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

The limit of $1+\dfrac{1}{x}$ raised to the power of $x$ as $x$ approaches infinity is a standard form in limits, often appeared in evaluating the limits of exponential functions. The limit of this special exponential function as its input tends to infinity is equal to $e$.

This standard rule is used as a formula in calculus and let’s prove this property of limits in mathematics firstly before using it in limits problems of exponential functions.

Expand the function as per Binomial Theorem

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

The algebraic function in exponential form is same as the Binomial Theorem. So, it can be expanded by 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, just replace $x$ by $\dfrac{1}{x}$ and $n$ by $x$ in the expansion of Binomial Theorem.

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

Simplify the Infinite series

The infinite series of the expansion of the exponential algebraic function can be further simplified and it helps us in the next few steps for evaluating the limit of exponential function.

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

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

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

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

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

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

The infinite series can be further simplified from the third term on wards by taking one $x$ common from each factor of the term.

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

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

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

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

Evaluate the Limit of the Infinite Series

Now, evaluate the limit of the infinite series as $x$ approaches infinity to find the limit of the exponential function.

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

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

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

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

The value of the infinite series can be evaluated by the expansion of natural logarithmic function.

According to the expansion of natural logarithmic function.

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

Take $x = 1$

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

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

The value of the infinite series is denoted by Napier constant $e$ in mathematics.

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

Therefore, it is proved that the limit of $x$-th power of the binomial $1+\dfrac{1}{x}$ as $x$ approaches infinity is equal to $e$

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