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

The limit of mathematical constant $e$ raised to the power of $x$ minus one divided by $x$ as the value of $x$ approaches $0$ is an exponential limit rule in calculus.

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

Let us learn how to prove the limit of $e$ raised to the power of $x$ minus one divided by $x$ as the value of $x$ approaches zero is equal to one in mathematics.

Expand the Natural exponential function

According to the Taylor series, the natural exponential function in $x$ can be expanded as the power series in terms of a variable $x$.

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

Transforming expansion into Required form

Firstly, subtract the number one from the expressions on both sides of the mathematical equation.

$\implies$ $e^{\displaystyle x}-1$ $\,=\,$ $\bigg(1$ $+$ $\dfrac{x}{1!}$ $+$ $\dfrac{x^2}{2!}$ $+$ $\dfrac{x^3}{3!}$ $+$ $\cdots\bigg)$ $-$ $1$

It is time to simplify the mathematical expressions on both sides of the equation. There is nothing to simplify on left hand side of the equation. So, let us focus on simplifying the expression on the right hand side of the equation.

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

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

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

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

Look at the expression in numerator of each term in the infinite series on the right hand side of the equation. A variable $x$ commonly appears in the expression of numerator in every term and it can be taken out common from all the terms for simplifying the series further.

$\implies$ $e^{\displaystyle x}-1$ $\,=\,$ $x \times \bigg(\dfrac{1}{1!}$ $+$ $\dfrac{x}{2!}$ $+$ $\dfrac{x^2}{3!}$ $+$ $\cdots\bigg)$

Now, divide the expressions on both sides of the equation by a variable $x$ to take a major step and it is useful to us in obtaining the function in required form.

$\implies$ $\dfrac{e^{\displaystyle x}-1}{x}$ $\,=\,$ $\dfrac{x}{x} \times \bigg(\dfrac{1}{1!}$ $+$ $\dfrac{x}{2!}$ $+$ $\dfrac{x^2}{3!}$ $+$ $\cdots\bigg)$

$\implies$ $\dfrac{e^{\displaystyle x}-1}{x}$ $\,=\,$ $\dfrac{\cancel{x}}{\cancel{x}} \times \bigg(\dfrac{1}{1!}$ $+$ $\dfrac{x}{2!}$ $+$ $\dfrac{x^2}{3!}$ $+$ $\cdots\bigg)$

$\implies$ $\dfrac{e^{\displaystyle x}-1}{x}$ $\,=\,$ $1 \times \bigg(\dfrac{1}{1!}$ $+$ $\dfrac{x}{2!}$ $+$ $\dfrac{x^2}{3!}$ $+$ $\cdots\bigg)$

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

The natural exponential function is successfully transformed into our required rational function form. Now, let’s concentrate on finding the limit as $x$ tends to zero.

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

Find the Limit by the Direct substitution

The above mathematical equation expresses that the limit of $e$ raised to the power of $x$ minus $1$ divided by $x$ can be evaluated by finding the limit of infinite power series as the value of $x$ approaches zero, and it can be done by the direct substitution method.

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

$\implies$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^{\displaystyle x}-1}{x}}$ $\,=\,$ $\dfrac{1}{1}$ $+$ $\dfrac{0}{2}$ $+$ $\dfrac{0}{6}$ $+$ $\cdots$

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

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

Therefore, it is derived that the limit of Napier constant $e$ raised to the power of $x$ minus one divided by $x$ is equal to one as the value of $x$ approaches $0$. Now, it can be used as a formula to find the limit of a rational function in which the natural exponential function is involved.

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