Math Doubts

Limit of $\Big(1+\dfrac{1}{x}\Big)^x$ as $x$ approaches infinity

Formula

$\displaystyle \large \lim_{x \,\to\, \infty} \, {\Big(1+\dfrac{1}{x}\Big)}^{\displaystyle x} \,=\, e$

Proof

$x$ is a literal and the sum of one and reciprocal of $x$ is $1+\dfrac{1}{x}$. Then $1+\dfrac{1}{x}$ raised to the power of $x$ is expressed in mathematical form as ${\Big(1+\dfrac{1}{x}\Big)}^{\displaystyle x}$. The limit of the function is expressed in the below mathematical form when $x$ tends to infinity.

$\displaystyle \large \lim_{x \,\to\, \infty} \, {\Big(1+\dfrac{1}{x}\Big)}^{\displaystyle x}$

01

Expansion of the function

The function is in the form of Binomial Theorem. So, it can be expanded by applying the Binomial Theorem.

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

In this case, replace $x$ by $\dfrac{1}{x}$ and $n$ by $x$.

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

Apply, quotient rule of exponents to simplify this expansion.

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

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

02

Simplification of the Series

Take $x$ common from every multiplicative binomial in the numerator.

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

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

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

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

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

03

Evaluate the expansion

Substitute $x = \infty$ to get the series in simplified form when the value of $x$ tends to infinity.

$\,=\,\,\,$ $\Bigg[1 + \dfrac{1}{1!}$ $+$ $\dfrac{\Big(1-\dfrac{1}{\infty}\Big)}{2!}$ $+$ $\dfrac{\Big(1-\dfrac{1}{\infty}\Big)\Big(1-\dfrac{2}{\infty}\Big)}{3!} + \cdots \Bigg]$

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

$\,=\,\,\,$ $\Bigg[1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{(1)(1)}{3!} + \cdots \Bigg]$

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

04

Evaluate the series

The exponential function $e^{\displaystyle x}$ can be expanded in terms of $x$ mathematically as follows.

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

Substitute $x = 1$

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

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

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

Therefore, the value of the function ${\Big(1+\dfrac{1}{x}\Big)}^{\displaystyle x}$ when the limit $x$ approaches infinity is equal to $e$.



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