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

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    A proof would depend on the definition of $e^x$. What is yours? – 2017-01-20
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    set $t=e^x-1$ to solve the problem – 2017-01-20
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    What is your definition of $e$? Because acutally this could be used as the defintion of e. – 2017-01-20
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    What is your definition of $e^x$? Is it $$e^x := \sum_{k = 0}^{\infty} \frac{x^k}{k!}$$ or is it something else? The correct argument depends on which definition is in use. – 2017-01-20
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    @Daniel Fischer Yes, using Maclaurin series that should be. – 2017-01-20
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    Then we have $$\frac{e^x-1}{x} = \sum_{m = 0}^{\infty} \frac{x^m}{(m+1)!}$$ and $$\biggl\lvert \frac{e^x-1}{x} - 1\biggr\rvert \leqslant \sum_{m = 1}^{\infty} \frac{\lvert x\rvert^m}{(m+1)!}$$ for $x\neq 0$. If $0 < \lvert x\rvert \leqslant 1$, we can estimate the bound with $$\sum_{m = 1}^{\infty} \frac{\lvert x\rvert^m}{(m+1)!} \leqslant \lvert x\rvert \sum_{m = 1}^{\infty} \frac{1}{(m+1)!} = \lvert x\rvert\cdot (e-2).$$ – 2017-01-20

3 Answers 3

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

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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 $k$.

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    This is a nice trick. Thank you very much. – 2017-01-20
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    its not exactly epsilon delta, but it can be modified to be one by first selecting $\frac{1}{k}<\epsilon$. – 2017-01-20
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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?

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    Sonnhard Graubner Yes, I can proceed. I got it. Thanks for the answer. – 2017-01-20