Let $f(x)=\frac{\log _e(1+x)^{1+x}-x}{x^2}$, then find the value of f(0) so that the f(x) is continuous at x = 0.

$\frac{1}{2}$

Since f(x) is continuous at x = 0, we must have

$f(0) =\lim\limits_{x \rightarrow 0} f(x)$

$=\lim\limits_{x \rightarrow 0} \frac{(1+x) \log _e(1+x)-x}{x^2}$               (form $\frac{0}{0}$)

$=\lim\limits_{x \rightarrow 0} \frac{\log _e(1+x)+1-1}{2 x}$        (Using L' Hospital Rule)

$=\frac{1}{2} \lim\limits_{x \rightarrow 0} \frac{\log _e(1+x)}{x}$

$=\frac{1}{2}$