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

Question

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 1

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

Repeated use of L'Hospital's Rule reveals

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

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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!