Having problems with this limit

1. Jan 18, 2005

twoflower

Hi all,

I'm having problems with this limit:

$$\lim_{n \rightarrow \infty} n^8 \left( 2\cos \left( \frac{1}{n^2} \right) - 2 + \frac{1}{n^4} \right)$$

I adjusted it to the fraction so that I can use l'Hospital, but it didn't get simpler...I know I should use the limit

$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$

$$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$

, the arguments are ready for that but I can't get it even after third l'Hospital...Is there any smart adjustment I can't see at the moment?

Thank you.

Last edited: Jan 18, 2005
2. Jan 18, 2005

Hurkyl

Staff Emeritus
One of those limits you said you should use is wrong.

Last edited: Jan 18, 2005
3. Jan 18, 2005

twoflower

I miswrote the limit with cos, sorry. So here is the point I got to:

$$- \frac{1}{2} \lim \frac{x^2.\sin \left( \frac{1}{x^2} \right) - 1}{\frac{1}{x^4}}$$

4. Jan 18, 2005

Hurkyl

Staff Emeritus
If I understand you right, you figured you should apply L'hôpital here too, right? (which is fine, once you prove both numerator and denominator go to 0) What did you get?

Last edited: Jan 18, 2005
5. Jan 18, 2005

Curious3141

$$\lim_{n \rightarrow \infty} n^8 \left( 2\cos \left( \frac{1}{n^2} \right) - 2 + \frac{1}{n^4} \right) = \lim_{n \rightarrow \infty}\left( n^8\left(2\cos(\frac{1}{n^2}) - 2)\right) + n^4 \right)$$

Does that help to express things better ? The cosine part approaches one as n becomes larger and larger, and the form $$2 \cos\theta - 2$$ will approach zero. You don't even have to wonder what happens when you multiply that by $$n^8$$ because you're adding to $$n^4$$ which unequivocally tends to infinity, so that's the limit.

6. Jan 18, 2005

Hurkyl

Staff Emeritus
Yes, 2 cos (1/n^2) - 2 does tend to zero, but n^8 tends to infinity, and 0 * infinity is an indeterminate form. Furthermore, that term is always nonpositive.

7. Jan 18, 2005

Curious3141

That's true. I'm wrong, sorry very sleepy at the moment (after 2 am here).

8. Jan 18, 2005

twoflower

Well, after next l'Hospitaling I got

$$\frac{1}{4} \lim \frac{ - \cos \left( \frac{1}{x^2} \right) + x^2.\sin \left( \frac{1}{x^2} \right)}{\frac{4}{x^6}}$$

which is 0/0 again...

9. Jan 18, 2005

Curious3141

I have an interesting idea. Try the Taylor's expansion for cosine. I get the final limit as 1/12, but I really am sleepy and could be wrong.

EDIT : The limit is $$\frac{1}{12}$$. Basically express the cosine part as a series of powers of n, multiply out by $n^8$, simplify and take the limit. Off to bed now.

Last edited: Jan 18, 2005
10. Jan 18, 2005

twoflower

You're right

11. Jan 18, 2005

twoflower

One more question, please, did you do the Taylor expansion right from the original limit or did you use l'Hospital and not before then did you use Taylor?

12. Jan 18, 2005

twoflower

Ok, now I read the edit in your post. I'll try it, unfortunately I'm not familiar with Mr. Taylor much

13. Jan 18, 2005

Galileo

If you go Taylor, the -2 and the 1/(n^4) cancel out against the first 2 terms in the series beautifully. Looks like it was made for this purpose.

14. Jan 18, 2005

twoflower

Probably yes. Maybe I should take a look at that magic Taylor before tommorow exam

15. Jan 18, 2005

Hurkyl

Staff Emeritus
Yes, that does look more complicated than the original, when you applied L'hôpital's rule. Can you think of any way to rewrite the fraction so that when you differentiate, it won't become more complicated?

16. Jan 18, 2005

twoflower

Well Hurkyl, in my opinion, when we know now that this limit was probably intended to be solved using Taylor, there may be no chance to l'Hospital-it into a more simple form...maybe I'm wrong, but I'm afraid sometimes the limits can be solved in only one way...

Or do you have the way how to make it simpler and are you just asking me whether I have found it too?

17. Jan 18, 2005

learningphysics

If you do a variable substitution in the beginning, then things get easier. Do you see it? Hint: differentiating cos(1/x^2) is really messy.

18. Jan 18, 2005

twoflower

Could you help me please? I'm having troubles with the Taylor expansion. I got:

$$\lim x^8 \left( \frac{1}{24x^8} + \frac{1}{2x^4} - 1 \right)$$

but it isn't much useful...What am I doing wrong with the expansion?

Thank you.

19. Jan 18, 2005

Hurkyl

Staff Emeritus
There is a way of making it simpler, and I am hoping to hint you towards seeing how.

Using Taylor series is certainly the easier way to do the limit, though.

20. Jan 18, 2005

Yegor

Cost=1-t^2/2!+t^4/4!-t^6/6!+t^8/8!+...