# Limit of $\dfrac{a^x-1}{x}$ as $x$ approaches $0$

## Formula

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

### Proof

$a$ and $x$ are two literal numbers but $a$ represents a constant and $x$ represents a variable. The ratio of subtraction of one from the $a$ raised to the power of $x$ to literal number $x$ is written as follows.

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

The value of this function is requisite when the value of $x$ tends to zero. It is expressed in limit by the calculus in mathematical form in the following way.

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

01

### Transformation of exponential term

ax is an exponential term and it can be transformed in another form by the relation between exponentiation and logarithms. It can be done by applying the fundamental logarithmic rule.

According to fundamental logarithmic identity, the exponential term $a^x$ can be written alternatively in the following form.

$e^{\displaystyle \log_{e}{a^{\displaystyle \normalsize x}}} = a^{\displaystyle \normalsize x}$

Replace the exponential term ${\displaystyle \normalsize x}$ by its equivalent value in the limit of the function.

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

02

### Expansion of the exponential function

According to the expansion of the exponential function.

$e^{\displaystyle x}$ $\,=\,$ $1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots$

In this case $\log_{e}{a^{\displaystyle x}}$ is the exponent of the mathematical constant $e$. So, replace the literal $x$ by $\log_{e}{a^{\displaystyle x}}$ in the expansion.

$\implies e^{\displaystyle \log_{e}{a^{\displaystyle \normalsize x}}}$ $\,=\,$ $1+\dfrac{\log_{e}{a^{\displaystyle x}}}{1!}$ $+$ $\dfrac{{(\log_{e}{a^{\displaystyle x}})}^2}{2!}$ $+$ $\dfrac{{(\log_{e}{a^{\displaystyle x}})}^3}{3!}+\cdots$

03

### Simplification of the exponential function

The expansion can be simplified by using the logarithmic fundamentals.

$e^{\displaystyle \log_{e}{a^{\displaystyle \normalsize x}}}$ $\,=\,$ $1+\dfrac{\log_{e}{a^{\displaystyle x}}}{1!}$ $+$ $\dfrac{{(\log_{e}{a^{\displaystyle x}})}^2}{2!}$ $+$ $\dfrac{{(\log_{e}{a^{\displaystyle x}})}^3}{3!}+\cdots$

$\implies e^{\displaystyle \log_{e}{a^{\displaystyle \normalsize x}}}$ $\,=\,$ $1+\dfrac{\log_{e}{a^{\displaystyle x}}}{1!}$ $+$ $\dfrac{(\log_{e}{a^{\displaystyle x}}) \times (\log_{e}{a^{\displaystyle x}})}{2!}$ $+$ $\dfrac{(\log_{e}{a^{\displaystyle x}}) \times (\log_{e}{a^{\displaystyle x}}) \times (\log_{e}{a^{\displaystyle x}})}{3!}+\cdots$

Apply the power rule of logarithms to each logarithmic term.

$\implies e^{\displaystyle \log_{e}{a^{\displaystyle \normalsize x}}}$ $\,=\,$ $1+\dfrac{x\log_{e}{a}}{1!}$ $+$ $\dfrac{(x\log_{e}{a}) \times (x\log_{e}{a})}{2!}$ $+$ $\dfrac{(x\log_{e}{a}) \times (x\log_{e}{a}) \times (x\log_{e}{a})}{3!}+\cdots$

$\implies e^{\displaystyle \log_{e}{a^{\displaystyle \normalsize x}}}$ $\,=\,$ $1+\dfrac{x\log_{e}{a}}{1!}$ $+$ $\dfrac{x^2 {(\log_{e}{a})}^2}{2!}$ $+$ $\dfrac{x^3 {(\log_{e}{a})}^3}{3!}+\cdots$

Comeback to the derivation and replace the by its simplified expansion.

$\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{e^{\displaystyle \log_{e}{a^{\displaystyle \normalsize x}}}-1}{x}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1+\dfrac{x\log_{e}{a}}{1!} + \dfrac{x^2 {(\log_{e}{a})}^2}{2!} + \dfrac{x^3 {(\log_{e}{a})}^3}{3!}+\cdots-1}{x}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\cancel{1}+\dfrac{x\log_{e}{a}}{1!} + \dfrac{x^2 {(\log_{e}{a})}^2}{2!} + \dfrac{x^3 {(\log_{e}{a})}^3}{3!}+\cdots-\cancel{1}}{x}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\dfrac{x\log_{e}{a}}{1!} + \dfrac{x^2 {(\log_{e}{a})}^2}{2!} + \dfrac{x^3 {(\log_{e}{a})}^3}{3!}+\cdots}{x}$

04

### Simplification

$x\log_{e}{a}$ is a common term in each term of the expansion in numerator. So, take it common from them and it makes both $x$ terms get cancelled mathematically.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{x\log_{e}{a} \Bigg(\dfrac{1}{1!} + \dfrac{x\log_{e}{a}}{2!} + \dfrac{x^2 {(\log_{e}{a})}^2}{3!}+\cdots\Bigg)}{x}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\cancel{x}\log_{e}{a} \Bigg(\dfrac{1}{1!} + \dfrac{x\log_{e}{a}}{2!} + \dfrac{x^2 {(\log_{e}{a})}^2}{3!}+\cdots\Bigg)}{\cancel{x}}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \log_{e}{a} \Bigg(\dfrac{1}{1!} + \dfrac{x\log_{e}{a}}{2!} + \dfrac{x^2 {(\log_{e}{a})}^2}{3!}+\cdots\Bigg)$

05

### Evaluation

Evaluate the infinite series by substituting $x = 0$.

$=\,\,\,$ $(\log_{e}{a}) \Bigg(1 + \dfrac{(0)\log_{e}{a}}{2!} + \dfrac{{(0)}^2 {(\log_{e}{a})}^2}{3!}+\cdots\Bigg)$

$=\,\,\,$ $(\log_{e}{a}) \Bigg(1 + \dfrac{0}{2!} + \dfrac{0}{3!}+\cdots\Bigg)$

$=\,\,\,$ $(\log_{e}{a})(1+0+0+\cdots)$

$=\,\,\,$ $(\log_{e}{a})(1)$

$=\,\,\,$ $\log_{e}{a}$

It is simply written as $\ln{a}$.