L'Hopitals Rule and Standard Limits

1. May 12, 2007

Illusionist

1. The problem statement, all variables and given/known data
Hello, I'm trying to find the limit(as n approaches 0) of [1-cos(n)]/(n^2). I have done a few of these before and haven't had to much trouble, but they all have been as n approaches infinity.

2. Relevant equations
I think the n approaches 0 is confusing me.

3. The attempt at a solution
I let n=0 and was left with 0/0, which would justify the use of L'Hopital's Rule. I then differentiated the numerator and denominator and was left with sin(n)/2n. Now do I let n=0, if so I would just be left with 0/0 again. I don't know where to go with this.

I am also having trouble with a another similar question, being finding the limit(as n approaches 0) of (n^2)sin^2(1/n). What sort of equations could I use as the less than and more than equations and how would the standard limits rules for n approaches infinity come into play?

2. May 12, 2007

danago

Try applying L'Hopital's rule again, to Sin(n)/2n

Last edited: May 12, 2007
3. May 12, 2007

Illusionist

Ok I see, thank you for that. Unfortunately I'm still having trouble with the second question.

4. May 12, 2007

Illusionist

I'm getting absolutely nowhere with the second question. I'm trying to find the limit of (n^2)sin^2(1/n).dx as n approaches 0 by using the Sandwich theorem.
IM a novice and inexperienced with the Sandwich thereom but I do understand it's principles. What sort of equations would be appropriate to use as the less than and greater than for the sandwich?
Any sort of help would be great and I'd appreciate it a bunch.

5. May 12, 2007

Mindscrape

What a minute, you got what happened with sin(n)/2n?

Hint:
Is (1-cos(n))/n^2 something that you can apply L'Hopital's rule to? What is cos(0)?

Sandwich thm is nice to know, but not needed here.

6. May 12, 2007

Illusionist

No I'm trying to do another question this time. I got the L'Hopital's Rule question but I'm now trying to solve the limit of (n^2)sin^2(1/n).dx as n approaches 0, by using the Sandwich Theorem.
I know for the Sandwich Theorem you must put a equation less than to the original on the left and greater than on the right, but I'm not sure what sort of equations to put in.
I tried putting 0 on the left, this would only work if the limit is zero I think, and 2(n^2)sin^2(1/n) on the right but dont know if this is right or where to go from here.
Sorry for the confusion.

7. May 12, 2007

Mindscrape

Oh, okay.

So the key to squeeze principle with trig functions is to start off with the basic trig inequalities such as
-1<= sin n <=1
and then add on to it

for example if you wanted to find the lim as n->infty of sin^2(5x)/(5-x) then you would start of with
-1<= sin5x <=1
0 <= sin^2(5x) <=1

then divide by the (5-x) and since x approaches infinity it will be negative and reverse the equalities

0/(5-x) >= sin^2(5x)/(5-x) >= 1/(5-x)

Then the limit as x->inft of 1/(5-x) = 0, so sin^2(5x)/(5-x) = 0.

Your problem is sort of similar, but has a multiplication instead of a division. Start off with
-1 <= sin(1/n) <= 1 (there is a discontinuity at n=0, but that is expected since we want to know that limit)
then build

8. May 12, 2007

Illusionist

Ok thanks for that. Let's see how I go:
-1<= sin n <=1
For my example I would start off with:
-1<= sin(1/n) <=1
0 <= sin^2(1/n) <=1
then multiply by n^2, gives:
0*(x^2) <= (x^2)sin^2(1/n) <=1*(n^2)

Now does this mean as n->inft of (n^2)=0?
Sorry I dont see how the right hand side limit would equal 0, did I go wrong somewhere?
Thanks again for that Mindscrape

9. May 12, 2007

VietDao29

The two last lines are wrong. You are throwing the x's out of nowhere. It should read:
0*(n2) <= (n2)sin^2(1/n) <=1*(n2)

Of course, the RHS limit is 0. Your n tends to 0, right? Not infinity. (It's what the problem stated, have a a closer look at what the problem says.) :)

10. May 12, 2007

Mindscrape

Maybe I should have made an example that even more closely resembled his problem, I think I might have confused him. I couldn't think of any good x->0 probs though.

Anyway, yes, with a few minor changes you got it.

11. May 12, 2007

Gib Z

I would directly calculate the taylor series from the taylor formula, then try work it out. Seems like it could help, but im not sure.

12. May 12, 2007

Illusionist

Of course, my apologies. What a stupid mistake. Thank you so much everyone, very helpful.