Finding lim x→0+ x^(x²) using L'Hôpital'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'hospital's 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.