Math Doubts

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


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


$e$ is a Napier mathematical constant and $x$ is a variable. The quotient of subtraction of one from $e$ raised to the power of $x$ by $x$ is expressed mathematically as $\dfrac{e^{\displaystyle x}-1}{x}$. The limit of this special exponential function as input approaches $0$ is often appeared in calculus. Some times, it’s essential to use this standard form as formula in evaluating the limits of functions in which natural exponential functions are involved.

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

The limit of this special exponential function is equal to one.

Other forms

The standard result of this limit of exponential function can be expressed in several ways in calculus.

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

$(2) \,\,\,$ $\displaystyle \large \lim_{v \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle \normalsize v}-1}{v}}$ $\,=\,$ $1$


Let us prove the limit of quotient of difference of $e^{\displaystyle x}$ and $1$ by $x$ as $x$ approaches $0$ is equal to $1$.

Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Math Problems

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.

Learn more