How can we prove that
limx→0ex−1−xx2=12 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.
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