Limit as X approaches 0 from the right of X^X

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[adlinanofal]

You can do this but can't spell indeterminate.. lmao best video ever! Thanks a lot, this helped me 👍

[georgefilos]

You can also use the identity (a^b)^c=a^(b•c) so that e^(xlnx)=(e^x)^(lnx) so its (e^0)^(-∞)=1/1^∞=1

[richardfrangie]

It's amazing!!!... How do you do that? How do you turn the pen like that? 00:36 I've always wanted to do that.

[connorjordan]

I'm in college, and this helped a lot. Thanks!

[neuralwarp]

I'd love to see you draw out the full graph of z=x^x : four separate regions: [1..+inf), (0..1), (-inf..0], and (0).
I call it the Eddie the Eagle curve.

[jesushernandez]

Thank you very much, for the idea to develop that Limit ...

[BriTheMathGuy]

3 Integral Tricks Teachers Don't Tell You!

[nikolatesla]

Sir actually i m confused that LHR is applied over the whole term i.e not only on lnx/1/x but to whole term e^lnx/1/x

[khinwin]

Thanks for your help I really appreciate this video ❤❤

[bestu]

if limit approaches to infinity then what is the answer???

[paramhanskumar]

What happen when x approaches to left side of 0

[Logan_]

X^X
Lim (x^x) x-> +0
Let x be n for the base
Let x be m for the power
=> n^m
Lim (n^m): n -> +0
+0^m -> 0
Lim (n^m) n^m: m-> +0
n^(+0) -> 1

We know:
Power is stronger than base.
Let n=m => n = m = x
=>
lim(x^x) -> +0
0^0 = 1

[neildey]

How to evaluate limit x->0+ x^(x^x)

[dope_gadget]

Can you tell me how to find lim([x]/(2x)) (x->0) ?

[madhabinath]

Why about tends to 0 from negative side

[mathewmiller]

I just had this on my test today and I put 1 so I got it right

[aaronbredon]

When you cancel out an x from -x²/x, you have to specify that x is not equal to 0 to avoid dividing by 0. That means your result becomes undefined at 0, and your math then fails, because you substitute 0 into the final equation, which means you substitute in a divide by 0.

[Manojkumar-cuch]

Best Analysis sir! & I subscribe your channel... Thank you...

[someperson]

(1) x^x = e^(ln(x^x)) is clearly true. But, contrary to
the video @2:21,
Lim(x approaches 0)(x^x) isn't = e^(ln(x^x)).
(2) Contrary to the statement in the video @4:17,
Lim(x approaches 0)(1/x) isn't infinity. However,
Lim(x approaches 0+)(1/x) = infinity.
(3) By L'Hospital's Rule, it's true that
Lim(x approaches 0+) [ln(x)/(1/x)] = 0.
But, the reason that @5:51
Lim(x approaches 0+) e^[ln(x)/(1/x)] = e^0
is the continuity of e^x at x = 0, not "we plug in 0."
I don't like being so pedantic, but many who have
taken calculus in high school have imprecise ideas
when they take a rigorous calculus course in college.

[arnabdas]

Hay bro, it will be more easy if you use the logarithmic infinite series.