Evaluate $\lim_{x\to0}\frac{\ln((1+x)^{1+x})}{x^2}-\frac1x$ $$L=\lim_{x\to0}\frac{\ln((1+x)^{1+x})}{x^2}-\frac1x$$
I am not sure of my answer. Please help me.
$$L=\lim_{x\to0}\frac{1+x}x\frac{\ln(1+x)}x-\frac1x=\lim_{x\to0}\frac{1+x-1}x=1$$

You are missing a factor of $\frac12$, so I recommend you be careful and write your steps out.
Using a MacLaurin Expansion: $\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$
$$\lim_{x \to 0} \frac{1+x}{x} \frac{\ln(1+x)}{x} - \frac1{x} = $$
$$\lim_{x \to 0} \frac{1+x}{x} \frac{x - \frac{x^2}2 + O(x^3)}{x} - \frac{1}{x} =$$
$$\lim_{x \to 0} \frac{(1+x)(1 - \frac{x}2 + O(x^2)) - 1}{x}=$$
$$\lim_{x \to 0} \frac{(1 - \frac{x}2 + O(x^2)) + (x - O(x^2)) - 1}{x}$$
$$ \lim_{x \to 0} \frac{\frac{x}2 + O(x^2)}{x} =$$
$$\lim_{x \to 0} (\frac12 + O(x)) =\frac12$$
