Proving a second order special limit without derivatives

Question

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?

Answer 1

Let $$f(x)=e^{\sqrt{x}}-\sqrt{x}$$

Then your limit is

$$\lim_{X\to0^+}\frac{f(X)-f(0)}{X-0}$$

Answer 2

$$\frac{e^x-1-x}{x^2}=\int_{0}^{1}(1-y)e^{xy}\,dy $$ hence by the dominated convergence theorem $$ \lim_{x\to 0}\frac{e^x-1-x}{x^2}=\int_{0}^{1}(1-y)\,dy = \frac{1}{2}.$$

Answer 3

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

• Are all limits solvable without L'Hôpital Rule or Series Expansion