The limit of (a^x-1)/x as x approaches 0 is equal to 1. The lim_{x -> 0} (a^{x}-1)/x formula is given by

$\lim\limits_{x \to 0} \dfrac{a^x-1}{x}=\ln a$

where ln a = log_{e}a. In this post, we will learn to prove the limit of (a^{x}-1)/x when x tends to zero.

## Lim_{x → 0} (a^{x} -1)/x

Prove that lim_{x → 0} (a^{x} -1)/x = ln a. |

**Answer:**

As a^{x }= e^{x lna}, the given limit lim_{x → 0} (a^{x} -1)/x can be written as follows.

$\lim\limits_{x \to 0} \dfrac{e^{x \ln a}-1}{x}$

= $\ln a \lim\limits_{x \to 0} \dfrac{e^{x \ln a}-1}{x\ln a}$

[ Let xlna =t. Then t→0 when x→0 ]

= $\ln a \lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$

= $\ln a \times 1$ by the limit formula lim_{x→0} (e^{x}-1)/x = 1.

= ln a

So the limit of (a^{x} -1)/x is equal to lna when x→0.

More Limits:

## FAQs

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

Answer: The limit of (a^{x} -1)/x when x tends to 0 is equal to lna, that is, lim_{x→0} (a^{x} -1)/x =lna.