The limit of $1/x$-th power of $1+x$ as $x$ approaches $0$ is a standard result in limits and it is used as a rule to evaluate the limits of algebraic functions which are exponential form. So, let’s us first prove this in calculus to use it as a formula in mathematics.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$
The exponential function in algebraic form is in the form of Binomial Theorem. So, it can be expanded infinitely on the basis of this 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 $\dfrac{1}{x}$.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{\Bigg(\dfrac{1}{x}\Bigg)}{1!}{x}$ $+$ $\dfrac{\Bigg(\dfrac{1}{x}\Bigg)\Bigg(\dfrac{1}{x}-1\Bigg)}{2!}{x^2}$ $+$ $\dfrac{\Bigg(\dfrac{1}{x}\Bigg)\Bigg(\dfrac{1}{x}-1\Bigg)\Bigg(\dfrac{1}{x}-2\Bigg)}{3!}{x^3} + \cdots \Bigg]$
Now, simplify each term in this series and it helps us to evaluate the limit of this exponential function in the next few steps.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{\Bigg(\dfrac{1}{x}\Bigg)}{1!}{x}$ $+$ $\dfrac{\Bigg(\dfrac{1}{x}\Bigg)\Bigg(\dfrac{1-x}{x}\Bigg)}{2!}{x^2}$ $+$ $\dfrac{\Bigg(\dfrac{1}{x}\Bigg)\Bigg(\dfrac{1-x}{x}\Bigg)\Bigg(\dfrac{1-2x}{x}\Bigg)}{3!}{x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{\Bigg(\dfrac{1}{x}\Bigg)}{1!}{x}$ $+$ $\dfrac{\Bigg(\dfrac{1 \times (1-x)}{x^2}\Bigg)}{2!}{x^2}$ $+$ $\dfrac{\Bigg(\dfrac{1 \times (1-x) \times (1-2x)}{x^3}\Bigg)}{3!}{x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{\Bigg(\dfrac{1}{x}\Bigg)}{1!}{x}$ $+$ $\dfrac{\Bigg(\dfrac{1-x}{x^2}\Bigg)}{2!}{x^2}$ $+$ $\dfrac{\Bigg(\dfrac{(1-x)(1-2x)}{x^3}\Bigg)}{3!}{x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{1}{1! \times x}{x}$ $+$ $\dfrac{1-x}{2! \times x^2}{x^2}$ $+$ $\dfrac{(1-x)(1-2x)}{3! \times x^3}{x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{x}{1! \times x}$ $+$ $\dfrac{(1-x)(x^2)}{2! \times x^2}$ $+$ $\dfrac{(1-x)(1-2x)(x^3) }{3! \times x^3}$ $+$ $\cdots$ $\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\require{cancel} \Bigg[1 + \dfrac{\cancel{x}}{1! \times \cancel{x}}$ $+$ $\require{cancel} \dfrac{(1-x)(\cancel{x^2})}{2! \times \cancel{x^2}}$ $+$ $\require{cancel} \dfrac{(1-x)(1-2x)(\cancel{x^3}) }{3! \times \cancel{x^3}}$ $+$ $\cdots$ $\Bigg]$
$\therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \,$ $\Bigg[1 + \dfrac{1}{1!}$ $+$ $\dfrac{(1-x)}{2!}$ $+$ $\dfrac{(1-x)(1-2x)}{3!}$ $+$ $\cdots$ $\Bigg]$
Evaluate the limit of the infinite series as $x$ approaches $0$ and it is equal to the limit of exponential function in algebraic form as $x$ tends to $0$.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $1$ $+$ $\dfrac{1}{1!}$ $+$ $\dfrac{(1-(0))}{2!}$ $+$ $\dfrac{(1-(0))(1-2(0))}{3!}$ $+$ $\cdots$
$\,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1 \times 1}{3!} + \cdots$
$\therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!} + \cdots$
The infinite series represents the expansion of natural exponential function when its exponent is equal to one.
$e^{\displaystyle x} \,=\,$ $1 + \dfrac{x}{1!}$ $+$ $\dfrac{x^2}{2!}$ $+$ $\dfrac{x^3}{3!} + \cdots$
Put $x = 1$
$e^1 \,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1^2}{2!}$ $+$ $\dfrac{1^3}{3!} + \cdots$
$e \,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!} + \cdots$
The value of $e$ in infinite series and the limit of exponential function as $x$ approaches $0$ are same.
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $1$ $+$ $\dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!}$ $+$ $\cdots$ $\,=\,$ $e$
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $e$
Therefore, the limit of the exponential function ${(1+x)}^{\frac{1}{x}}$ as $x$ approaches zero is equal to $e$.
A best free mathematics education website for students, teachers and researchers.
Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.
Learn how to solve the maths problems in different methods with understandable steps.
Copyright © 2012 - 2022 Math Doubts, All Rights Reserved