)}}−{tan}\left({x}\right)−\mathrm{1}\right]\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{1}−\mathrm{1}−\mathrm{1}\right]=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{{sec}^{\mathrm{2}} \left({x}\right)−\mathrm{2}\sqrt{{tan}\left({x}\right)}}{\mathrm{2}{cos}\left({x}\right)−\mathrm{2}{sin}\left({x}\right)}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Commented by cortano12 last updated on 13/Apr/23
$$\mathrm{ans}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$