Limiting Behavior of xsin(1/x)

[utkarshakash]

Homework Statement

\lim_{x->0} x\sin\frac{1}{x}

The Attempt at a Solution

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

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

 

[LCKurtz]

Homework Statement

\lim_{x->0} x\sin\frac{1}{x}

The Attempt at a Solution

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

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

Hint: How big can $|\sin \frac 1 x|$ be?

 

[utkarshakash]

Hint: How big can $|\sin \frac 1 x|$ be?

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

 

[LCKurtz]

Hint: How big can $|\sin \frac 1 x|$ be?

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

So...?

 

[Yanick]

I would use \lim_{x\rightarrow a} (f(x) \cdot g(x)) = \lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x) to tackle this problem personally.

 

[LCKurtz]

I would use \lim_{x\rightarrow a} (f(x) \cdot g(x)) = \lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x) to tackle this problem personally.

Bad idea.

 

[Yanick]

Oops! Sorry, you can't use that property. I will go hang my head in shame.

 

[utkarshakash]

So...?

But what about 1/x contained in the denominator? It tends to infinity as x approaches zero. How do I find limit then?

 

[LCKurtz]

Your original problem is $x\sin\frac 1 x$. Use that.

 

[ChrisVer]

it's better to think about:
lim_{x\rightarrow 0^{+}} f(x) = lim_{x\rightarrow 0^{-}} f(x)=A
then lim_{x\rightarrow 0} f(x)=A

 

[utkarshakash]

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?

 

[Saitama]

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.

 

[utkarshakash]

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.

 

[Saitama]

But the correct answer is 1 according to the solution.

I don't see how it should be 1.

W|A agrees with me: http://www.wolframalpha.com/input/?i=lim(x->0)+xsin(1/x)&dataset=

 

[Curious3141]

But the correct answer is 1 according to the solution.

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?

 

[utkarshakash]

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.

 

[LCKurtz]

@utkarshakash: The correct answer is zero, but you haven't given a good argument for it yet. Write down some inequalities.

 

[vthem]

Just take abs(xsin(1/x))=<abs (x).

 

[utkarshakash]

@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:

-1 \leq \sin \frac{1}{x} \leq 1 \\<br /> -x \leq \sin \frac{1}{x} \leq x \\<br /> lim_{x \to 0} -x = lim_{x \to 0} x = 0

The limit should therefore be 0.

Is my reasoning correct?

 

[LCKurtz]

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

-1 \leq \sin \frac{1}{x} \leq 1 \\<br /> -x \leq \color{red}{x}\sin \frac{1}{x} \leq x

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

<br /> lim_{x \to 0} -x = lim_{x \to 0} x = 0

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.

 

[lurflurf]

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

 

[utkarshakash]

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.

0 \leq \left|\sin(\frac 1 x)\right| \leq 1 \\<br /> 0 \leq \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 .

 

[vthem]

0 \leq \left|\sin(\frac 1 x)\right| \leq 1 \\<br /> 0 \leq \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 general, whene you have a clogged function (like \sin or \cos) times 0 the \lim 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: |x \sin(\frac 1 x) | \leq |x| ⇔ -|x| \leq x \sin(\frac 1 x) \leq |x|
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 \lim_{x \to α} f(x) instead of \lim_{x \to α} |f(x)|

 

[utkarshakash]

In general, whene you have a clogged function (like \sin or \cos) times 0 the \lim 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: \abs{x \sin(x) } \left \abs{x} ⇔ -\abs{x} \left x \sin(x) \left \abs{x}
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

Can you please format your equations correctly? I can't understand it.

 

[vthem]

Can you please format your equations correctly? I can't understand it.

I just edited it!

 

[utkarshakash]

I just edited it!

Thanks! You explained it pretty well.

 

[LCKurtz]

0 \leq \left|\sin(\frac 1 x)\right| \leq 1 \\<br /> 0 \leq \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 .

Good. Much better than your earlier argument.