What is the limit of $\dfrac{\lim_{h\to 0} \sqrt{h}}{\lim_{x\to\infty} 1/x}$ I'd like to know the result of the following quotient:
$\dfrac{\lim_{h\to 0} \sqrt{h}}{\lim_{x\to\infty} 1/x}$.
I think the result is $0$ because the first derivative of $\sqrt{h}$ is $\dfrac{1}{2 \sqrt{h}}$ and $- \dfrac{1}{2 x^2}$ for $1/x$. Thus, $\sqrt{h}$ approaches 0 faster and faster the closer $h$ gets to $0$. Whereas $1/x$ approaches 0 slower and slower the more $x$ grows to infinity. 
In case my reasoning is true, I still think it is not formal enough. How could this be done more formally? 
 A: We have $\lim_{h \to 0} \sqrt h = 0$, and $\lim_{x \to \infty} 1/x = 0$.  Therefore, your quotient is simply $\frac{0}{0}$, which is undefined.
In general, the equality
$$
\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}
$$
will only hold if both $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ are finite, and $\lim_{x \to a}g(x) \neq 0$.

It looks like you're trying to evaluate something like
$$
\lim_{h \to \infty} \frac{\sqrt h}{1/f(h)}
$$
where $f$ is continuous and $\lim_{h \to 0} f(h) = \infty$.  Note that the limit depends on which $f$ you happen to choose.  That is, it matters how fast $x$ approaches infinity relative to how fast $h$ approaches zero, and we get no information about this from the problem as you've presented it.

On the other hand, a limit like
$$
\lim_{(h,x) \to (0,\infty)} \frac{\sqrt h}{1/x} = \lim_{(x,h)} x\sqrt h
$$
would fail to exist.
A: This quotient simply does not make sense because the denominator is $0$. If you are trying to apply l'Hopital's rule or something like that, this rule applies to the limit of a quotient (in some circumstances); not to the quotient of two limits.
A: As has been pointed out already, the problem does not make sense as stated, since you are dividing by zero.
We can make a few changes that might be towards what you are looking for. Note that $\lim_{x\to \infty} 1/x$ is the same as $\lim_{x\to 0} x$. Replacing $1/x$ with $x$ and moving the limit outside we find:
$$\lim_{x\to 0^+} \frac{\sqrt{x}}{x} = \lim_{x\to 0^+} \frac{1}{\sqrt{x}} = \infty$$
