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

But it's not necessarily true that\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) \times \lim_{x \rightarrow a} g(x) unless you know that the limits of f(x) and g(x) both exist.But if they do both exist, then it's true.In summary, the limit of x^2sin(1/x) as x approaches 0 can be solved using the sandwich or squeeze rule, where -x^2 ≤ x^2sin(1/x) ≤ x^2. This can also be solved by factoring x out of the expression and using the fact that the limit of a

## 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?

Use the sandwich rule.

-x2 ≤ x2 sin(1/x) ≤ x2

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?

Yes, the product law is the way …

just say it's lim{AB}, and |B| ≤ 1 …

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

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

Yes, thanks matt. I corrected it.

dx said:
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.

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.

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.

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.

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|).

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.

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.

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

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

rock.freak667 said:
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.