For $\mathrm{x}>0,\underset{\mathrm{x}\to 0}{lim} \left[(\sin \mathrm{x}{)}^{1/\mathrm{x}}+{\left(\frac{1}{\mathrm{x}}\right)}^{\sin \mathrm{x}}\right]$ is equal to

Solution:

Here ,

$\begin{array}{r}\underset{\mathrm{x}\to 0}{lim} (\sin \mathrm{x}{)}^{1/\mathrm{x}}+\underset{\mathrm{x}\to 0}{lim} {\left(\frac{1}{\mathrm{x}}\right)}^{\sin \mathrm{x}}=0+\underset{\mathrm{x}\to 0}{lim} {\mathrm{e}}^{\log {\left(\frac{1}{\mathrm{x}}\right)}^{\sin \mathrm{x}}}\\ \left[\because \underset{\mathrm{x}\to 0}{lim} (\sin \mathrm{x}{)}^{1/\mathrm{x}}\to 0\text{ as }0<\sin \mathrm{x}<1\right]\end{array}$

       $={\mathrm{e}}^{\mathrm{x}\to 0}\frac{\log (1/\mathrm{x})}{\mathrm{cosec}\mathrm{x}}={\mathrm{e}}^{\mathrm{x}\to 0}\frac{\mathrm{x}\left(-\frac{1}{{\mathrm{x}}^2}\right)}{-\mathrm{cosec}\mathrm{xcot}\mathrm{x}}$    [using L' Hospital's rule]

       $={\mathrm{e}}^{\underset{\mathrm{x}\to 0}{lim} \frac{\sin \mathrm{x}}{\mathrm{x}}\tan \mathrm{x}}={\mathrm{e}}^0=1$