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

The limit of the quotient of a raised to the power of x minus 1 by x as the value of $x$ approaches $0$ is a formula in limits and its limit is equal to natural logarithm of $a$. So, let’s learn how to prove this limit rule mathematically in calculus.

### Express the function in Natural exponential form

According to fundamental law of logarithms, the $a$ raised to the $x$-th power can be written in the form of natural exponential function as follows.

$a^\normalsize x} \,=\, e^\log_{e}{\big(a^\normalsize x}\big)}$

Now, replace the exponential function in the limit function by its equivalent form in natural exponential function.

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

### Expand the exponential function

The natural exponential function can be expanded as the following power series.

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

In the same way, it can be written in terms of a variable $y$ as follows.

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

Take $y \,=\, e^\log_{e}{\big(a^\normalsize x}\big)}$ and substitute the value of $y$ in the expansion of natural exponential function.

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

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

### Calculate the value of Rational function

The value of $a$ raised to the power of $x$ is evaluated in a power series in the above step and it is time to find the value of rational function by using the power series.

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

Write each logarithmic function in factor form for our convenience.

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

Apply the power rule of logarithms to each logarithmic term in the infinite series.

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

Now, multiply the factors in each term to express their product in simple form.

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

In each term of the power series, $x$ is a common factor. So, let’s take out the factor $x$ common from all the terms.

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

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

The common factor $x$ multiplies the expression on right hand side of the equation and it divides the expression on the left-hand side of the equation.

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

### Evaluate the Limit of Rational function

In the above step, the value of the rational function is calculated as a power series and it can be replaced by the power series for finding the limit of $a$ to the $x$-th power minus $1$ by $x$ as $x$ is closer to zero.

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

Use the direct substitution method to find the limit of the power series.

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

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

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

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

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

Therefore, it is proved that the limit of $a$ raised to the power of $x$ minus $1$ by $x$ as the value of $x$ tends to $0$ is equal to natural logarithm of constant $a$. It can also be written in the following form.

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

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