Math Doubts

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

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


lim of $\dfrac{a^x-1}{x}$ as x approaches 0 rule is derived in calculus naturally by considering the concepts of exponents, logarithms and exponential series.

Transformation of exponential term

$a^x$ is an exponential term and it can be expressed as a natural exponential function by the fundamental logarithmic identity.

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

Now, replace the exponential term $a^x$ as a natural exponential function in numerator of the expression.

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

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.

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

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.

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

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

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


$x\log_{e}{a}$ 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.

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

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

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


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 $\ln{a}$.

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

Therefore, it is proved that lim x approaches 0 $\dfrac{a^{x}-1}{x}$ is equal to natural logarithm of $a$.

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