Math Doubts

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

$x$ is a variable and $a$ is a constant. The $a$ raised to the power of $x$ is written as $a^x$ in mathematics. The limit of quotient of subtraction of $1$ from $x$-th power of $a$ by $x$ as $x$ approaches $0$ is equal to natural logarithm of $a$ and this standard result can be derived in calculus.

Express Exponential function as Natural Exponential function

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

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

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

Expand the Natural exponential function

According to the expansion of the natural 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 for the mathematical constant $e$. So, replace the variable $x$ by the log function $\log_{e}{a^{\displaystyle x}}$ in the expansion of natural exponential function.

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

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

Simplify the expansion of Exponential function

Let’s try to simplify this infinite series of expansion of natural logarithmic function.

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

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

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

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

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

$x$ is a common factor in each term of infinite series and it can be taken common from all of them.

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

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

Evaluate Limit of Exponential function

Now, the limit of this exponential function in algebraic form can be evaluated when $x$ approaches zero.

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

Now, evaluate the limit of exponential function $\dfrac{a^{\displaystyle x}-1}{x}$ as $x$ approaches $0$ by finding the limit of the infinite series as $x$ approaches $0$.

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

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

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

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

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

The limit of this exponential function can be written simply in the following mathematical form by the natural logarithmic system.

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

Therefore, it’s proved that the limit of ratio of difference of $x$-th power of $a$ and $1$ to $x$ as $x$ tends to $0$ is equal to natural logarithm of $a$. It’s often used as a formula while evaluating limits of exponential functions.

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