Find-domain-of-1-1-x-x-Also-prove-thatL-x-0-1-1-x-x-1-

Question Number 39121 by rahul 19 last updated on 02/Jul/18

Find domain of {(1 + \frac{1}{x})}^{x} ?

Also prove that \underset{x \to 0^{+}}{L} {(1 + \frac{1}{x})}^{x} = 1 ?

Commented by math khazana by abdo last updated on 03/Jul/18

D_{f} =] - \infty, - 1 [\cup] 0, + \infty [.

Commented by prof Abdo imad last updated on 02/Jul/18

l e t f (x) = {(1 + \frac{1}{x})}^{x} w e h a v e

f (x) = e^{x l n (1 + \frac{1}{x})} s o x \in D_{f} \Leftrightarrow 1 + \frac{1}{x} > 0 a n d x \neq 0

\Rightarrow \frac{x + 1}{x} > 0 a n d x \neq 0 \Rightarrow x (x + 1) > 0 a n d x \neq 0 \Rightarrow

D_{f} =] - \infty, - 1] \cup] 0, + \infty [

w e h a v e {l i m}_{x \to 0^{+}} x l n (1 + \frac{1}{x})

= {l i m}_{x \to 0^{+}} x l n (1 + x) - x l n (x) = 0 \Rightarrow

{l i m}_{x \to 0^{+}} f (x) = e^{0} = 1 .

Commented by rahul 19 last updated on 03/Jul/18

What is the problem in domainif x lies between

- 1 & 0 ? ? ? let x = - \frac{1}{3} \dots \dots . pls comment .

Commented by MJS last updated on 03/Jul/18

mistake in line 3 ?

\frac{x + 1}{x} > 0 ⇏ x (x + 1) > 0 but :

\frac{x + 1}{x} > 0 \Rightarrow x \neq 0

case 1

x > 0 \Rightarrow (x + 1 > 0 \Rightarrow x > - 1) \Rightarrow x > 0

case 2

x < 0 \Rightarrow (x + 1 < 0 \Rightarrow x < - 1) \Rightarrow x < - 1

x \in [- 1; 0] \Rightarrow \ln \frac{x + 1}{x} \notin R

x = - \frac{1}{3} \Rightarrow \ln \frac{x + 1}{x} = \ln \frac{\frac{2}{3}}{- \frac{1}{3}} = \ln - 2

Commented by rahul 19 last updated on 04/Jul/18

Sir i have checked on desmos and the

ans . given by Prof Abdo is correct!

(Domain one) although i^{'} m not getting it