Clarification of L'hospital's rule I have a question regarding L'hospital's rule.  
Why can I apply L'hospital's rule to $$\lim_{x\to 0}\frac{\sin 2x}{ x}$$ and not to $$\lim_{x\to 0} \frac{\sin x}{x}~~?$$
 A: The reason you cannot use L'Hopital on the $\sin(x)/x$ limit has nothing to do with calculus, and more with logic, and the problem is subtle.
To use L'Hopital you need to know the derivative of $\sin(x)$. What is that derivative? You'd say $\cos(x)$ on reflex, and then you'd miss the problem. See, to calculate the derivative of $\sin(x)$ in the first place, you need to calculate
$$
\lim_{h \to 0}\frac{\sin(x + h) - \sin(x)}{h}
$$
We use the formula for $\sin(u + v)$ and simplify the fraction:
$$
\frac{\sin(x + h) - \sin(x)}{h} = \frac{\sin(x)\cos(h) + \sin(h)\cos(x) - \sin(x)}{h}\\\\
= \frac{\cos(h) - 1}{h}\sin(x) + \frac{ \sin(h)}{h}\cos(x)
$$
Thus to justify the use of L'Hopital, you need to know the limit of the two fractions as $h \to 0$, one of which is the limit we tried to use L'Hopital on in the first place.
If someone says that you cannot use it on $\sin(x)/x$, but you can use it on $\sin(2x)/x$, then that is because the former is in some way needed to justify L'Hopital in the first place, while when solving the latter you implicitly assume somehow that the derivative of $\sin(x)$ is already known. It all comes down to trying to read the mind of whoever poses the problem and try to see how heavy machinery they allow you to use.
A: You can apply l'Hôpital's rule in both cases! 
You can apply l'Hôpital's rule whenever you have an indeterminate form.
A: You can apply L'Hospital's rule just fine to $\lim_{x\to 0}\frac{\sin x}{x}$. It just doesn't tell you anything you didn't already know, because it concludes that the limit is
$$ \frac{\sin'(0)}{1} = \sin'(0) = \lim_{h\to 0}\frac{\sin(0+h)-\sin(0)}{h} = \lim_{h\to 0}\frac{\sin h - 0}{h} = \lim_{h\to 0}\frac{\sin(h)}{h} $$
which is exactly the same as the limit you started out with, except for the name of the variable.
This problem shows up whenever the denominator is $x$. However, as a practical matter, applying L'Hospital in this case can still be quicker and easier than recognizing that the limit is itself a differential quotient, so you can use your standard knowledge of derivatives on it.
