Limit Evaluation Given $a>1$ and $f:\mathbb{R}\backslash{\{0}\} \rightarrow\mathbb{R}$ defined $f(x)=a^\frac{1}{x}$ 
how do I show that $\lim_{x \to 0^+}f(x)=\infty$? 
Also, is the following claim on sequences correct and can it be used somehow on the question above(by using Heine and the relationship between sequences and fucntions)? Given two sequences {$a_n$},{$b_n$} that converge to $a$ and $b$ respectively, then $a_n^{b_n}$ converges to $a^b$ as $n$ approces to $\infty$. 
I think about this claim because in respect to the original question I know that $a^\frac{1}{x}$ is composed of a constant function $a$ which at any point converges to a and $\frac{1}{x}$ which converges to infinity as x approaches to zero, however I'm not sure how to make the transition between sequences and fucntions in this particular case.
I hope the question is clear.
(By the way, this isn't an homework assignment).
Thanks.
 A: You could do this:
$$\begin{align}
\lim_{x \to 0^{+}} a^{\frac{1}{x}} &= \lim_{u \to \infty} a^u = \infty.
\end{align}
$$
EDIT: Just to add a bit more to my answer. What I have written is all fine and good. The idea is that as $x$ goes to $0$ from the right, then $u = \frac{1}{x}$ goes to infinity. And when you take $a>1$ and raise it to increasing powers, then the result will also go to infinity.
You mention in your question something about sequences. In general, I think, that is a good way to think about it. But in your example the functions (sequence) in the exponent is not convergent (does not have a limit as $x$ goes to $0$ from the right) as it goes to infinity (as so by definition is not convergent). Hence you can't really use the result that you are referring to.
So what can you do? It depends on what you already "know". My answer of course assumed that you know that $\lim_{x\to 0^{+}} = \infty$. And it assumes that you can bring the limit up in the exponent. But if you want to know why these things are so - if you want to prove that the limit is infinity going back to the  the definition, then you have to do as in Mark's answer.
A: My thought was to go back to the definition of the limit. Choose $N$ large and positive. You want to show that you can choose (a small positive) $x$ so that $f(x)>N$
So we want $$a^{\frac1x}>N$$
Take logs (note log is an increasing function) $$\frac1x \log a > \log N$$
and this is true if $$x < \frac{\log a}{\log N}$$
This inequality is the right way round, and we can reverse the logic. $N$ was arbitrary, so for $x$ small enough $f(x)$ is greater than $N$.
Note that we can take logs to any base - I did not specify which above, and it does not make a difference to the ratio - $e$ would be normal, but if we note that $a>1$ we can use $a$ as a base for logarithms with $\log_a a=1$ the second equation becomes:
$$\frac 1x>\log_a N$$ and the third beomes:
$$x<\frac 1 {\log_a N}$$
These equations look simpler, but contain the same information - to compute them using tables or a calculator requires a base which is tabulated or calculated.
A: I think it is easier to use this:
$$\lim_{x \to 0^+} a^{1/x}=\exp\lim_{x \to 0^+} \log a^{1/x}$$
$$=\exp\lim_{x \to 0^+} \left( \frac{1}{x}\log a \right)$$
$$=\exp \left( +\infty\cdot\log a \right)=\exp +\infty =+\infty$$
This is all legitimate because the exponential and logarithmic functions are continuous in $(0,+\infty)$, and we're working with positive quantities. (Note that if $0<a<1$, the result would be $0$, since the logarithm of a number in $(0,1)$ is negative.)
