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

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

limx→0 (ex-1)/x = 1.

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

Limit of (e^x-1)/x when x tends to 0

Limit of (e^x-1)/x as x→0

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

Proof:

Note that when we put x=0 in the function (ex-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:

limx→0 $\dfrac{e^x-1}{x}$

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

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

= limx→0 $\dfrac{e^x-0}{1}$ as the derivative of ex is ex and the derivative of xn is nxn-1 (power rule of derivatives).

= limx→0 $e^x$

= e0

= 1.

This proves that the limit of (ex-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`.

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Limit of x/sinx when x→0

Limit of sin(1/x) when x→0

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

Solution:

limx→0 $\dfrac{e^{2x}-1}{x}$

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

= 2 limx→0 $\dfrac{e^{2x}-1}{2x}$

= 2 limt→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 (e2x-1)/x, when x tends to 0, is equal to 2.

You can read: Sum rule of limits: Proof

Product rule of limits: Proof

Quotient rule of limits: Proof

FAQs

Q1: What is the limit of (ex-1)/x when x tends to 0?

Answer: The limit of (ex-1)/x is equal to 1, when x tends to 0

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