The limit of (e^{x}-1)/x, when x tends to zero, is equal to 1. That is, we have

lim_{x→0} (e^{x}-1)/x = 1.

In this post, we will learn to find the limit of the function (e^{x}-1)/x when x approaches to zero.

## Limit of (e^x-1)/x as x→0

**Prove that** lim_{x→0} $\dfrac{e^x-1}{x}$ = 1.

**Proof:**

Note that when we put x=0 in the function (e^{x}-1)/x, we get 0/0, that is, the numerator and denominator both approach 0 as x tends to 0. So the given limit is an indeterminate form.

By l’Hopital’s rule, we have:

lim_{x→0} $\dfrac{e^x-1}{x}$

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

= lim_{x→0} $\dfrac{\frac{d}{dx}(e^x)-\frac{d}{dx}(1)}{\frac{d}{dx}(x)}$ by the difference rule of derivatives.

= lim_{x→0} $\dfrac{e^x-0}{1}$ as the derivative of e^{x }is e^{x} and the derivative of x^{n }is nx^{n-1} (power rule of derivatives).

= lim_{x→0} $e^x$

= e^{0}

= 1.

This proves that the limit of (e^{x}-1)/x, when x goes to 0, is equal to 1. ♣

For the list of all limit formulae, click on the page `List of Limit Formulas`.

**ALSO READ:**

Limit of (1+$\frac{1}{n}$)^{n} when n→∞

**Question:** Evaluate lim_{x→0} $\dfrac{e^{2x}-1}{x}$

*Solution:*

lim_{x→0} $\dfrac{e^{2x}-1}{x}$

= lim_{x→0} $(\dfrac{e^{2x}-1}{2x} \times 2)$

= 2 lim_{x→0} $\dfrac{e^{2x}-1}{2x}$

= 2 lim_{t→0} $\dfrac{e^{t}-1}{t}$ where t=2x (so that t→0 when x→0)

= 2 × 1, by the above formula.

= 2

So the limit of (e^{2x}-1)/x, when x tends to 0, is equal to 2.

You can read: Sum rule of limits: Proof

Quotient rule of limits: Proof

## FAQs

**Q1: What is the limit of (e ^{x}-1)/x when x tends to 0?**

Answer: The limit of (e^{x}-1)/x is equal to 1, when x tends to 0