$x$ is a variable. The natural logarithm of summation of $1$ and $x$ is written as $\log_{e}{(1+x)}$ or simply $\ln{(1+x)}$. The limit of quotient of logarithm of $1+x$ to base $e$ by $x$ as $x$ approaches zero is expressed in mathematical form as follows.

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

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

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

Now, replace the logarithm of $1+x$ to base $e$ by its series.

$= \,\,\,$ $\displaystyle \lim_{x \,\to\, 0} \dfrac{x -\dfrac{x^2}{2} + \dfrac{x^3}{3} -\dfrac{x^4}{4} + \ldots}{x}$

In the denominator, there is an $x$ term. Similarly, each term in the expression in the numerator also has $x$ as a factor. So, it can be taken common from all of them for simplifying the whole function.

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

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

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

Now, find the limit of the function as $x$ approaches zero.

$= \,\,\,$ $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 $x$ approaches zero is equal to one.

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

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