real analysis - How do i evaluate this limit :$lim_{x to 0} frac{e^x-x-1}{x²}$ without using Taylor expansion

I would like to evaluate this limit without using Taylor expansion:

$$\lim_{x \to 0} \frac{e^x-x-1}{x^2}$$ .

Note: by Taylor Expansion i have got :$\frac{1}{2}$ .

Thank u for any help .!!!!

Answer

METHODOLOGY $1$: Use L'Hospital's Rule Successively

Repeated use of L'Hospital's Rule reveals

$$\begin{align}
\lim_{x\to 0}\frac{e^x-x-1}{x^2}&=\lim_{x\to 0}\frac{e^x-1}{2x}\\\\
&=\lim_{x\to 0}\frac12 e^x=\frac12
\end{align}$$

---

METHODOLOGY $2$: Integral representation of the numerator

Note that we can write the numerator as

$$\begin{align}
e^x-x-1&=\int_0^x \int_0^t e^s \,ds\,dt\\\\
&=\int_0^x \int_s^x e^s\,dt\,ds\\\\
&=\int_0^x (x-s)e^s\,ds
\end{align}$$

Now, we can use the Mean-Value-Theorem for integrals to reveal

$$\begin{align}

e^x-x-1&=e^{s^*(x)}\int_0^x(x-s)\,ds\\\\
&=\frac12 x^2e^{s^*(x)}
\end{align}$$

for some value of $s^*(x) \in (0,x)$.

Finally, exploiting the continuity of the exponential function yield the coveted limit

$$\begin{align}
\lim_{x\to 0}\frac{e^x-x-1}{x^2}&=\lim_{x\to 0}\frac{\frac12 x^2e^{s^*(x)}}{x^2}\\\\

&=\frac12
\end{align}$$

as expected!