TITLE: Doubt on limits evaluation.
QUESTION [8 upvotes]: Can anyone tell me what am I doing wrong here.
The limit provided is$$\lim_{x \to 0}\dfrac{xe^x-\ln(1+x)}{x²}$$
Method 1 (using standard limits)
$$= \ \ \ \  \dfrac{\displaystyle \lim_{x \to 0}\dfrac{xe^x}{x}-\lim_{x \to 0}\dfrac{\ln(1+x)}{x}}{\displaystyle \lim_{x \to 0}x}$$
$$= \ \ \ \  \dfrac{\displaystyle \lim_{x \to 0}e^x-\lim_{x \to 0} 1}{\displaystyle \lim_{x \to 0}x}$$
$$= \ \ \ \  \lim_{x \to 0} \dfrac{e^x-1}{x}$$
$$= \ \ \ \ 1$$

Method 2 (using  Maclaurin series )
$$\lim_{x \to 0}\dfrac{xe^x-\ln(1+x)}{x²}$$
$$= \ \ \ \ \dfrac{3}{2}$$
Even with L'Hopital rule I get $\dfrac{3}{2}$. Then what's wrong with method 1.
Some limit's properties

REPLY [10 votes]: The error is quite simply described.

Theorem. If $\lim_{x\to a}f(x)=l$ and $\lim_{x\to a}g(x)=m$ and $m\ne0$, then
  $$
\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{l}{m}
$$

This is a correct statement. But you cannot apply it when not all the hypotheses are satisfied.
In your case, $\lim_{x\to0}x=0$, so the theorem cannot be applied.
Why can't it? Because division by $0$ is not allowed.