Math Doubts

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

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

Proof

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.

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

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

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

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.

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

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

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

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 $\ln{a}$.

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

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



Follow us
Email subscription
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