Limit of x^x as x goes to 0+

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Limit of x^x as x goes to 0+,
0 to the 0 power,
Indeterminate form 0^0,

blackpenredpen,
math for fun,
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Could you do a video on the whole thing about "we can bring the limit into the exponent because e^something is a continuous function"?

TheGeneralThings
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You just saved my new assignments and taught me something new! (Didn't know about l'Hôpital's rule, researching that rn). Thank you!!!

awesomemilkshake
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Lovely. You should make a video to summarize all main the basic tricks (transformation) ppl can do to solve limits (i.e. X = 1/1/X or A=e^ln(A) and son on)

Zonnymaka
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I REALLY love your work ! I learn so many things. Thank you very much !!

fanapik
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I just got out of the class where I learned this.

Econcrafter
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You’re just like my professor but he retired, so I started watching you, man you’re great!

masteroogwayisntdead__
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Very impressive but you can show the proof of L'Hopital's rule?

MyEyedol
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We solved this exact problem when I was in high school. It was one of the coolest such problems.

In fact, that inspired me to make the hypothesis that if the functions f(x) and g(x) are both smooth around a value of A, and f(A)=0 and g(A)=0, then the limit when x->A of f(x)^g(x) is 1.

DjVortex-w
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Beautiful my man, very easy to understand proof.

rjc
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What is the limit of this function as it approaches 0 from the left? Wolfram Alpha says that it is 1, but you've made me skeptical of that calculator. Btw, thanks for the video. You're videos never fail to entertain and to blow my mind. :)

null
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"Just to be legitimate..." You must note that x = e^(ln x) only when x>0, which in this case is true.

DougCube
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Really... You are looking smart Sir.
I'm very impress with you because your teaching style is super. 👌👌👌👌👌👌👌👌👏👏👏👏👍👍👍👍

Magic
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Two classic tricks. limit of continuous function, and L'Hospitals. Neat.

simonburgess
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I have a good challenge for you.
Test for convergence ∑(n=1 to inf) 1/(n^3 * (sin(n))^2).

yannepomnyashi
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You are the best teacher I've already known.

lwckash
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Can you do a follow up video showing why the low-point of this of this function is (1/e, e^(-1/e))??

Also on why there are two solutions for this function between x= 0 < 1/e < 1

tannerzuleeg
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Can you help to solve this?
x^y = y^x
Where x and y are different.
Ex: 2^4 = 4^2

DO
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is there a similar demonstration for limit x going to 0- ? I know imaginary numbers could get involved, but I think that would also go towards real 1 in the end...

tharnjaggar
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loving your videos! please keep going :^)

Kloedelchen
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Is there a way to calculate the limit without De l'Hopital's rule and without Taylor expansions?

michelef