Limit of $\dfrac{e^x-1}{x}$ as $x$ approaches $0$

Formula

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

Proof

$x$ is a literal number and represents a real number. The ratio of the subtraction of number $1$ from $e$ raised to the power of $x$ to the literal number $x$.

$\dfrac{e^{\displaystyle \normalsize x}-1}{x}$

The value of the function in fraction form is required to evaluate as $x$ approaches zero and it is mathematically written in the following form.

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

01

Expansion of the exponential function

According to the expansion of the exponential function.

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

Apply the expansion of the exponential function.

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

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

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

02

Simplification

The literal $x$ is a common term in each term of the numerator. So, take $x$ common from all the terms to eliminate the $x$ term from the denominator.

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

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

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

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

03

Evaluation

Substitute, $x = 0$ in this infinite series to obtain the solution of this limit of algebraic exponential function as $x$ approaches zero.

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

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

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


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