Math Question and Answers


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 80690 by ahmadshahhimat775@gmail.com last updated on 05/Feb/20

Commented by jagoll last updated on 05/Feb/20

$\lim_{x \to e} \frac{x l n x - x}{2 - 2 l n x} = \lim_{x \to e} \frac{x (l n x - 1)}{- 2 (l n x - 1)} =$ $- \frac{e}{2}$
Commented by abdomathmax last updated on 05/Feb/20
$let f(x)=((ln(x^x )−x)/(2−ln(x^2 ))) ⇒f(x)=((xln(x)−x)/(2−2lnx)) let x=e+u ⇒f(x)=g(u)=(((e+u)ln(e+u)−u−e)/(2{1−ln(e+u)})) x→e ⇒u→0 ⇒ g(u)=(((e+u)(1+ln(1+(u/e)))−u−e)/(2{1−ln(e)−ln(1+(u/e))})) =(((e+u)ln(1+(u/e)))/(−2ln(1+(u/e)))) ⇒g(u)∼ (((e+u)(u/e))/(−2×(u/e))) (u ∈V(0)) ⇒g(u)∼− ((u+(u^2 /e))/2)×(e/u)=−(e/2)(1+(u/e))→−(e/2) (u→0) ⇒lim_(x→e) f(x)=−(e/2)$
$l e t f (x) = \frac{l n (x^{x}) - x}{2 - l n (x^{2})} \Rightarrow f (x) = \frac{x l n (x) - x}{2 - 2 l n x}$ $l e t x = e + u \Rightarrow f (x) = g (u) = \frac{(e + u) l n (e + u) - u - e}{2 {1 - l n (e + u)}}$ $x \to e \Rightarrow u \to 0 \Rightarrow$ $g (u) = \frac{(e + u) (1 + l n (1 + \frac{u}{e})) - u - e}{2 {1 - l n (e) - l n (1 + \frac{u}{e})}}$ $= \frac{(e + u) l n (1 + \frac{u}{e})}{- 2 l n (1 + \frac{u}{e})} \Rightarrow g (u) \sim \frac{(e + u) \frac{u}{e}}{- 2 \times \frac{u}{e}} (u \in V (0))$ $\Rightarrow g (u) \sim - \frac{u + \frac{u^{2}}{e}}{2} \times \frac{e}{u} = - \frac{e}{2} (1 + \frac{u}{e}) \to - \frac{e}{2} (u \to 0)$ $\Rightarrow {l i m}_{x \to e} f (x) = - \frac{e}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum