L'Hopitals Rule and Standard Limits

In summary, In attempting to solve the limit as n approaches 0 of (n^2)sin^2(1/n), the author tried using L'Hopital's rule, differentiated numerator and denominator, and was left with sin(n)/2n. However, when n=0, he was left with 0/0 and was not sure where to go from there. When trying to solve the same question but with sin(1/n), the author tried applying L'Hopital's rule again, to Sin(n)/2n and was also left with 0/0. He then tried applying the sandwich theorem, which led to finding the limit as x->infty of 1/(5-
  • #1
Illusionist
34
0

Homework Statement


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.


Homework Equations


I think the n approaches 0 is confusing me.


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?

Thanks in advance for any help or advice.
 
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  • #2
Try applying L'Hopital's rule again, to Sin(n)/2n
 
Last edited:
  • #3
Ok I see, thank you for that. Unfortunately I'm still having trouble with the second question.
 
  • #4
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
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
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 don't know if this is right or where to go from here.
Sorry for the confusion.
 
  • #7
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
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 don't see how the right hand side limit would equal 0, did I go wrong somewhere?
Thanks again for that Mindscrape
 
  • #9
Illusionist said:
...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 don't see how the right hand side limit would equal 0, did I go wrong somewhere?
Thanks again for that Mindscrape

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
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
I would directly calculate the taylor series from the taylor formula, then try work it out. Seems like it could help, but I am not sure.
 
  • #12
VietDao29 said:
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.) :)

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

1. What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical theorem that allows us to find the limit of a function that cannot be easily evaluated by finding the limits of the numerator and denominator separately.

2. When can L'Hopital's Rule be applied?

L'Hopital's Rule can only be applied when we have a limit of the form "0/0" or "infinity/infinity".

3. What are Standard Limits?

Standard Limits are the common limits that are used in calculus, such as the limit as x approaches 0 of sin(x)/x = 1 or the limit as x approaches infinity of 1/x = 0.

4. How do I use L'Hopital's Rule to find a limit?

To use L'Hopital's Rule, we take the derivative of the numerator and denominator separately and evaluate the limit again. If we still have a "0/0" or "infinity/infinity" form, we repeat the process until we get a value for the limit.

5. Are there any limitations to using L'Hopital's Rule?

Yes, L'Hopital's Rule can only be used for limits involving real numbers. It also does not work for all limits, so it is important to check the conditions for its application before using it.

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