Question

Find Lt (1 + x)
x →0​

Answer 1

Consider,

$\bullet \: \blue{ \tt \:\displaystyle\lim_{x \: \to \: 0} {(1 + x)}^{ \displaystyle\frac{1}{x} } }$

The exponential function in algebraic form is in the form of Binomial Theorem.

So,

it can be expanded on the basis of this theorem.

$\rm\bullet\displaystyle {(1 + x)}^{n} = 1 + nx + \dfrac{n(n - 1) {x}^{2} }{2!} + \dfrac{n(n - 1) (n - 2){x}^{3} }{3!} + - - -$

Now,

$\displaystyle \rm \: replace \: n \: by \: \dfrac{1}{x}$

we get

$\rm\bullet\displaystyle {(1 + x)}^{\displaystyle \: \dfrac{1}{x} } = 1 + \dfrac{1}{x} \cdot \: x + \dfrac{\dfrac{1}{x} (\dfrac{1}{x} - 1) {x}^{2} }{2!} + \dfrac{\dfrac{1}{x} (\dfrac{1}{x} - 1) (\dfrac{1}{x} - 2){x}^{3} }{3!} + - - -$

$\rm\bullet\displaystyle {(1 + x)}^{\displaystyle \: \dfrac{1}{x} } = 1 +1 + \dfrac{(1 - x)}{2!} + \dfrac{(1 - x)(1 - 2x)}{3!} + - - -$

On applying limits x tends to 0, we get

$\bullet \: \tt \:\displaystyle\lim_{x \: \to \: 0} {(1 + x)}^{ \displaystyle\frac{1}{x} } = \displaystyle\lim_{x \: \to \: 0}(1 + 1 + \dfrac{1 - x}{2!} + \dfrac{(1 - x)(1 - 2x)}{3!} + - -$

$\bullet \: \tt \:\displaystyle\lim_{x \: \to \: 0} {(1 + x)}^{ \displaystyle\frac{1}{x} } = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + - - -$

$\bullet \: \: \therefore \: \bf \:\displaystyle\lim_{x \: \to \: 0} {(1 + x)}^{ \displaystyle\frac{1}{x} } = \bf \: e$

$\boxed{ \pink{ \because \tt \: \: e \: = 1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + - - - }}$

Hence,

$\large \boxed{ \boxed{ \purple{ \rm \: \bullet \: \bf \:\displaystyle\lim_{x \: \to \: 0} {(1 + x)}^{ \displaystyle\frac{1}{x} } = e \: }}}$