lim-x-pi-x-pi-x-pi-x-pi-x-pi-

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

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

Facebook Tweet Pin