Solve Cosine IND Limit Without L'Hospital

[Hernaner28]

It's the following one:

$$\displaystyle\lim_{x \to{0}}{\frac{1-\cos(1-\cos x)}{3x^4}}$$

In case we have to apply L'Hospital, appart from it, how could I solve this without it?
Thanks!

 

[Dick]

You could use the Taylor series expansion of cos(x) around x=0.

 

[Hernaner28]

Mmm I haven't learned Taylor series expansions yet. So anyway, could you tell me how to apply L'Hospital here? There are a lot of steps! I keep getting indeterminations. I cannot figure it out yet...
Thanks for the reply :)

 

[Dick]

Mmm I haven't learned Taylor series expansions yet. So anyway, could you tell me how to apply L'Hospital here? There are a lot of steps! I keep getting indeterminations. I cannot figure it out yet...
Thanks for the reply :)

If you don't have Taylor series yet, then you'll probably want to stick with l'Hopital. But the idea is to use cos(x)=1-x^2/2!+ terms of higher order in x. It does make things easier.

 

[Hernaner28]

Sorry but I didn't understand the idea.. could you explain the first steps of the resolution? I keep getting IND 0/0 -- I know I've got to apply l'Hopitale every time I get the indtermination but there're just too many.. it never ends.

 

[Dick]

Sorry but I didn't understand the idea.. could you explain the first steps of the resolution? I keep getting IND 0/0 -- I know I've got to apply l'Hopitale every time I get the indtermination but there're just too many.. it never ends.

l'Hopital will end at the fourth derivative. It has to. Then the denominator becomes a constant. It is a little hard to keep track of the numerator, I will admit.

 

[Fastman99]

Yo just plug that mofo numerator equation into WolframAlpha:
http://www.wolframalpha.com/input/?i=fourth+derivative+of+1-cos(1-cosx)

I agree, it's a nasty numerator, but you can just plug in x = 0 now. Looking at it real quick, and it's looks like the numerator at 0 equals 3, so the limit is 3.

Edit: Oh wait, the limit wouldn't be 3, it would 3/(3*4*3*2*1) = 1/24