Limit of (e^x – 1)/x as x approaches 0

The Mathematician — Sun, 02 Oct 2022 13:00:48 +0000

The limit of

\frac{e^x - 1}{x}

x

approaches 0 is equal to 1. This is easy to see with the Maclaurin series.

Proof. If we apply the Bernoulli inequality and let

n \rightarrow \infty

, then we get

\begin{align*} 1 + x \leq e^x. \end{align*}

Now, for

\lvert x \rvert

1

, we get that

1 - x \leq e^{-x}

, which implies that

e^x \leq \frac{1}{1-x}

. So we have the following:

\begin{align*} 1 + x \leq e^x \leq \frac{1}{1-x} \end{align*}

Subtract both sides with

-1

and divide by

x

. Then we get:

\begin{align*} 1 \leq \frac{e^x - 1}{x} \leq \frac{1}{1-x} \end{align*}

Now, by the squeeze theory, we see that, indeed, the limit of

\frac{e^x - 1}{x}

x

approaches 0 is equal to 1.

Proof. This proof is more of circular reasoning, but still valuable to see how easy it works for in the future. We want to determine the next limit:

\begin{align*} \lim_{x \rightarrow 0} \frac{e^x - 1}{x} \end{align*}

We know that the Maclaurin series of

e^x

\begin{align*} e^x &= \sum_{n = 0}^{\infty} \frac{x^n}{n!} \\ &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \end{align*}

Further, we will subtract the Maclaurin series with 1:

\begin{align*} e^x - 1 &= \sum_{n = 1}^{\infty} \frac{x^n}{n!} \\ &= x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \end{align*}

Now we divide it out with

x

\begin{align*} \frac{e^x - 1}{x} &= \sum_{n = 1}^{\infty} \frac{x^{n-1}}{n!} \\ &= 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots \end{align*}

Wrapping up together, we get:

\begin{align*} \lim_{x \rightarrow 0} \frac{e^x - 1}{x} &= \lim_{x \rightarrow 0} \sum_{n = 1}^{\infty} \frac{x^{n-1}}{n!} \\ &= \lim_{x \rightarrow 0} (1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots) \\ &= 1 + \lim_{x \rightarrow 0} (\frac{x}{2!} + \frac{x^2}{3!} + \cdots) \\ &= 1 + 0 + 0 + 0 + \cdots \\ &= 1. \end{align*}

So, in conclusion, we have that

\begin{align*} \lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1 \end{align*}

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Limit of (e^x – 1)/x as x approaches 0