Does x sin(1/x) exist as x approaches zero?

[randomgamernerd]

Homework Statement

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lim_x->0xsin(1/x)

Homework Equations

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The Attempt at a Solution

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I feel the limit does not exist. Because sin(1/x) is largely changing value as x approaches 0,(since it is an oscillating function), and in limit, we check what happens in neighborhood of the point in concern. Now for x= say 10^-999 we might have y>0. But since sin(1/x) is rapidly changing values, so it might happen at x=10^-999 +0.000001, y<0. so i guess the function is rapidly changing altitude. So i guess limit does not exist. But I notice in the graph that the amplitude of the function is tending towards zero, and at x=0 it is exactly equal to zero. But I still disagree on the point that limit tends to zero. I don't know why. I have seen many proofs on the internet using sandwich theorem, but I just cannot get where am I going wrong..
Please Help me!

 

[Stephen Tashi]

But I still disagree on the point that limit tends to zero. I don't know why.

Do you mean that you don't know why you disagree?

I have seen many proofs on the internet using sandwich theorem, but I just cannot get where am I going wrong..

In the problem statement, you didn't say what the problem is. You only wrote $lim_{x \rightarrow 0} x\sin{x}$.

Are you asking how to write a formal proof that the value of that limit is zero?
Or are you asking for a intuitive way of understanding, to your own satisfaction, that the value of the limit is zero?

In your discussion, you seem to have forgotten that the definition of limit (as usually stated in textbooks) involves absolute values. So a change of sign can be irrelevant.

 

[DrStupid]

Maybe I'm wrong, but I would say it is equal to the limit

\mathop {\lim }\limits_{x \to 0} x \cdot y

with

- 1 \le y \le 1

 

[fresh_42]

Homework Statement

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lim_x->0xsin(1/x)

Homework Equations

: [/B]-

The Attempt at a Solution

:[/B]
I feel the limit does not exist. Because sin(1/x) is largely changing value as x approaches 0,(since it is an oscillating function), and in limit, we check what happens in neighborhood of the point in concern. Now for x= say 10^-999 we might have y>0. But since sin(1/x) is rapidly changing values, so it might happen at x=10^-999 +0.000001, y<0. so i guess the function is rapidly changing altitude. So i guess limit does not exist. But I notice in the graph that the amplitude of the function is tending towards zero, and at x=0 it is exactly equal to zero. But I still disagree on the point that limit tends to zero. I don't know why. I have seen many proofs on the internet using sandwich theorem, but I just cannot get where am I going wrong..
Please Help me!

In such cases, the intuition fails to be of help here. So what is left is, back to the roots. What does $\lim_{x \to 0} f(x)$ mean? The definition of limits?

 

[DrStupid]

In such cases, the intuition fails to be of help here. So what is left is, back to the roots. What does $\lim_{x \to 0} f(x)$ mean? The definition of limits?

I was referring to the definition for functions of a single variable as given in Wikipedia:

For every $\varepsilon > 0$ there is a $\delta > 0$ with $0 < \left| x \right| < \delta$ and

\left| {x \cdot \sin \left( {\frac{1}{x}} \right)} \right| &lt; \varepsilon

Due to

- 1 \le \sin \left( {\frac{1}{x}} \right) \le 1

this is always given for $\delta = \varepsilon$.

 

[fresh_42]

That's continuity, and as the function isn't defined at $x=0$ it cannot be continuous there. Now we could talk about the nature of this singularity, but that is a different question. Here we have only a limit, i.e. in every open neighborhood $U_\varepsilon(0)$ of $x=0$ there must be a point of $x\cdot \sin\frac{1}{x}$ which is the case, how small we choose $\varepsilon$ to be.

However, continuity would require $\lim_{x \to 0} (x\cdot \sin \frac{1}{x}) = f(0) = 0 \cdot \sin \frac{1}{0}$ which does not exist. So the limit exists, although the function isn't continuous. To be continuous, we would have to explicitly set $f(0)=0$.

 

[Mark44]

That's continuity, and as the function isn't defined at x=0 it cannot be continuous there.

If you're referring to DrStupid's post (#5), he wasn't saying that the function $f(x) = x \sin(1/x)$ is continuous at x = 0, but rather that the limit as x approaches zero exists. For the function to be continuous at x = 0, it must be true that ##\lim_{x \to 0} f(x) exists, and that f(0) equals this limit.

 

[randomgamernerd]

Or are you asking for a intuitive way of understanding, to your own satisfaction, that the value of the limit is zero?

I’m asking for this

 

[randomgamernerd]

In such cases, the intuition fails to be of help here. So what is left is, back to the roots. What does $\lim_{x \to 0} f(x)$ mean? The definition of limits?

lim_x->0f(x) means what value is f(x) approaching as x is approaching zero, i feel f(x) is not decreasing straightaway towards zero, i mean the wave like part has a decreasing amplitude, which is tending towards zero, but we cannot in general say the function is tending towards zero

 

[Mark44]

lim_x->0f(x) means what value is f(x) approaching as x is approaching zero, i feel f(x) is not decreasing straightaway towards zero, i mean the wave like part has a decreasing amplitude, which is tending towards zero, but we cannot in general say the function is tending towards zero

Yes we can. It is immaterial the 1/x is undefined at x = 0. For every other value of x, sin(1/x) is always between -1 and 1, so $x\sin(\frac 1 x) \to 0$ as $x \to 0$. Someone earlier in this thread mentioned the Sandwich Theorem. You should look it up.

 

[fresh_42]

lim_x->0f(x) means what value is f(x) approaching as x is approaching zero, i feel f(x) is not decreasing straightaway towards zero, i mean the wave like part has a decreasing amplitude, which is tending towards zero, but we cannot in general say the function is tending towards zero

Have you looked at it?
http://www.wolframalpha.com/input/?i=y=x+sin(1/x)
You said it yourself: the amplitude is closing down to zero, so the oscillation is simply running out of space. The function trembles to zero and if we set $f(x)=0$ we get even continuity, because that $x=0$ isn't in the domain is the only reason, why it isn't continuous. If we define it, your proof above is the way to "see" it.

 

[randomgamernerd]

lim_x->0f(x) means what value is f(x) approaching as x is approaching zero, i feel f(x) is not decreasing straightaway towards zero, i mean the wave like part has a decreasing amplitude, which is tending towards zero, but we cannot in general say the function is tending towards zero

ok, thanks

 

[randomgamernerd]

Have you looked at it?
http://www.wolframalpha.com/input/?i=y=x+sin(1/x)
You said it yourself: the amplitude is closing down to zero, so the oscillation is simply running out of space. The function trembles to zero and if we set $f(x)=0$ we get even continuity, because that $x=0$ isn't in the domain is the only reason, why it isn't continuous. If we define it, your proof above is the way to "see" it.

got it

 

[epenguin]

Convinced? Is there even anything unexpected here? I know it doesn't, can't, exactly really prove anything, but you said you'd seen proof, you just didn't believe it.

 

[randomgamernerd]

View attachment 220961

Convinced? Is there even anything unexpected here? I know it doesn't, can't, exactly really prove anything, but you said you'd seen proof, you just didn't believe it.

yah, lol..

 

[epenguin]

Or actually, this is brought up in e.g. #(5). However huge 1/x is, sin(1/x) is never outside the range -1 to 1! Whereas the more usual exercises are about the product of one thing that is zero at some point and something that is infinite there, I.e. increases without limit towards that point, here it is a product of something that is zero and something very tame. Probably you have got used to more difficult problems and are flummoxed by such an easy one.