lim_(x→π) ((x^π^x −π^x^π )/(x−π))=?


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Question Number 163650 by mathlove last updated on 09/Jan/22

$\lim_{x \to π} \frac{x^{π^{x}} - π^{x^{π}}}{x - π} = ?$
Commented by Zaynal last updated on 09/Jan/22

${is}^{'} t not value of limit$
Commented by mathlove last updated on 09/Jan/22

$w a y ?$
	Answered by mr W last updated on 09/Jan/22
	$generally: (a^(f(x)) )′=(e^(f(x) ln a) )′=a^(f(x)) (ln a)f′(x) (x^(f(x)) )′=(e^(f(x)ln x) )′=x^(f(x)) (((f(x))/x)+f′(x)ln x) lim_(x→a) ((x^a^x −a^x^a )/(x−a)) =lim_(x→a) (x^a^x −a^x^a )′ =lim_(x→a) {x^a^x ((a^x /x)+(ln x) (a^x )^′ )−a^x^a (ln a)(x^a )′} =lim_(x→a) {x^a^x [(a^x /x)+(ln x) (a^x )(ln a)]−a^x^a a(ln a)(x^(a−1) )} = {a^a^a [a^(a−1) +(ln a)^2 (a^a )]−a^a^a (ln a)(a^a )} = a^(a^a +a−1) {1+a(ln a)^2 −a(ln a)} = a^(a^a +a−1) [a (ln a)(ln a−1) +1] lim_(x→π) ((x^π^x −π^x^π )/(x−π))=π^(π^π +π−1) [π (ln π)(ln π−1) +1]$
	$g e n e r a l l y :$ ${(a^{f (x)})}^{'} = {(e^{f (x) l n a})}^{'} = a^{f (x)} (l n a) f^{'} (x)$ ${(x^{f (x)})}^{'} = {(e^{f (x) \ln x})}^{'} = x^{f (x)} (\frac{f (x)}{x} + f^{'} (x) \ln x)$ $\lim_{x \to a} \frac{x^{a^{x}} - a^{x^{a}}}{x - a}$ $= \lim_{x \to a} {(x^{a^{x}} - a^{x^{a}})}^{'}$ $= \lim_{x \to a} {x^{a^{x}} (\frac{a^{x}}{x} + (\ln x) {(a^{x})}^{^{'}}) - a^{x^{a}} (\ln a) {(x^{a})}^{'}}$ $= \lim_{x \to a} {x^{a^{x}} [\frac{a^{x}}{x} + (\ln x) (a^{x}) (\ln a)] - a^{x^{a}} a (\ln a) (x^{a - 1})}$ $= {a^{a^{a}} [a^{a - 1} + {(\ln a)}^{2} (a^{a})] - a^{a^{a}} (\ln a) (a^{a})}$ $= a^{a^{a} + a - 1} {1 + a {(\ln a)}^{2} - a (\ln a)}$ $= a^{a^{a} + a - 1} [a (\ln a) (\ln a - 1) + 1]$ $\lim_{x \to π} \frac{x^{π^{x}} - π^{x^{π}}}{x - π} = π^{π^{π} + π - 1} [π (\ln π) (\ln π - 1) + 1]$
	Commented by mathlove last updated on 09/Jan/22

	$t h a n k s m r W$
	Commented by Tawa11 last updated on 09/Jan/22

	$Great sir$
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