Limit of exponential function at infinity

In this video, I showed how to find the limit of an exponential log function

[@carolnoelle2468]

Great explanations and smooth chalk. Thanks!

[@ebrimagajaga4639]

Yeah thank you so much..... I have learned from your video..... Additionally the question can be solved by easily plugging large positive numbers e.g 100, 10000, 1000000, 100000000 etc, each of these would directly give answer as exactly 1. Thank you! I love the way you've delivered the lesson

[@masoudhabibi700]

Thank you master....very well

[@utuberaj60]

Great derivation using LH rule- but with the ln idea

[@takudzwajoezaza6467]

Great video❤

[@punditgi]

We need Prime Newtons! 😊

[@WagesOfDestruction]

nice, I solved it by inspection by saying e^-x when x-> inf is 0. so inf^0 ->1. => answer =1

[@rangaweerakkody165]

I would do it as a two part problem. First I will show that y = e^-x when lim->inf y ->0. Then x^y when y->0 would be x->1.

[@Ennio483]

Let e^(-x)=y then y goes to zero So lim when x goes to infinity of x^[e^(-x)] = lim when y goes to ZERO of (- ln(y))^y Since lim (-ln(y)) at 0 =+ infinity lim y at 0 =0 Then lim at 0 of (-ln y)^y = 1 It's a limit of a power with a base coming greater but the exponent is going to zero . Thank you for your efforts , but I hope you'll be convicted 😂I usually avoid L'HOSPITAL'S

[@PrimeNewtons]

I share your sentiment about L'Hospital's Rule. Sometimes you just use it to save time

[@Ennio483]

Yes of course , good explanation👍

[@suyunbek1399]

now do the derivative of p to the eye to the am to the p i like good kitty as i like good tea

[@smiley_exe4960]

Or just substitute… e^-x = 1/e^x => lim x -> ♾️ (x^e^-x) = lim x -> ♾️ (x^(1/e^x)) = ♾️^0 = 1.

[@salahbekhit2685]

The limit of the natural logarithm is not always equal to the natural log of the limit The conditions for equality are continuity ,and boundedness We should mention at least the continuity in our problem Good work,but not perfect