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.



Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more