Limit of $(1+x)^\frac{1}{x}$ when $x$ tends to $0$

Formula

$\displaystyle \large \lim_{x \,\to\, 0} \, {(1+x)}^{\huge \frac{1}{x}} \,=\, e$

Proof

$x$ is a literal number and the sum of one and $x$ is $1+x$. Then, $1+x$ whole power of the quotient of one by $x$ is written as ${(1+x)}^{\Large \frac{1}{x}}$. The limit of the function ${(1+x)}^{\Large \frac{1}{x}}$ is written in the following mathematical form as $x$ approaches zero.

$\displaystyle \large \lim_{x \,\to\, 0} \, {(1+x)}^{\huge \frac{1}{x}}$

01

Expand the function

The function can be expanded by using the expansion of the Binomial Theorem.

${(1+x)}^{\displaystyle n}$ $\,=\, 1 + \dfrac{n}{1!} x$ $+$ $\dfrac{n(n-1)}{2!} x^2$ $+$ $\dfrac{n(n-1)(n-3)}{3!} x^3 + \cdots$

There is no much difference between them but replace $n$ by $\dfrac{1}{x}$.

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

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

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

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

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

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

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

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

02

Evaluate the expansion

Substitute $x = 0$ to evaluate the function as the value of $x$ approaches zero.

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

$\,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1 \times 1}{3!} + \cdots$

$\,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!} + \cdots$

03

Evaluate the series

According to the expansion of the exponential function.

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

Put $x = 1$

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

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

$\therefore \,\,\,\,\,\, 1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!} + \cdots$ $\,=\, e$

Therefore, the value of the function ${(1+x)}^{\Large \frac{1}{x}}$ as the limit $x$ approaches zero is equal to $e$.

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