A limit without using L'hoptial's rule

Question

Define the natural number $e$ by $e=\lim_{x\to 0} (1+x)^{1/x}$.

Then, I can prove $\lim_{x\to 0} \frac{e^x-1}{x}=1$.

Let $z=e^x-1$. Then, $x=\ln(z+1)$ and $$\lim_{x\to 0} \frac{e^x-1}{x}=\lim_{z\to 0} \frac{z}{\ln(z+1)}=\frac{1}{\ln e}=1\text{.}$$

Using a similar trick (without L'Hoptial's rule), can we prove $\lim_{x \to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$?

Answer 1

$e^x= 1+x+\frac{x^2}{2}+\frac{x^3}{3!} + \cdots$

$\frac{e^x-1-x}{x^2} = \frac{1}{2} + \frac{x}{3!} + \frac{x^2}{4!}+ \cdots$

$\lim\limits_{x \to 0}\frac{e^x-1-x}{x^2} = \frac{1}{2}$