# 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.

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

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