Question Number 48581 by Kiran bendkoli last updated on 25/Nov/18
![f(x)=[tan( π/4+x)^(1/x) ] =k is conntinous at x=0 then k=?](Q48581.png)
$${f}\left({x}\right)=\left[\mathrm{tan}\left(\:\pi/\mathrm{4}+{x}\right)^{\mathrm{1}/{x}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{k} \\ $$$${is}\:{con}\mathrm{ntinous}\:\mathrm{at}\:\mathrm{x}=\mathrm{0}\:\mathrm{then}\:\mathrm{k}=? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18
![i think f(x)=[tan((π/4)+x)]^(1/x) f(x)=[((1+x)/(1−x))]^(1/x) ln{f(x)}=((ln(1+x)−ln(1−x))/x) ln{f(x)}=((ln(1+x))/x)+((ln(1−x))/(−x)) lim_(x→0) ln{f(x)}=lim_(x→0) {((ln(1+x))/x)+((ln(1−x))/(−x))} =1+1=2 so k=e^2](Q48587.png)
$${i}\:{think}\:{f}\left({x}\right)=\left[{tan}\left(\frac{\pi}{\mathrm{4}}+{x}\right)\right]^{\frac{\mathrm{1}}{{x}}} \\ $$$${f}\left({x}\right)=\left[\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right]^{\frac{\mathrm{1}}{{x}}} \\ $$$${ln}\left\{{f}\left({x}\right)\right\}=\frac{{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)}{{x}} \\ $$$${ln}\left\{{f}\left({x}\right)\right\}=\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}+\frac{{ln}\left(\mathrm{1}−{x}\right)}{−{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{ln}\left\{{f}\left({x}\right)\right\}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}+\frac{{ln}\left(\mathrm{1}−{x}\right)}{−{x}}\right\} \\ $$$$\:\:\:\:\:=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$${so}\:{k}={e}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$