<s>[INST] Finding the error in an evaluation of the limit $\lim_{x\to0} \frac{e^x-x-1}{x^2} $ \begin{align}
\lim_{x\to0} \frac{e^x-x-1}{x^2}
&= \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} \\
&= \lim_{x\to0} \frac{e^x-1}{x}\lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\ &= \lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\
&= 0
\end{align}
 [/INST] The error starts  here:
$$\lim_{x\to0} \frac{e^x-x-1}{x^2} $$ $$\color{red}{\neq\lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} 
=\dots}$$
See Basic Limit Laws 

Hint: Applaying L'Hopital twice gives $\frac 1 2 \lim\limits_{x\to 0} e^x$
</s>