Having problems with this limit

[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.

 

[Hurkyl]

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

Anyways, show your work and we can help you with it.

 

[twoflower]

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

Anyways, show your work and we can help you with it.

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

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

 

[Hurkyl]

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?

 

[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.

 

[Hurkyl]

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.

 

[Curious3141]

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.

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

 

[twoflower]

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?

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...

 

[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.

 

[twoflower]

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.

You're right

 

[twoflower]

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.

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?

 

[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?

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

 

[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.

 

[twoflower]

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.

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

 

[Hurkyl]

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?

 

[twoflower]

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?

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?

 

[learningphysics]

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?

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.

 

[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.

 

[Hurkyl]

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.

 

[Yegor]

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

 

[twoflower]

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

I finally have it, I did the expansion of cos properly, but forgot to multiply each term with 2...

 

[learningphysics]

I finally have it, I did the expansion of cos properly, but forgot to multiply each term with 2...

Well, since you've completed the problem, I thought I'd just mention how I did it, and what I think Hurkyl was hinting at. I substitued a=1/x^2, and then took the limit with a->0. This makes differentiating a little easier. Using l'Hopital's rule over and over you get 1/12.

 

[Hurkyl]

Actually, I was hinting to divide the numerator and denominator through by x^2, so that the "complication" of the product rule doesn't enter.