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

## Formula

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

### Introduction

Let $a$ be a constant and $x$ be a variable. $a$ raised to the power of $x$ forms an exponential function $a^{\displaystyle \normalsize x}$. The subtraction of one from the exponential function forms a mathematical expression $a^{\displaystyle \normalsize x}-1$. The quotient of the function in difference form by the variable $x$ is written mathematically as follows.

$\dfrac{a^{\displaystyle \normalsize x}-1}{x}$

The limit of this type of special mathematical function has to evaluate In calculus as the value of variable is closer to zero. In this case, the literal $x$ represents a variable. Therefore, the limit of this function as the value of $x$ approaches $0$ is written as follows.

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

The limit of the quotient of $x^{th}$ power of $a$ minus $1$ by $x$ as the value of $x$ tends to $0$ is equal to natural logarithm of constant $a$.

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

According to the natural logarithms, the logarithm of a to base neper constant is also simply written as natural logarithm of a.

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

This mathematical equation is used as a formula in calculus to find the limit of functions in which the exponential functions are involved.

#### Other forms

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

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

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

$(3). \,\,\,$ $\displaystyle \large \lim_{z \,\to\, 0}{\normalsize \dfrac{l^{\displaystyle \normalsize z}-1}{z}}$ $\,=\,$ $\ln{(l)}$

##### Proof

Learn how to derive the limit rule for the quotient of $a$ to the $x$th power minus $1$ by $x$ as $x$ approaches $0$.

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