Math Doubts

$\displaystyle \lim_{x \,\to\, 0}{\dfrac{a^x-1}{x}}$ formula

Formula

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle \normalsize x}-1}{x}}$ $\,=\,$ $\log_{e}{a}$

Introduction

If $a$ is a constant and $x$ is a variable, then quotient of subtraction one from $a$ raised to the power of $x$ by $x$ is written as $\dfrac{a^x-1}{x}$ and the limit of this special exponential function is written mathematically as follows.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle \normalsize x}-1}{x}}$

The limit of this special exponential function is equal to natural logarithm of $a$. It’s used as a formula in finding the limits of exponential functions in calculus.

Other forms

This standard result in limits can be written in several ways in calculus.

$(1) \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{c^{\displaystyle \normalsize y}-1}{y}}$ $\,=\,$ $\ln{(c)}$

$(2) \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{m^{\displaystyle \normalsize u}-1}{u}}$ $\,=\,$ $\ln{(m)}$

Proof

Let’s prove the limit of quotient of difference of $a^x$ and $1$ by $x$ as $x$ approaches $0$ is equal to natural logarithm of $a$.

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved