Limiting Behavior of xsin(1/x)

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In summary: I don't know what is wrong with the latex :confused::eek::grumpy::cry:Can you please format your equations correctly? I can't understand...Sure:$$\left|\sin(\frac 1 x)\right| \leq 1 \\ \left|x\sin(\frac 1 x)\right| \leq |x|$$Applying Sandwich Theorem now would give the answer 0. I think I'm correct this time .In summary, the limit of x times the sine of 1 over x as x approaches 0 is 0. This can be shown using the Sandwich Theorem and the inequality |xsin(1/x)| ≤ |x|,
  • #1
utkarshakash
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Homework Statement


[itex]\lim_{x->0} x\sin\frac{1}{x} [/itex]


The Attempt at a Solution



I can rewrite the function as [itex]\dfrac{\sin\frac{1}{x}}{1/x} [/itex]

But this can't be equal to 1 as the argument tends to infinity.
 
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  • #2
utkarshakash said:

Homework Statement


[itex]\lim_{x->0} x\sin\frac{1}{x} [/itex]


The Attempt at a Solution



I can rewrite the function as [itex]\dfrac{\sin\frac{1}{x}}{1/x} [/itex]

But this can't be equal to 1 as the argument tends to infinity.

Hint: How big can ##|\sin \frac 1 x|## be?
 
  • #3
LCKurtz said:
Hint: How big can ##|\sin \frac 1 x|## be?

It will oscillate between -1 and 1. So the maximum absolute value can be 1.
 
  • #4
LCKurtz said:
Hint: How big can ##|\sin \frac 1 x|## be?

utkarshakash said:
It will oscillate between -1 and 1. So the maximum absolute value can be 1.

So...?
 
  • #5
I would use [itex] \lim_{x\rightarrow a} (f(x) \cdot g(x)) = \lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x) [/itex] to tackle this problem personally.
 
  • #6
Yanick said:
I would use [itex] \lim_{x\rightarrow a} (f(x) \cdot g(x)) = \lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x) [/itex] to tackle this problem personally.

Bad idea.
 
  • #7
Oops! :redface: Sorry, you can't use that property. I will go hang my head in shame.
 
  • #8
LCKurtz said:
So...?

But what about 1/x contained in the denominator? It tends to infinity as x approaches zero. How do I find limit then?
 
  • #9
Your original problem is ##x\sin\frac 1 x##. Use that.
 
  • #10
it's better to think about:
[itex]lim_{x\rightarrow 0^{+}} f(x) = lim_{x\rightarrow 0^{-}} f(x)=A[/itex]
then [itex] lim_{x\rightarrow 0} f(x)=A[/itex]
 
  • #11
LCKurtz said:
Your original problem is ##x\sin\frac 1 x##. Use that.

When x approaches 0 sin(1/x) approaches 1. So the limit should be 0, right?
 
  • #12
utkarshakash said:
When x approaches 0 sin(1/x) approaches 1. So the limit should be 0, right?

The answer is correct but the reasoning is not, IMO.

As x approaches 0+ (or 0-), 1/x approaches infinity (or -infinity) but sin(1/x) approaches some value between -1 and 1 which is a finite number and a finite number multiplied with 0 is zero.
 
  • #13
Pranav-Arora said:
The answer is correct but the reasoning is not, IMO.

As x approaches 0+ (or 0-), 1/x approaches infinity (or -infinity) but sin(1/x) approaches some value between -1 and 1 which is a finite number and a finite number multiplied with 0 is zero.

But the correct answer is 1 according to the solution. :frown:
 
  • #15
utkarshakash said:
But the correct answer is 1 according to the solution. :frown:

Then that "correct answer" is incorrect. The limit of your expression is 0.

But if the limit is taken as x →∞, then the answer is indeed 1. You sure this wasn't the question?
 
  • #16
Curious3141 said:
Then that "correct answer" is incorrect. The limit of your expression is 0.

But if the limit is taken as x →∞, then the answer is indeed 1. You sure this wasn't the question?

I too think the given answer is incorrect.
 
  • #17
@utkarshakash: The correct answer is zero, but you haven't given a good argument for it yet. Write down some inequalities.
 
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  • #18
Just take abs(xsin(1/x))=<abs (x).
 
  • #19
LCKurtz said:
@utkarshakash: The correct answer is zero, but you haven't given a good argument for it yet. Write down some inequalities.

I tried to apply Sandwich Theorem. Here's my attempt:

[itex]-1 \leq \sin \frac{1}{x} \leq 1 \\
-x \leq \sin \frac{1}{x} \leq x \\
lim_{x \to 0} -x = lim_{x \to 0} x = 0 [/itex]

The limit should therefore be 0.

Is my reasoning correct?
 
