calculus - Prove this limit without using these techniques, and for beginner students: $lim_{xto0} frac{e^x-1-x}{x^2}=frac12$

How can we prove that

$$\lim_{x\rightarrow 0}\cfrac{e^x-1-x}{x^2}=\cfrac{1}{2}$$ Without using L'hopital rule, and Taylor expansions?

Thanks

Answer

Use the limit laws and the binomial theorem: you have $$\frac{e^x - (1+x)}{x^2} = \lim_{n \to \infty} \left( \frac{ (1+ \frac xn)^n - (1+x)}{x^2} \right) = \lim_{n \to \infty} \sum_{k=2}^n \binom nk \frac{x^{k-2}}{n^k} \\ = \frac 12 + x \left( \lim_{n \to \infty} \sum_{k=3}^n \binom nk \frac{x^{k-3}}{n^k} \right)$$
provided that the limit $ \displaystyle e^x = \lim_{n \to \infty} \left(1 + \frac xn \right)^n$ is assumed to exist.

As a by-product of this computation you get that $$\lim_{n \to \infty} \sum_{k=3}^n \binom nk \frac{x^{k-3}}{n^k}$$ exists too. With $x=1$ this implies $$\sup_n \sum_{k=3}^n \binom nk \frac{1}{n^k} < \infty$$
and consequently if $|x| \le 1$ then $$\sup_{n} \left| \sum_{k=3}^n \binom nk \frac{x^{k-3}}{n^k} \right| \le \sup_{n} \sum_{k=3}^n \binom nk \frac{|x^{k-3}|}{n^k} \le \sup_n \sum_{k=3}^n \binom nk \frac{1}{n^k} < \infty.$$

So, if $|x| \le 1$ then $$\left| \frac{e^x - (1+x)}{x^2} - \frac 12 \right| \le |x| \sup_n \sum_{k=3}^n \binom nk \frac{1}{n^k}.$$

Now let $x \to 0$.