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.
$$L=\lim_{x\to0}\frac{1+x}x\frac{\ln(1+x)}x-\frac1x=\lim_{x\to0}\frac{1+x-1}x=1$$
 A: You are missing a factor of $\frac12$, so I recommend you be careful and write your steps out.
Using a MacLaurin Expansion: $\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$
$$\lim_{x \to 0} \frac{1+x}{x} \frac{\ln(1+x)}{x} - \frac1{x} = $$
$$\lim_{x \to 0} \frac{1+x}{x} \frac{x - \frac{x^2}2 + O(x^3)}{x} - \frac{1}{x} =$$
$$\lim_{x \to 0} \frac{(1+x)(1 - \frac{x}2 + O(x^2)) - 1}{x}=$$
$$\lim_{x \to 0} \frac{(1 - \frac{x}2 + O(x^2)) + (x - O(x^2)) - 1}{x}$$
$$ \lim_{x \to 0} \frac{\frac{x}2 + O(x^2)}{x} =$$
$$\lim_{x \to 0} (\frac12 + O(x)) =\frac12$$
A: Replacing $(\log(1+x))/x$ by $1$ is simply wrong and the mistake is committed frequently by many students. The reason that it is wrong is simply because the ratio is not equal to $1$ rather the limit of ratio is $1$ as $x\to 0$. Hence if you see something like $\lim_{x\to 0}\dfrac{\log(1+x)}{x}$ then you can replace it with $1$. To put the things in a concise manner $\dfrac{\log(1+x)}{x}$ is not the same as its limit $\lim_{x\to 0}\dfrac{\log(1+x)}{x}$.
To evaluate the limit in question you will need to make use of more advanced tools like Taylor series or L'Hospital's Rule. The expression under limit can be written as $$\frac{(1+x)\log(1+x)-x}{x^{2}}=\frac{\log(1+x)}{x}+\frac{\log(1+x)-x}{x^{2}}$$ Now the first term tends to $1$ and the second term tends to $-1/2$ (using either Taylor series or L'Hospital's Rule) so that the desired limit is $1/2$.
