That's correct!
Now, to introduce L'Hopital's rule, let's consider h(x)=f(x)/g(x), and search for h(X) when f(X)=g(X)=0 (X being the point we let x approach.
Now, according to the mean value theorem, there exists an y in the interval (X,x) so that f(x)=f(X)+f'(y)(x-X) and similarly, a z in (X,x), so that g(x)=g(X)+g'(z)(x-X) (y and z need not be the same number!)
Thus, we have:
[tex]h(x)=\frac{f(X)+f'(y)(x-X)}{g(X)+g'(z)(x-X)}=\frac{f'(y)}{g'(z)},y,z\in(X,x)[/tex]
since f(X)=g(X)=0, and [itex]X\neq{x}[/itex]
Now, as you let x approach X, then evidently we must have that y approaches X, and z approaches X.
Thus, you get:
[tex]\lim_{x\to{X}}h(x)=\frac{f'(X)}{g'(X)}[/itex]<br />
insofar as this is defined.<br />
<br />
Thus, L'Hopital's rule says that if h(X) is a 0/0-expression, you can find its limit value by computing the ratio between the the derivatives of f and g, evaluated at X.<br />
<br />
In your example:<br />
[tex]\lim_{x\to\frac{\pi}{4}}\frac{\cos(2x)}{2\cos(x)-\sqrt{2}}=\lim_{y,z\to\frac{\pi}{4}}\frac{-2\sin(2y)}{-2\sin(z)}=\frac{\sin(2\frac{\pi}{4})}{\sin(\frac{\pi}{4})}=\sqrt{2}[/tex][/tex]
