Does this limit exist?

The Limit is in the Indeterminate Form $\frac{0}{0}$

Applying L'Hopital's Rule Yields :

$\lim_{x\to 0}\frac{(1+x)^x-1}{x^2}$

= $\frac{[(1+x)^x]'}{2x}$

= $\lim_{x\to 0}\frac{(1+x)^x\ln (1+x)+x(1+x)^{x-1}}{2x}$

= $\lim_{x\to 0}(1+x)^x\cdot\frac{1}{2}\cdot\frac{\ln (1+x)}{x}+\lim_{x\to 0}\frac{(1+x)^{x-1}}{2}$

= $1\cdot\frac{1}{2}\cdot 1+\frac{1}{2}=1$

Thus Answer Is $\boxed{1}$

$\cdots$ $\cdots$ $\cdots$ $\cdots$ $\cdots$

Note That

$[(1+x)^x]' = [e^{x\ln (1+x)}]'$

= $e^{x\ln (1+x)}\cdot [x\ln (1+x)]'$

= $(1+x)^x\left(\ln (1+x)+\frac{x}{x+1}\right)$

= $(1+x)^x\ln (1+x)+(1+x)^{x-1}x$

$(1+x)^x-1=e^{\ln(1+x)x}-1=e^{x^2+o(x^2)}-1=x^2+o(x^2)$ so that the limit is $\boxed{1}$

We are using the Taylor series $\ln(1+x)=x+o(x)$ and $e^x=1+x+o(x)$

Here we say that a function $f(x)$ is $o(x^k)$ if $\lim_{x\to 0}\frac{f(x)}{x^k}=0$

(1+x)n =1+nx if x is very small ∴ (1+x)x =1+x2 ∴lim x->0 ((1+x)x -1)/x2 =x2/x2=1