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Question Number 201443 by sonukgindia last updated on 06/Dec/23

Commented by mr W last updated on 06/Dec/23

$i a n d m a n y o t h e r s d o o u r b e s t t o$ $a n s w e r y o u r q u e s t i o n s . b u t w h y$ ${d o n}^{'} t y o u a n s w e r a n y q u e s t i o n o f$ $m i n e a n d o t h e r s ?$
Commented by mr W last updated on 06/Dec/23

$w h e n i a s k y o u i f y o u g o t t h e r i g h t$ $a n s w e r o f a s p e c i a l q u e s t i o n, t h i s$ $i s n o t p e r s o n a l . y o u m a y a n s w e r$ $y e s o r n o . b u t y o u j u s t i g n o r e . t h i s$ $i s n o t r e s p e c t f u l, a t l e a s t n o t p o l i t e,$ $i t h i n k a l s o i n y o u r c o u n t r y .$
Commented by sonukgindia last updated on 06/Dec/23

$i w i l l g i v e a n s w e r$
Commented by mr W last updated on 06/Dec/23

$t h a n k s f o r r e p l y!$
Commented by mr W last updated on 10/Dec/23

$w h a t a p i t y! y o u o n l y p r o m i s e,$ $b u t {d o n}^{'} t a c t u a l l y d o .$ $s i n c e t h i s p o s t, y o u o n c e a g a i n$ ${h a v n}^{'} t g i v e n a n y f e e d b a c k .$
	Answered by mr W last updated on 07/Dec/23

	$\lim_{x \to 0^{+}} {(x^{\frac{1}{x}})}^{x}$ $= \lim_{x \to 0^{+}} (x)$ $= 0$
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