# Proof of Natural logarithmic Limit rule

The natural logarithmic limit rule is a most useful formula for dealing the logarithmic functions in the limits. According to this limit formula, the limit of a rational expression in the following form is equal to one.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+x)}}{x}}$ $\,=\,$ $1$

Now, let us learn how to derive the natural logarithmic limit rule in calculus mathematically in three steps.

### Expand the Logarithmic function

According to Taylor or Maclaurin series, the logarithm of $1+x$ to base $e$ can be expanded as an infinite series in terms of $x$ as follows.

$\ln{(1+x)}$ $\,=\,$ $x$ $-$ $\dfrac{x^2}{2}$ $+$ $\dfrac{x^3}{3}$ $-$ $\dfrac{x^4}{4}$ $+ \ldots$

Now, we can replace the logarithmic function by its equivalent infinite series in the rational function for evaluating its limit.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+x)}}{x}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\ldots}{x}}$

### Simplify the Rational function

There is a common factor $x$ in each term of the infinite series in the numerator. So, the common factor can be taken out from all the terms.

$= \,\,\,$ $\displaystyle \large \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\Bigg(1-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\ldots \Bigg)}{x}}$

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cancel{x}\Bigg(1-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\ldots \Bigg)}{\cancel{x}}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(1-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\ldots \Bigg)}$

### Find the limit of the function

Now, we can evaluate the limit of the infinite series as $x$ tends to zero by the direct substitution.

$= \,\,\,$ $1-\dfrac{(0)}{2}+\dfrac{{(0)}^2}{3}-\dfrac{{(0)}^3}{4}+\ldots$

$= \,\,\,$ $1-0+0-0+\ldots$

$= \,\,\,$ $1$

Therefore, it’s proved that the limit of ratio of $\ln{(1+x)}$ to $x$ as $x$ closer to zero is equal to one.

$\therefore\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\ln{(1+x)}}{x}}$ $\,=\,$ $1$

Latest Math Topics
Email subscription
Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects. Know more
Follow us on Social Media
###### 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