Question

Evaluate ${lim}_{x\to 0}\frac{{(1+x)}^{\frac{1}{x}}-e+\frac{1}{2}ex}{x^2}$

Answer 1

We have ${lim}_{x\to 0}{(1+x)}^{\frac{1}{x}}=e$

Let $L={lim}_{x\to 0}{(1+x)}^{\frac{1}{x}}$

Taking $\ln$ both sides, we get

$\ln L=\ln \left[{lim}_{x\to 0}{(1+x)}^{\frac{1}{x}}\right]$

$={lim}_{x\to 0}\ln {(1+x)}^{\frac{1}{x}}$

$={lim}_{x\to 0}\frac{\ln (1+x)}{x}$

Now, we use L'Hospital's Rule,

$\ln L={lim}_{x\to 0}\frac{\frac{1}{1+x}}{1}=1$

$\therefore L={lim}_{x\to 0}{(1+x)}^{\frac{1}{x}}=e$

so,${lim}_{x\to 0}\frac{{(1+x)}^{\frac{1}{x}}-e+\frac{1}{2}ex}{x^2}$

$={lim}_{x\to 0}\frac{e-e+\frac{1}{2}ex}{x^2}$

$={lim}_{x\to 0}\frac{e}{2x}=\infty$