Evaluating $\lim_{x\to 0} \left( \frac{e^x - 1 }{x} \right) $ without L'Hopital's rule

Question

I have to solve this limit:

$$\lim\limits_{x\to 0} \left( \dfrac{e^x - 1 }{x} \right) $$

I can easily do it by using hospital's rule, numerator becomes $e^0$ and denominator becomes $1$.

How should I solve that limit by using definition of $e$ as $e= \lim\limits_{n \to \infty } {\left( {1 + \frac 1n } \right)}^n$

Answer 1

Method 1: Defintion of derivative Hint:

$$\lim_{x\to 0}\frac{e^x-e^0}{x-0}.$$

This is nothing but $\frac{d}{dx}e^x|_{x=0}$.

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Method 2:Taylor Series

You could also plug in the definition of $e^x$ as a Taylor series and collect all higher order terms as $O(x^2)$: $e^x=1+x+O(x^2)$.

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Method 3: Euler

The last method was the method Euler used, which is not rigorous but still very nice.

Look at $$\lim_{x\to 0}\frac{a^x-1}{x}$$, Euler first showed convergence (not sure if he even cared about it :D) and then said that we want to choose $a$ such that the limit is $1$. So approximately $$\frac{a^x-1}{x}\approx 1\implies a^x\approx 1+x \implies a \approx (1+x)^{1/x}.$$ Now, substitute $n=1/x$, hence $n\to \infty$ and we get $$a\approx \lim_{n\to \infty}\left(1+\frac{1}{n} \right)^n=e$$

Answer 2

You can see that:

$$\frac{e^x-1}{x}=\frac{e^x-e^0}{x-0}.$$

So since $x\mapsto e^x$ is differentiable,

$$\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}\frac{e^x-e^0}{x-0}=\lim_{x\to 0}\frac{d}{dx} (e^x)=\lim_{x\to 0}e^x=1.$$

Answer 3

Here is an attempt to use only integers.

To show $\lim_{x \to 0} \dfrac{e^x-1}{x} = 0 $ using $e = \lim_{n \to \infty} \left(1+\frac1{n}\right)^n $.

From this, $e^x = \lim_{n \to \infty} \left(1+\frac1{n}\right)^{nx} $.

This is a double limit, which is often problematical.

For general $x$, this seems to require the generalized binomial theorem. To avoid this, I will assume $x = \frac1{m}$ as integer $m \to \infty$, and let $n=km$ as integer $k \to \infty$.

Then $e^{1/m} = \lim_{k \to \infty} \left(1+\frac1{mk}\right)^{k} $. Now we can use the binomial theorem.

$\begin{array}\\ \left(1+\frac1{mk}\right)^{k} &=\sum_{j=0}^k \binom{k}{j}\frac1{(mk)^j}\\ &=\sum_{j=0}^k \frac{\prod_{i=0}^{j-1}(k-i)}{j!}\frac1{(mk)^j}\\ \text{so}\\ f(m, k) &=\dfrac{\left(1+\frac1{mk}\right)^{k}-1}{1/m}\\ &=m\sum_{j=1}^k \frac{\prod_{i=0}^{j-1}(k-i)}{j!}\frac1{(mk)^j}\\ &\le m\sum_{j=1}^k \frac{1}{j!}\frac1{m^j}\\ &= \sum_{j=1}^k \frac{1}{j!}\frac1{m^{j-1}}\\ &=1+ \sum_{j=2}^k \frac{1}{j!}\frac1{m^{j-1}}\\ &\le 1+ \frac1{m}\sum_{j=2}^k \frac{1}{j!}\frac1{m^{j-2}}\\ &= 1+ \frac1{m}\sum_{j=0}^{k-2} \frac{1}{(j+2)!}\frac1{m^j}\\ \end{array} $

Since $\sum_{j=0}^{k-2} \frac{1}{(j+2)!}\frac1{m^j}$ is easily shown to be bounded by $1$ (actually, by $\sum_{j=0}^{k-2} \frac{1}{(j+2)!} = e-2$), $f(m, k) \le 1+\frac1{m}$.

Since $f(m, k) \ge 1$ (from its first term), $\lim_{m \to \infty} f(m, k) = 1$.