Can I use L'Hopital's rule when finding a derivative using the limit definition? When finding the derivative of $f(x) = \sqrt x$ via the limit definition, one gets
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x}{h}$$
For this, I could get the answer from applying L'Hopital's rule... but to me, this line of reasoning is a bit circular. I'm computing a derivative from first principles while needing to use derivative rules (from L'Hopital's) to compute this derivative.
Can I use L'Hopital's rule when finding a derivative using the limit definition?
 A: The line of reasoning is in fact circular; you need to know the derivative of the function to apply L'Hopital's rule in the first place.
One can (infamously) likewise apply this to
$$\lim_{x \to 0} \frac{\sin(x)}{x}$$
but
$$\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \sin(x) &= \lim_{h\to 0}\dfrac{\sin(x+h)-\sin x}{h} \\
&= \lim_{h\to 0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h} \\
&= \cos x\cdot\left(\lim_{h\to 0}\dfrac{\sin h}{h}\right)+ \sin x\cdot\left(\lim_{h\to 0}\dfrac{\cos h -1}{h}\right)
\end{align*}$$
which requires knowledge of the limit we originally sought! And if you don't know the derivative, how can you apply L'Hopital's rule to our original limit, which uses that definition?
That is, to find the original limit this way is to assume - without basis - what the derivative of sine is. (Of course, who's to say this is the only way you can find the derivative? If you can justify that the derivative is cosine some other way, that doesn't make use of this limit, then there is no issue.)
To find the derivative in your case with $f(x) = \sqrt x$, using L'Hopital's rule, is to assume what the derivative is before you've even found it.
Some discussion is on this MSE post here for the $\sin(x)/x$ limit.
A: The circular use of the knowledge of $f'(x)$ in its computation using the definition is not reasonable and proves nothing. Your actual conclusion would just be that "If $f'(x) = \frac{1}{2 \sqrt{x}}$ then  $f'(x) = \frac{1}{2 \sqrt{x}}$". The way to go in this case is just multiplying and dividing by the conjugate expression:
$$
\lim_{h\to 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h} = \lim_{h \to 0}\dfrac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} = \dfrac{1}{\sqrt{x}+\sqrt{x}}=\frac{1}{2\sqrt{x}}.
$$

Just for laughs, suppose that you wrongly assume $(\sqrt{x})' = \frac 1x$. Now using the definition + L'Hôpital's rule would give you
$$
(\sqrt{x})' = \lim_{h\to 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h} = \lim_{h\to 0} \dfrac{\frac{1}{x+h}}{1} = \frac 1x ...
$$
Voila, confirmed! See the problem here?
A: L’Hopital’s rule has been proven, many, many times in the past. It’s true, because you believe hundreds of mathematicians who told you so. You don’t need to know details of the proof of the rule.
So you can use the rule to prove something. And I use what you proved to prove L’Hopital. Yes, there is a circle. Nevertheless, L’Hopital’s rule is true. As is everything proved by using it.
