If $\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} {\left(2-\frac{\mathrm{\lambda }}{\mathrm{x}}\right)}^{\mathrm{\lambda tan}\left(\frac{\mathrm{\pi x}}{2\mathrm{\lambda }}\right)}=\frac{1}{\mathrm{e}}$  then $\lambda$ is equal to

If $\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} {\left(2-\frac{\mathrm{\lambda }}{\mathrm{x}}\right)}^{\mathrm{\lambda tan}\left(\frac{\mathrm{\pi x}}{2\mathrm{\lambda }}\right)}=\frac{1}{\mathrm{e}}$ 

$=\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} {\left[2-\frac{\mathrm{\lambda }}{\mathrm{x}}\right]}^{\mathrm{\lambda tan}\left(\frac{\mathrm{\pi x}}{2\mathrm{\pi }}\right)}={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} \mathrm{\lambda tan}\left[\frac{\mathrm{\pi x}}{2\mathrm{\pi }}\right]\left[2-\frac{\mathrm{\lambda }}{\mathrm{x}}-1\right]}$

$={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} \mathrm{\lambda tan}\left[\frac{\mathrm{\pi x}}{2\mathrm{\lambda }}\right]\left[1-\frac{\mathrm{\lambda }}{\mathrm{x}}\right]}={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} \frac{\mathrm{\lambda }\left[1-\frac{\mathrm{\lambda }}{\mathrm{x}}\right]}{\cot \left[\frac{\mathrm{\pi x}}{2\mathrm{\pi }}\right]}}$

$={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} \frac{{\mathrm{\lambda }}^2/{\mathrm{x}}^2}{-{\mathrm{cosec}}^2\left[\frac{\mathrm{\pi x}}{2\mathrm{\pi }}\right]\cdot \frac{\mathrm{\pi }}{2\mathrm{\lambda }}}}={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} \left(\frac{{\mathrm{\lambda }}^2}{{\mathrm{x}}^2}\right)(-1)\frac{{\sin }^2\left[\frac{\mathrm{\pi x}}{2\mathrm{\lambda }}\right]}{\left(\frac{\mathrm{\pi }}{2\mathrm{\lambda }}\right)}}$

$={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} \frac{-{\mathrm{\lambda }}^2{\sin }^2\left[\frac{\mathrm{\pi x}}{2\mathrm{\lambda }}\right]}{\left(\frac{2\mathrm{\pi }}{\mathrm{\pi }}\right){\left(\frac{\mathrm{\pi }}{2\mathrm{\lambda }}\right)}^2{\mathrm{x}}^2}}={\mathrm{e}}^{{ }^{\underset{\mathrm{x}\to \mathrm{\lambda }}{lim} \left(\frac{{\displaystyle -2\mathrm{\lambda }}}{{\displaystyle \mathrm{\pi }}}\right)}\frac{{\displaystyle {\sin }^2\left[\frac{{\displaystyle \mathrm{\pi x}}}{{\displaystyle 2\mathrm{\lambda }}}\right]}}{{\displaystyle {\left(\frac{{\displaystyle \mathrm{\pi x}}}{{\displaystyle 2\mathrm{\lambda }}}\right)}^2}}}$

$={\mathrm{e}}^{\frac{-2\mathrm{\lambda }}{\mathrm{\pi }}}={\mathrm{e}}^{-1}$

$\begin{array}{l}\Rightarrow -\mathrm{\lambda }\cdot \left(\frac{2}{\mathrm{\pi }}\right)=-1\\ \therefore \mathrm{\lambda }=\frac{\mathrm{\pi }}{2}\end{array}$