  • #20
utkarshakash said:
I tried to apply Sandwich Theorem. Here's my attempt:

[itex]-1 \leq \sin \frac{1}{x} \leq 1 \\
-x \leq \color{red}{x}\sin \frac{1}{x} \leq x [/itex]

That only follows from the first inequality if ##x\ge 0##.

[itex]
lim_{x \to 0} -x = lim_{x \to 0} x = 0 [/itex]

So that would only work for ##x\to 0^+##.

The limit should therefore be 0.

Is my reasoning correct?

It is possible to fix that argument, but much easier to work with ##\left| x\sin(\frac 1 x)\right|##.

[Edit] Note lurflurf's following comment about the missing ##x## which I just added.
 
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  • #21
That is good except for a typo, it should read
$$-1 \leq \sin \frac{1}{x} \leq 1 \\
-x \leq x\, \sin \frac{1}{x} \leq x \\
lim_{x \to 0} -x = lim_{x \to 0} x = 0$$

edit: The inequality changes sign at x=0
 
Last edited:
  • #22
LCKurtz said:
That only follows from the first inequality if ##x\ge 0##.



So that would only work for ##x\to 0^+##.



It is possible to fix that argument, but much easier to work with ##\left| x\sin(\frac 1 x)\right|##.

[Edit] Note lurflurf's following comment about the missing ##x## which I just added.

[itex]0 \leq \left|\sin(\frac 1 x)\right| \leq 1 \\
0 \leq \left|x\sin(\frac 1 x)\right| \leq |x| [/itex]

Applying Sandwich Theorem now would give the answer 0. I think I'm correct this time .
 
  • #23
utkarshakash said:
[itex]0 \leq \left|\sin(\frac 1 x)\right| \leq 1 \\
0 \leq \left|x\sin(\frac 1 x)\right| \leq |x| [/itex]

Applying Sandwich Theorem now would give the answer 0. I think I'm correct this time .

In general, whene you have a clogged function (like [itex]\sin[/itex] or [itex]\cos[/itex]) times 0 the [itex]\lim[/itex] will be 0. It is usefull to think like this! Taking the inequalities and sandwich theorem will allways help!

I usually face these limits like this: [itex] |x \sin(\frac 1 x) | \leq |x| ⇔ -|x| \leq x \sin(\frac 1 x) \leq |x| [/itex]
Then you apply the sandwich theorem exactly as you did. Allthough I best like your method this is supposed to work a lot more times I think. Plus you take the [itex]\lim_{x \to α} f(x)[/itex] instead of [itex]\lim_{x \to α} |f(x)|[/itex]
 
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  • #24
vthem said:
In general, whene you have a clogged function (like [itex]\sin[/itex] or [itex]\cos[/itex]) times 0 the [itex]\lim[/itex] will be 0. It is usefull to think like this! Taking the inequalities and sandwich theorem will allways help!

I usually face these limits like this: [itex]\abs{x \sin(x) } \left \abs{x} ⇔ -\abs{x} \left x \sin(x) \left \abs{x} [/itex]
Then you apply the sandwich theorem exactly as you did. Allthough I best like your method this is supposed to work a lot more times I think.

I don't know what is wrong with the latex :confused::eek::grumpy::cry:

Can you please format your equations correctly? I can't understand it.
 
  • #25
utkarshakash said:
Can you please format your equations correctly? I can't understand it.

I just edited it!
 
  • #26
vthem said:
I just edited it!

Thanks! You explained it pretty well.
 
  • #27
utkarshakash said:
[itex]0 \leq \left|\sin(\frac 1 x)\right| \leq 1 \\
0 \leq \left|x\sin(\frac 1 x)\right| \leq |x| [/itex]

Applying Sandwich Theorem now would give the answer 0. I think I'm correct this time .

Good. Much better than your earlier argument.
 

1. What is the limit of xsin(1/x) as x approaches 0?

The limit of xsin(1/x) as x approaches 0 is 0. This can be seen by using L'Hopital's rule or by graphing the function.

2. Does the function xsin(1/x) have a limit at x = 0?

Yes, the function xsin(1/x) does have a limit at x = 0. The limit is 0.

3. What happens to the graph of xsin(1/x) as x approaches infinity?

The graph of xsin(1/x) oscillates infinitely between -1 and 1 as x approaches infinity. This behavior is known as a "wild" or "pathological" function.

4. Can the limit of xsin(1/x) be found algebraically?

No, the limit of xsin(1/x) cannot be found algebraically. This is because the function oscillates infinitely and does not approach a single value as x approaches 0.

5. How can the limiting behavior of xsin(1/x) be used in real-world applications?

The limiting behavior of xsin(1/x) can be used in fields such as engineering and physics to model oscillatory systems or to analyze the behavior of certain functions. It can also be used as a counterexample in calculus to demonstrate the limitations of algebraic methods in finding limits.

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