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Find the limit of the sequence x * sin(1/x)

  • Thread starter Ryker
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  • #1
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Homework Statement


Find
[tex]\displaystyle\lim_{x\to\infty}x\sin(\frac{1}{x})[/tex]

where x is a natural number, and this is a sequence, not a real function.


Homework Equations





The Attempt at a Solution


I know the answer is 1, and that supposedly one solution is to introduce a dummy variable y = 1/x. However, I was wondering whether there is another way to solve the problem without doing that, perhaps by somehow applying the squeeze law. The best I can do is squeeze it between 0 and 1, but that's not good enough, of course. I've tried all sorts of trig identity rearrangements, and it gets me nowhere.

Also, if you do introduce that dummy variable, can you really just do that without having to explain why exactly the limit of the sequence (as x goes to infinity) is equal to the limit of the function you get, as y approaches 0? Or can you accomodate that by saying that a sequence is a function itself?

Thanks in advance.
 

Answers and Replies

  • #2
gb7nash
Homework Helper
805
1

Homework Statement


Find
[tex]\displaystyle\lim_{x\to\infty}x\sin(\frac{1}{x})[/tex]

where x is a natural number, and this is a sequence, not a real function.
L'hopital's Rule. Rewrite as

[tex]\displaystyle\lim_{x\to\infty}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}[/tex]

After applying and cancellation, we get

[tex]\displaystyle\lim_{x\to\infty}\cos(\frac{1}{x})[/tex]

= 1

edit: I didn't see the part where you said it was a sequence and not a function. I don't think it matters though.
 
Last edited:
  • #3

Homework Statement


Find
[tex]\displaystyle\lim_{x\to\infty}x\sin(\frac{1}{x})[/tex]

where x is a natural number, and this is a sequence, not a real function.


Homework Equations





The Attempt at a Solution


I know the answer is 1, and that supposedly one solution is to introduce a dummy variable y = 1/x. However, I was wondering whether there is another way to solve the problem without doing that, perhaps by somehow applying the squeeze law. The best I can do is squeeze it between 0 and 1, but that's not good enough, of course. I've tried all sorts of trig identity rearrangements, and it gets me nowhere.

Also, if you do introduce that dummy variable, can you really just do that without having to explain why exactly the limit of the sequence (as x goes to infinity) is equal to the limit of the function you get, as y approaches 0? Or can you accomodate that by saying that a sequence is a function itself?

Thanks in advance.
Squeezing theorem won't. It doesn't work well with indeterminate forms.

As for the substitution, it is an easy matter to prove that if [tex] \frac{1}{a_n} \rightarrow 0 [/tex] then [tex]a_n \rightarrow \infty[/tex]. just use the definition of a limit approaching zero and infinity.

I am afraid there is not a "neat" way of doing this.
 
Last edited:
  • #4
1,086
2
Thanks for the responses, both of you.
L'hopital's Rule. Rewrite as

[tex]\displaystyle\lim_{x\to\infty}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}[/tex]

After applying and cancellation, we get

[tex]\displaystyle\lim_{x\to\infty}\cos(\frac{1}{x})[/tex]

= 1

edit: I didn't see the part where you said it was a sequence and not a function. I don't think it matters though.
Oh yeah, I forgot to mention we haven't done L'hopital's rule yet, so I guess this one is out of the question.
Squeezing theorem won't. It doesn't work well with indeterminate forms.
Yeah, like I said I've tried all kinds of stuff, and squeezing really didn't do the trick. Still, I wasn't quite sure if it can't be done at all or there's just something I was missing.
As for the substitution, it is an easy matter to prove that if [tex] \frac{1}{a_n} \rightarrow 0 [/tex] then [tex]a_n \rightarrow \infty[/tex]. just use the definition of a limit approaching zero and infinity.
Yeah, I have no trouble proving that, and we've done that in class, so I wouldn't even need to do it. The reason I was a bit apprehensive of this is due to the fact that we are dealing with a sequence, and you only take limits of sequences as n (or x or whatever the variable is called) approaches infinity. Although I guess they are functions, as well, just that their domain is the set of natural numbers.
 
  • #5
Yeah, like I said I've tried all kinds of stuff, and squeezing really didn't do the trick. Still, I wasn't quite sure if it can't be done at all or there's just something I was missing.
The only approximation I can come up with is sin(1/x) ~= 1/x when x is large. But you still need an inequality on the left side to squeeze the limit.
Yeah, I have no trouble proving that, and we've done that in class, so I wouldn't even need to do it. The reason I was a bit apprehensive of this is due to the fact that we are dealing with a sequence, and you only take limits of sequences as n (or x or whatever the variable is called) approaches infinity. Although I guess they are functions, as well, just that their domain is the set of natural numbers.
I think I see what you mean. But you have to remember that there is a very close relationship between limits of sequences and limits of functions. The relationship is too close for them to mutually exclusive. Sure when we talk about sequences we take the limit as n approaches infinity but, in essence, what we are doing is studying the characteristics of elementary functions that resemble the sequence. In fact, a sequence is a map(or function) on the natural numbers.
I. E

Lim sin(1/n) is zero but how do we know this ? Well we know the behaviour of 1/n as n gets large and we also know the behaviour of sin(x) when x is around zero. These facts help us to propose a limit of zero. But it is important to notice that the limit is only partially dependent on how 1/n... a more important matter is the behaviour of sinx.

What I am trying to illustrate is the idea that limits of sequences and functions are practically the same thing the only difference is how the independent variable changes. In the sequence case the independent variable approaches in a discrete, well behaved, manner; while in the case of functions the I.V approaches in a wild manner from different "paths" ( rational and irrational).

I feel like I am ranting XD. I hope that help remove some doubts from your mind.
 
  • #6
Char. Limit
Gold Member
1,204
13
It's easy. If y=1/x, then your limit reduces to sin(y)/y, which has a well known solution.
 
  • #7
hunt_mat
Homework Helper
1,724
16
Or write out the power series of sin(1/x), that will work just as well
 
  • #8
1,086
2
Alright, we did a bit of a review in class today and amidst all of that we did this problem, as well. As it turns out, you really can't just solve it by introducing a new variable 1/x, because we are dealing with sequences, and you also can't just skip from a sequence to a function without explaining why you're able to do that. So the way we did it is we started out with a fact proven in class, namely that the lim (sinx/x) as x approaches 0 is 1, so by the theorem that reduces the limit of a function to the limit of a sequence, if xn is any sequence that tends to 0, then sinxn/xn tends to 1. Then take the sequence 1/x, where x is a natural number, and you get lim sin(1/x)/(1/x) = 1. So this was exactly the reason for my apprehensiveness towards just plugging in 1/x, but I just couldn't put it in proper words. Introducing a new variable, however, then really does seem to be crucial, and there isn't a nice way around it, but just doing that isn't enough to constitute a formal proof.

So thanks again for all the help and suggestions, everyone.
 

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