Mastering Limits: Solving \lim_{x\rightarrow 0+} x^{x^2} Using L'Hopital's Rule

[laker88116]

\lim x \rightarrow 0+ x^x^2. This should read x^x^2 if that isn't clear. I am not sure where to start. I know that I need to use L'Hopitals Rule and I know that I need to get this in a form of \frac{f(x)}{g(x)}. Any suggestions on what else to do?

 

[VietDao29]

Hmm, if you take logs of both sides, what could you get?
Let y = \lim_{x \rightarrow 0 ^ +} x^{(x ^ 2)}
So:
\ln y = \lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x = z
Now, to find:\lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x, you can use L'Hopital rule. Rewrite it as:
z = \lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x = \lim_{x \rightarrow 0 ^ +} \frac{\ln x}{\frac{1}{x ^ 2}}
So y = e ^ z.
Viet Dao,

 

[laker88116]

But how do you know that its isn't (x^x)^2? I understand what you are saying though and it makes perfect sense, I just wasn't sure where to break it up.

 

[whozum]

Because (x^x)^2 = x^{2x} and it would seem silly to write it that way.

 

[laker88116]

Ok, well I understand now I was able to interpret it after VietDao29 did most of the work, but anyways, thanks for the help I see it now.