Question

$\displaystyle \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^x = e$
​

Answer 1

$\large{ \underline{\sf{Solution :-}}}$

We have,

$\rm \lim \limits_{x \to \infty} \left( 1 + \dfrac{1}{x} \right)^x$

Let's assume that,

$\implies \rm y = \lim \limits_{x \to \infty} \left( 1 + \dfrac{1}{x} \right)^x$

Taking natural log both sides.

$\implies \rm \: \ln y = \lim \limits_{x \to \infty} \ln\left( 1 + \dfrac{1}{x} \right)^x$

${ \implies \rm \: \ln y = \lim \limits_{x \to \infty} \dfrac{ \ln\left( 1 + \dfrac{1}{x} \right)}{ {x}^{ - 1} }}$

Now applying L'Hopital rule.

${ \implies \rm \: \ln y = \lim \limits_{x \to \infty} \dfrac{ \dfrac{d}{dx} \ln\left( 1 + \dfrac{1}{x} \right)}{ \dfrac{d}{dx} {x}^{ - 1} }}$

${ \implies \rm \: \ln y = \lim \limits_{x \to \infty} \dfrac{ \dfrac1{\left( 1 + \frac{1}{x} \right)} \times ( - {x}^{ - 2} )}{ { - x}^{ - 2} }}$

${ \implies \rm \: \ln y = \lim \limits_{x \to \infty} \dfrac1{\left( 1 + \dfrac{1}{x} \right)}}$

${ \implies \rm \: \ln y = \dfrac1{\left( 1 + \dfrac{1}{ \infty} \right)}}$

${ \implies \rm \: \ln y = \dfrac1{\left( 1 + 0\right)}}$

${ \implies \rm \: \ln y = 1}$

Or in terms of exponential,

${ \implies \rm \: {e}^{\ln y }= {e}^{1} }$

${ \implies \rm \: y= {e}^{1}}$

${ \implies \rm \: y= {e}}$

Therefore the required result is proved that,

$\underline{ \boxed{ \rm \lim \limits_{x \to \infty} \left( 1 + \dfrac{1}{x} \right)^x = e}}$

Additional Information :-

$\boxed{\begin{array}{c|c} \tt f(x)& \tt f'(x)\\\underline{\qquad\qquad}&\underline{\qquad\qquad}\\ \\ \sf x^n&\sf nx^{n-1}\\\\\sf {ln \: a}&\sf{\dfrac{1}{x}}\\\\\sf {e^x}&\sf {e^x} \\ \\ \sf log_{a}x & \sf\dfrac{1}{x}\log_ea\\\\ \sf a^x&\sf a^x\log_ea\\\\\sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\end{array}}$

Answer 2

[ I will solve this question in another method ]

${\blue{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: we \: have \: to \: prove \: : \\ \\ \sf \hookrightarrow \: \displaystyle \lim_{x \to \infty} \rm \: {(1 + \frac{1}{x} })^{x} = e \end{array}}}}}$

${ \boxed{ \boxed{ \begin{array}{cc} \rm \: L.H.S = \displaystyle \lim_{x \to \infty} \rm {(1 + \frac{1}{x} })^{x} \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: we \: know : \\ \: {(1+ x)}^{n } = 1 + n \: x + \frac{n(n - 1)}{2!} {x}^{2} + \frac{n(n - 1)(n - 2)}{3!}. {x}^{3} + ... \end{array}}}} \\ \\ \small{ =\displaystyle \lim_{x \to \infty} \rm \{1 + \frac{x}{1!} . \frac{1}{x} + \frac{x(x - 1)}{2!} . {( \frac{1}{x} })^{2} + \frac{x(x - 1)(x - 2)}{3!}. {( \frac{1}{x} })^{3} + ... \} } \\ \\ = \displaystyle \lim_{x \to \infty} \rm \{1 + \frac{1}{1!} + \frac{ {x}^{2}( 1 - \frac{1}{x}) }{2!}. \frac{1}{ {x}^{2} } + \frac{ {x}^{3} (1 - \frac{1}{x} )(1 - \frac{2}{x}) }{3!} . \frac{1}{ {x}^{3} } + ... \} \end{array}}}}$

${ \boxed{ \boxed{ \begin{array}{cc} =\displaystyle \lim_{x \to \infty} \rm \{ 1 + \frac{1}{1!} + \frac{(1 - \frac{1}{x}) }{2!} + \frac{(1 - \frac{1}{x})(1 - \frac{2}{x}) }{3!} + ... \} \\ \\ = \rm1 + \frac{1}{1!} + \frac{ (1 - \frac{1}{ \infty} )} {2!} + \frac{(1 - \frac{1}{ \infty} ( 1 - \frac{2}{ \infty} ) }{3!} + ... \\ \\ \rm = 1 + \frac{1}{1!} + \frac{(1 - 0)}{2!} + \frac{(1 - 0)(1 - 0)}{3!} + ... \\ \\ \rm = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: we \: know \: that: \\ \\ \rm \: {e}^{x} = 1 + \frac{x}{1!} + \frac{ {x}^{2} }{2!} + \frac{ {x}^{3} }{3!} + ... \\ \\ \therefore \rm \: e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ...\end{array}}}} \\ \\ \rm = e \\ \\ \rm = R.H.S\end{array}}}}$