Math Doubts

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

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.

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

Expand the Binomial function

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 exponential function

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 $0$.

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

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

$\therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $1 + \dfrac{1}{1!}$ $+$ $\dfrac{1}{2!}$ $+$ $\dfrac{1}{3!} + \cdots$

Evaluate the series

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 $e$.



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