Proving $\lim_{x\to 0}\frac{e^x-1}{x}=1$

Question

Without using the l'hopital's rule, how can we prove that $$\lim_{x\to 0}\frac{e^x-1}{x}=1?$$

Answer 1

With the series definition of the exponential function, we compute

$$\frac{e^x-1}{x} = \sum_{m = 0}^{\infty} \frac{x^m}{(m+1)!}$$

for $x\neq 0$ and hence obtain the bound

$$\biggl\lvert\frac{e^x-1}{x}-1\biggr\rvert \leqslant \sum_{m = 1}^{\infty} \frac{\lvert x\rvert^m}{(m+1)!} \leqslant \lvert x\rvert \sum_{m = 1}^{\infty} \frac{1}{(m+1)!} = \lvert x\rvert\cdot (e-2)$$

for $0 < \lvert x\rvert \leqslant 1$.

Thus with $\delta = \min \bigl\{ 1 , \frac{\varepsilon}{e-2}\bigr\}$ we have

$$\biggl\lvert \frac{e^x-1}{x} - 1\biggr\rvert < \varepsilon$$

for $0 < \lvert x\rvert < \delta$.

Answer 2

then we have $$\frac{t}{\ln(t+1)}=\frac{1}{\frac{1}{t}\ln(t+1)}=\frac{1}{\ln(t+1)^{1/t}}$$ can you proceed?

Answer 3

Since since $\left(1+\frac{x}{k}\right)^{k+1}$ is decreasing for $x\leq 1$.

Thus we have that $$1\leq \frac{e^{x}-1}{x} \leq \frac{1}{x}\left[\left( 1+\frac{x}{k} \right)^{k+1} -1\right]$$ This implies that

$$1\leq \lim_{x\to 0}\frac{e^x-1}{x}\leq \frac{k+1}{k}$$ for all $k$.