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Question Number 53805 by hassentimol last updated on 26/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19

$f r o m g r a p h$ $\lim_{x \to 0} x^{x} = 1$ $\lim_{x \to 0} x^{x} (1 + l n x)$ $\lim_{x \to 0} x^{x} \times \lim_{x \to 0} (1 + l n x)$ $1 \times (- \infty) = - \infty$
Commented by maxmathsup by imad last updated on 26/Jan/19
$we have for x>0 x^x =e^(xln(x)) ⇒(x^x )^′ =(xln(x))^′ e^(xln(x)) =(lnx +1)x^x =x^x +ln(x).x^x we have lim_(x→0^+ ) {x^x +x^x ln(x)} =lim_(x→0^+ ) x^x +lim_(x→0^+ ) x^x ln(x) but lim_(x→0^+ ) x^x =lim_(x→0^+ ) e^(xln(x)) =e^0 =1 but x^x =e^(xln(x)) ∼1+xln(x) ⇒x^x ln(x) ∼ln(x)+x(ln(x))^2 =ln(x){1+xln(x)} →−∞ (x→0^+ ) ⇒ lim_(x→0^+ ) {x^x +x^x ln(x)} =−∞ .$
$w e h a v e f o r x > 0 x^{x} = e^{x l n (x)} \Rightarrow {(x^{x})}^{^{'}} = {(x l n (x))}^{^{'}} e^{x l n (x)}$ $= (l n x + 1) x^{x} = x^{x} + l n (x) . x^{x}$ $w e h a v e {l i m}_{x \to 0^{+}} {x^{x} + x^{x} l n (x)} = {l i m}_{x \to 0^{+}} x^{x} + {l i m}_{x \to 0^{+}} x^{x} l n (x) b u t$ ${l i m}_{x \to 0^{+}} x^{x} = {l i m}_{x \to 0^{+}} e^{x l n (x)} = e^{0} = 1$ $b u t x^{x} = e^{x l n (x)} \sim 1 + x l n (x) \Rightarrow x^{x} l n (x) \sim l n (x) + x {(l n (x))}^{2}$ $= l n (x) {1 + x l n (x)} \to - \infty (x \to 0^{+}) \Rightarrow$ ${l i m}_{x \to 0^{+}} {x^{x} + x^{x} l n (x)} = - \infty .$
Commented by hassentimol last updated on 29/Jan/19

$Thank you so much!$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19

	$y = x^{x}$ $l n y = x l n x$ $\frac{1}{y} \frac{d y}{d x} = x \frac{d l n x}{d x} + l n x \frac{d x}{d x}$ $\frac{d y}{d x} = y (x \times \frac{1}{x} + l n x) = x^{x} (1 + l n x)$
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