Existence of anti-derivative for the function $f(x)=x^\frac{1}{\sqrt x}$ 
I'm given this function, and asked to find whether it is bounded or unbounded.
$$f(x)=x^\frac{1}{\sqrt x}$$

I tried integrating this function with the limits from $0$ to $\infty$, but I got nowhere. I looked at the graph of this function, which I attach below:

Looking at this graph, I've concluded (with no proper explanation) that the value of the function is tending to $0$ as the value of $x$ tends to $\infty$.
Shouldn't this imply that the area bounded by the curve and the $x$-axis is finite?
I would appreciate all explanations pointing out the mistakes in my query. Thank you.
 A: The function $f(x)=x^{1/\sqrt{x}}$ is bounded.  $\lim_{x \rightarrow 0} f(x) = 0$ and  $\lim_{x \rightarrow \infty} f(x) = 1$ (as you've graphed, and shown by applying L'Hopital's rule to $\log f(x)$, for example.
Nevertheless, the area under the graph is unbounded (the integral from zero to infinity diverges).
$$\int_0^M f(x) \, dx > M/2$$ for large $M$. 
A: To see if $f(x)$ is bounded we notice that it contains no poles and discontinuities and thus it suffices to see what the behavior of the function is as $x\to\infty$. We start by taking the logarithm of $f(x)$:
$$\ln\left(f(x)\right)=\frac{\ln(x)}{\sqrt{x}}$$
We then take the limit as $x\to\infty$:
$$\lim_{x\to\infty} \ln(f(x))=\ln\left(\lim_{x\to\infty}f(x)\right)=\lim_{x\to\infty} \frac{\ln(x)}{\sqrt{x}}=\lim_{x\to\infty}\frac{2\sqrt{x}}{x}=\lim_{x\to\infty}\frac{2}{\sqrt{x}}=0$$
$$\ln\left(\lim_{x\to\infty}f(x)\right)=0$$
$$\lim_{x\to\infty} f(x)=1$$
By the same methods we can prove the behavior of $f(x)$ as $x\to0$ is$f(x)\to0$. Thus by continuity of $f(x)$ and its limiting behavior as $x\to\infty$ we prove that $f(x)$ is bounded, that is $|f(x)|\le M$ for some finite constant $M$.
