Couple of L'Hospital simplifications and concept clarification. I have 2 questions and a question about my text. In my text I just read this about indeterminate powers:

and then this:

So why is it that this example 10 is an indeterminate power? $x^x$ is in the form $0^0$ right? Why do they have to say this:

Notice that this limit is indeterminate since $0^x = 0$ for any $x > 0$ but $x^0 = 1$ for any $x \ne 0$ 

Can't they just say that the limit is $0^0$ and that's why it's indeterminate?
Questions


*

*$$\lim_{x \to \infty} x^3 \cdot e^{-x^2}$$
$$ = \lim_{x \to \infty} \frac{x^3}{e^{x^2}}$$
Is this right? From here I can just keep applying L'Hospital's rule until . I get 0 right?
2.
$$\lim_{x \to \infty} x^{\frac{1}{x}}$$
$$=\lim_{x \to \infty} e^{\ln{x} ^{\frac{1}{x}}}$$
$$= \lim_{x \to \infty} e^{\ln{1}}$$
$$ = 1$$
Is this correct?
 A: $0^0$ is indeterminate because we have two conflicting properties:


*

*Anything to the $0$th power is $1$

*$0$ to any power is $0$


Thus, we restrict the first property to apply for bases other than $0$, and the second property to powers greater than $0$. That means, there really is no way of evaluating this: $0^0$—it's an indeterminate form.
In the context of limits, you have to realize, by blindly plugging in you get $0^0$, but different functions may grow at different rates and may give you unexpected results. A good example is:
$$\lim_{n\to\infty} \left(1+{1\over n}\right)^n = e$$
When you blindly plug in $\infty$ you get $1^\infty$ which may seem like it should be $1$, but the power $n$ approaches infinity as $1 \over n$ approaches $0$, which gives a value of $e$.

As for your other concerns: yes you can consecutively apply L'Hopital to find that the polynomial grows slower than the exponential, giving a limit of $0$. 
Your second answer is correct, though I'm not exactly sure how you got $\ln1$. I would use log properties to bring the $1\over x$ out and find the limit $\lim_{x\to \infty} {\ln x\over x}$ through L'Hopital.
