The special limit $$ \lim_{x \to 0} \frac{e^x-x-1}{x^2}=\frac 1 2 $$ can be proved by Taylor expansion or with L'Hôpital's rule. Is it possoble to prove it without using derivatives?
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2Is there a reason that neither of these methods can be used? – Henry Lee Dec 02 '18 at 18:54
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1If we define $e^x$ as $\sum_{n\geq 0}\frac{x^n}{n!}$ that is trivial. – Jack D'Aurizio Dec 02 '18 at 19:07
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See https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee Dec 05 '18 at 08:45
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$$\frac{e^x-1-x}{x^2}=\int_{0}^{1}(1-y)e^{xy}\,dy $$ hence by the dominated convergence theorem $$ \lim_{x\to 0}\frac{e^x-1-x}{x^2}=\int_{0}^{1}(1-y)\,dy = \frac{1}{2}.$$
Jack D'Aurizio
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We have by $x=2y$
$$\frac{e^x-x-1}{x^2}=\frac{e^{2y}-2y-1}{4y^2}=\frac{(e^y-1)^2+2e^y-2y-2}{4y^2}=\frac14\left(\frac{e^y-1}{y}\right)^2+\frac12\frac{e^y-y-1}{y^2}$$
therefore assuming that the limit exists we have
$$L=\frac14+\frac12L \implies L=\frac12$$
Refer also to
user
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Since L'Hôpital and Taylor expansion are explicitly ruled out, you might want to mention how $(e^y-1)/y \to 1$ is obtained. – Martin R Dec 02 '18 at 21:25
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@MartinR My interpretation is that we can't use l'Hopital and Taylor to solve the limit but I think we can use the well known standard limit which, as you can know I suppose, can be derived without Hopital and Taylor. I'll add a reference to that anyway. Thanks – user Dec 02 '18 at 21:38
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Let $$f(x)=e^{\sqrt{x}}-\sqrt{x}$$
Then your limit is
$$\lim_{X\to0^+}\frac{f(X)-f(0)}{X-0}$$
hamam_Abdallah
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