Having trouble solving this limit? Need some hints?

[mohlam12]

Hi
I've been trying to solve a limit, and I found some difficulties doing that. Any help would be appreciated!


lim cos(2x) / 2cosx - √2
x→ (π/4)

so, what I tried to do is expand that cos(2x) but I still get 0/0
Any hints ?
Thank cyou

 

[arildno]

First, get your parentheses right!

Secondly, have you learned L'Hopital's rule yet?

 

[mohlam12]

Sorry... but I don't see anything wrong with them!
No we have not learned it yet, and I would like to learn it.

 

[arildno]

Oh, really?
What you've ACTUALLY written means:
$$\frac{\cos(2x)}{2\cos(x)}-\sqrt{2}$$

But that wasn't the main point, though.

 

[mohlam12]

That's right, sorry!

 

[arildno]

In order to proceed without L'Hopital's rule, multiply with the conjugate expression:
$$\frac{\cos(2x)}{2\cos(x)-\sqrt{2}}=\frac{\cos(2x)}{2\cos(x)-\sqrt{2}}*\frac{2\cos(x)+\sqrt{2}}{2\cos(x)+\sqrt{2}}=\frac{\cos(2x)}{4\cos^{2}(x)-2}(2\cos(x)+\sqrt{2})=\frac{\cos(2x)}{2\cos^{2}(x)-1}\frac{2\cos(x)+\sqrt{2}}{2}$$
Now, any thoughts of simplifying this further?

 

[mohlam12]

I can see that (cos(2x))/(2cos²(x)-1) equals to 1.
Does that mean, the limit is equal to sqrt(2) ?
If that's right, how can it be solved by that L'Hopital rule ?

 

[arildno]

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:
$$h(x)=\frac{f(X)+f'(y)(x-X)}{g(X)+g'(z)(x-X)}=\frac{f'(y)}{g'(z)},y,z\in(X,x)$$
since f(X)=g(X)=0, and $X\neq{x}$

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]
insofar as this is defined.

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.

In your example:
$$\lim_{x\to\frac{\pi}{4}}\frac{\cos(2x)}{2\cos(x)-\sqrt{2}}=\lim_{y,z\to\frac{\pi}{4}}\frac{-2\sin(2y)}{-2\sin(z)}=\frac{\sin(2\frac{\pi}{4})}{\sin(\frac{\pi}{4})}=\sqrt{2}$$

 

[mohlam12]

Ah I understood a little bit... because we haven't learned derivatives yet :-/
Thanks anyway!