Math Doubts

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

$x$ 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.

Expand the exponential function

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

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

Get the Required Mathematical form

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 $0$.

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

Evaluate Limit of Exponential function

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.

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