Math Doubts

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

Formula

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

Introduction

$e$ 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.

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

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

Proof

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

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