Math Doubts

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.

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

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

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



Follow us
Email subscription
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
Mobile App for Android users Math Doubts Android App
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