limits

[JeckShirauw]

Could someone please explain to me why
lim x^x
x->0
=
lim ln(x^x)
x->0

?

Thank you in advance.

 

[tkhunny]

Are you sure they are the same? Perhaps the second is the logarithm of the first?

The logarithm transformation preserves relationships. In other words, if a > b, and the logarithms exist, then log(a) > log(b). In still other words, successfully introducing the logarithm doesn't destroy what used to be. The square root transformation behaves similarly - when it exists. IT's just a nice, continuous, well-behaved transformation.

It is a classic proof, demonstrating that the log of the limit is the limit of the log.

 

[lookagain]

lim x^x
x->0

This should be

\(\displaystyle \lim_{x \to \ 0^+} (x^x)\)


It can be approached from the right, but not from the left.

 

[JeckShirauw]

Yes that is correct. But does the logarithm ln has to do something with this limit? Like an easier way to calculate it maybe?

 

[pka]

Could someone please explain to me why
lim x^x
x->0

\(\mathop {\lim }\limits_{x \to {0^ + }} {x^x} = \mathop {\lim }\limits_{x \to {0^ + }} \exp \left( {\log \left( {{x^x}} \right)} \right) = \exp \left( {\mathop {\lim }\limits_{x \to {0^ + }} \log \left( {{x^x}} \right)} \right) = \exp \left( {\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\log \left( x \right)}}{{1/x}}} \right) = ?\)

 

[Steven G]

Let y = x^x,
Then ln y = ln x^x=xln x

Now compute the limit of the rhs and suppose it approaches the number a

Then as x->0⁺, ln y approaches a

But if ln y approaches a, then y is approaching e^a.

So we find the limit of ln x^x because it is doable and then the answer is e^{the limit we just found}

 

[Steven G]

For the record,
lim x^x
x->0
=
lim ln(x^x)
x->0
is NOT true. Those two limits are not equal.

 

[JeckShirauw]

For the record,
lim x^x
x->0
=
lim ln(x^x)
x->0
is NOT true. Those two limits are not equal.

Thank you!