Question

prove that lim x → 0 ( 1 + x ) 1/ x = e​

Answer 1

We have to prove that :  $\lim_{x \to 0} (1+x)^{\dfrac{1}{x}}$ = e.

L.H.S. = $\lim_{x \to 0} (1+x)^{\dfrac{1}{x}}$

Using the binomial expansion of $(1+x)^{\dfrac{1}{x}}$

= $\lim_{x \to 0} (1+\dfrac{1}{x}+\dfrac{\dfrac{1}{x}(\dfrac{1}{x}-1) }{2!} ).x^2+\dfrac{\dfrac{1}{x}(\dfrac{1}{x}-1) (\dfrac{1}{x}-2) }{2!} ).x^3+$ ..... + ∞)

Put x = 0, we get

= 1 + 1 + $\dfrac{1}{2!}$ + $\dfrac{1}{3!}$

= e [ ∵ $e^x$ = 1 + x + $\dfrac{x^2}{2!}$ + $\dfrac{x^3}{3!}$ + ..... ]

= R.H.S., proved.

Thus, $\lim_{x \to 0} (1+x)^{\dfrac{1}{x}}$ = e, proved.