$x$ is a variable and corresponding natural exponential function is $e^{\displaystyle x}$. The subtraction of one from natural exponential function is $e^{\displaystyle x}-1$. The ratio of subtraction of one from natural exponential function to variable forms a special exponential function.

The limit of this special exponential function as $x$ approaches zero is written mathematically as follows.

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

The natural exponential function can be expanded in terms of variable by the expansion of natural exponential function.

$e^{\displaystyle x}$ $\,=\,$ $1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots$

So, replace the natural exponential function by its expansion and it helps us to simplify this special form exponential function.

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

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

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

The variable $x$ is a common factor in each term of expression in numerator. So, take $x$ common from all of them because the same factor is appearing in the denominator. Obviously, they both get cancelled each other mathematically.

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

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

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

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

Substitute $x$ is equal to zero to obtain the limit of the quotient of subtraction of one from natural exponential function by variable as the variable approaches zero.

$=\,\,\,$ $1+\dfrac{0}{2!}+\dfrac{{(0)}^2}{3!}+\cdots$

$=\,\,\,$ $1+0+0+\cdots$

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

Therefore, it is proved that the limit of quotient of subtraction of one from natural exponential function ($e^x$) by variable $x$ as $x$ tends to $0$ is equal to one.

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

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