Calculating Limit of sin(1/x) - Product Law of Limits

[username12345]

Homework Statement

Calculate \mathop {\lim }\limits_{x \to 0 } x^2 \sin \frac{1}{x}

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 x^2 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?

 

[dx]

Use the sandwich rule.

-x² ≤ x² sin(1/x) ≤ x²

 

[tiny-tim]

Calculate \mathop {\lim }\limits_{x \to 0 } x^2 \sin \frac{1}{x}
…
"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!

Yes, the product law is the way …

just say it's lim{AB}, and |B| ≤ 1 …

 

[matt grime]

I think you're tex'ing or mathml'ing is off, dx. Did you mean to write

-x^2 \le x^2\sin(1/x) \le x^2

 

[dx]

Yes, thanks matt. I corrected it.

 

[username12345]

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.

 

[dx]

See this page if you don't remember the statement of the theorem: http://en.wikipedia.org/wiki/Sandwich_Rule

 

[username12345]

OK I am trying to use this sandwhich rule, but it doesn't make sense to me.

\mathop {\lim }\limits_{x \to 1 } |x-1|\sin \frac{1}{x-1}

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

-|x-1| \leq \sin \frac{1}{x-1} \leq |x-1|
0 \leq \sin \frac{1}{x-1} \leq 0

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

What is odd to me is that as x tends to 1 \sin \frac{1}{x-1} is not 0. For example when x is .999999999999 the result is 0.98.

 

[dx]

Is this a different question?

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

\lim_{x \rightarrow 0} \sin \frac{1}{x}

doesn't exist.

 

[username12345]

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.

 

[dx]

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|).

 

[username12345]

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.

 

[Random Variable]

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.

 

[rock.freak667]

Shouldn't doing this work the same

x^2sin(\frac{1}{x})=x\frac{sin(\frac{1}{x})}{\frac{1}{x}}

and then just use the fact that

\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) \times \lim_{x \rightarrow a} g(x)

 

[Random Variable]

Shouldn't doing this work the same

x^2sin(\frac{1}{x})=x\frac{sin(\frac{1}{x})}{\frac{1}{x}}

and then just use the fact that

\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) \times \lim_{x \rightarrow a} g(x)

That's very clever.