# How do you solve the following limit (e^x-1)/x as x approaches zero?

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68
May 9, 2015

It is a remarkable limit, but, if you want to demonstrate it, you have to know the fundamental limit:

${\lim}_{x \rightarrow \infty} {\left(1 + \frac{1}{x}\right)}^{x} = e$ (number of Neper), and also this limit:

${\lim}_{x \rightarrow 0} {\left(1 + x\right)}^{\frac{1}{x}} = e$ that it is easy to demonstrate in this way:

let $x = \frac{1}{t}$, so when $x \rightarrow 0$ than $t \rightarrow \infty$ and this limit becomes the first one.

So:

let ${e}^{x} - 1 = t \Rightarrow {e}^{x} = t + 1 \Rightarrow x = \ln \left(t + 1\right)$

and if $x \rightarrow 0 \Rightarrow t \rightarrow 0$

${\lim}_{x \rightarrow 0} \frac{{e}^{x} - 1}{x} = {\lim}_{t \rightarrow 0} \frac{t}{\ln} \left(t + 1\right) = {\lim}_{t \rightarrow 0} \frac{1}{\ln \frac{t + 1}{t}} =$

$= {\lim}_{t \rightarrow 0} \frac{1}{\frac{1}{t} \ln \left(t + 1\right)} = {\lim}_{t \rightarrow 0} \frac{1}{\ln {\left(t + 1\right)}^{\frac{1}{t}}} =$

(for the second limit)$= \frac{1}{\ln e} = \frac{1}{1} = 1$.

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25
mason m Share
Apr 9, 2017

The limit is in the indeterminate form $\frac{{e}^{0} - 1}{0} = \frac{0}{0}$, so l'Hopital's rule applies:

${\lim}_{x \rightarrow 0} \frac{{e}^{x} - 1}{x} = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} \left({e}^{x} - 1\right)}{\frac{d}{\mathrm{dx}} \left(x\right)} = {\lim}_{x \rightarrow 0} {e}^{x} / 1 = {e}^{0} = 1$

Then teach the underlying concepts
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#### Explanation

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23
Sep 12, 2017

${\lim}_{n \setminus \to 0} \frac{{e}^{x} - 1}{x} = 1$

#### Explanation:

We have,

$f \left(x\right) = \frac{{e}^{x} - 1}{x}$

The series expansion of ${e}^{x}$ is,

e^x = \sum_(n=1)^(\infty) x^n/(n!) = 1 + x +x^2/(2!) + x^3/(3!) + x^4/(4!) + \cdots

Hence,

f(x) = (e^x - 1)/x = ((1 + x +x^2/(2!) + x^3/(3!) + x^4/(4!) + \cdots)-1)/x

color(crimson)\Rightarrow = 1 +x/(2!) + x^2/(3!) + x^3/(4!) + \cdots

Applying limit,

lim_(x \to 0) f(x) = lim_(x \to 0) (1 +x/(2!) + x^2/(3!) + x^3/(4!) + \cdots) = 1

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