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Limit of sin(1/x)

  • #1

Homework Statement



Calculate [tex]\mathop {\lim }\limits_{x \to 0 } x^2 \sin \frac{1}{x}[/tex]


Homework Equations



"product law" of limits I think


The Attempt at a Solution



I took a guess at 0, as 1/x would become very large as x goes to 0, so sin would become a maximum of 1 but [tex]x^2[/tex] would become 0.

Is this actually the correct way to solve this limit problem? How should I actually express the answer, just put 0 or is there some intermediate working that can be shown?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
dx
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Use the sandwich rule.

-x2 ≤ x2 sin(1/x) ≤ x2
 
  • #3
tiny-tim
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Calculate [tex]\mathop {\lim }\limits_{x \to 0 } x^2 \sin \frac{1}{x}[/tex]

"product law" of limits I think

Is this actually the correct way to solve this limit problem? How should I actually express the answer, just put 0 or is there some intermediate working that can be shown?
Hi username12345! :smile:

Yes, the product law is the way …

just say it's lim{AB}, and |B| ≤ 1 … :wink:
 
  • #4
matt grime
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I think you're tex'ing or mathml'ing is off, dx. Did you mean to write

[tex] -x^2 \le x^2\sin(1/x) \le x^2[/tex]
 
  • #5
dx
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Yes, thanks matt. I corrected it.
 
  • #6
Use the sandwich rule.

-x2 ≤ x2 sin(1/x) ≤ x2
The text I have has a "squeeze law" which I think is the same. Looks the same.
 
  • #7
dx
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  • #8
OK I am trying to use this sandwhich rule, but it doesn't make sense to me.

[tex]\mathop {\lim }\limits_{x \to 1 } |x-1|\sin \frac{1}{x-1}[/tex]

By the product rule |x-1| we'd get 0 so the limit would be zero, but for fun I try squeeze.

[tex]-|x-1| \leq \sin \frac{1}{x-1} \leq |x-1|[/tex]
[tex]0 \leq \sin \frac{1}{x-1} \leq 0[/tex]

so, [tex]\mathop {\lim }\limits_{x \to 1 } |x-1|\sin \frac{1}{x-1} = 0[/tex]

What is odd to me is that as x tends to 1 [tex]\sin \frac{1}{x-1}[/tex] is not 0. For example when x is .999999999999 the result is 0.98.
 
  • #9
dx
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Is this a different question?

For the original question, technically you cannot use the product rule because

[tex] \lim_{x \rightarrow 0} \sin \frac{1}{x} [/tex]

doesn't exist.
 
  • #10
Yes, sorry, it is a different question.

You suggested to use sandwhich so rather than make a new thread I tried to apply it to a similar problem.
 
  • #11
dx
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Ok,

Do you understand why

-|x-1|≤ |x-1|sin(1/|x-1|) ≤ |x-1|?

You forgot the |x-1| in the middle multiplying sin(1/|x-1|).
 
  • #12
With |x-1| in the middle it makes more sense. My understanding though is limited to simply substituting x = 1 and evaluating the expression. I couldn't really explain it.

Thanks dx and to tiny-tim aswell.
 
  • #13
I would just say that the limit is zero because x^2 goes to zero and the function sin(1/x) is bounded.

And I don't know if you're interested, but you can show that the limit of sin(1/x) as x->0 doesn't exist by letting f(x)=sin(1/x) and finding a sequence (Sn) that goes to zero but f(Sn) doesn't converge.
 
  • #14
rock.freak667
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Shouldn't doing this work the same

[tex]x^2sin(\frac{1}{x})=x\frac{sin(\frac{1}{x})}{\frac{1}{x}}[/tex]

and then just use the fact that

[tex]\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) \times \lim_{x \rightarrow a} g(x)[/tex]
 
  • #15
Shouldn't doing this work the same

[tex]x^2sin(\frac{1}{x})=x\frac{sin(\frac{1}{x})}{\frac{1}{x}}[/tex]

and then just use the fact that

[tex]\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) \times \lim_{x \rightarrow a} g(x)[/tex]
That's very clever.
 

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