$x$ is a variable and corresponding natural exponential function is $e^{\displaystyle x}$. The quotient of subtraction of $1$ from $e$ raised to the power of $x$ by $x$ as $x$ approaches $0$ is often appeared while finding the limits of exponential functions. So, this standard result in limits is used as a formula in calculus.

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

The limit of this special exponential function as its input approaches zero is equal to one. Let’s prove this rule before using it as a formula in calculus.

According to expansion of natural exponential function, the function $e^{\displaystyle x}$ can be expanded as follows.

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

Try to transform the expansion of natural exponential function as our required mathematical form.

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

$x$ is a common factor in each term of the infinite series and it can be taken common from all the terms.

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

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

Now, evaluate the limit of the exponential function as $x$ approaches $0$.

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

The limit of the exponential function $\dfrac{e^{\displaystyle x}-1}{x}$ as $x$ approaches $0$ can be evaluated from its equivalent infinite series function.

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

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

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

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

This standard result expresses that the limit of ratio of difference of natural exponential function and one to variable as the input approaches zero is always equal to one.

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

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.