Limit of xsin(1/x)

**Timbuk2**

I'm having difficulty understanding this. Only a first year in college, so the question is very basic

If I want to evaluate [latex]\lim_{x\to0}x\sin(\frac{1}{x})[/latex]
(How do I make the x->0 appear under the lim in latex on boards?)

From just looking at the graph on Wolfram Alpha, it looks like it will be 0.

My first instinct would be to say that
[latex]x\sin(\frac{1}{x}) = \displaystyle \frac{sin(\frac{1}{x})}{\frac{1}{x}} $ provided $ x\neq0[/latex], which is (I think!) a fair assumption, as we are looking for the limit as x tends to 0 (from both sides), not AT x=0.

But it's a known identity that [latex]\lim_{t\to0}\frac{\sin(t)}{t} = 1[/latex], which is not the result I expected from looking at the graph.

So if I try something different.
Obviously [latex]-1 \leq \sin(\frac{1}{x}) \leq 1[/latex]
If I multiply across by x>0, I get [latex]-x \leq x\sin(\frac{1}{x}) \leq x[/latex]

By the Squeeze/Sandwich Theorem, which states that if [latex]g(x) \leq f(x) \leq h(x)[/latex] and [latex]\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L.[/latex] then [latex]\lim_{x \to a} f(x) = L.[/latex], we can clearly see that as x tends to 0, xsin(1/x) tends to 0 also.

(Multiplying the original inequality by x<0 gives me the same result, so the limit exists and = 0).

Why the contradiction in the two methods? The Squeeze theorem method gives me the answer I expected, but I can't see anything wrong with the sin(1/x) / (1/x) method, although I suspect that the denominator not being defined at x=0 may have something to do with it.

LeixlipRed

As far as I'm aware we only know that functions of the form (sin(ax))/ax have the limit 1 as x tends to 0.

**Timbuk2**

Oh dear. that makes perfect sense!

What I failed to realise is that sin(1/x) / (1/x) doesn't even satisfy what I was trying to get it to satisfy, because it's the limit as x->0, not as (1/x)->0.

I suppose it's better to make the mistake now, rather than in the exam (where I'll have no WolframAlpha to quickly sketch graphs )

Thanks

Eliot Rosewater

Well, you've inadvertently have proven a nice limit, that
[LATEX]\displaystyle \lim_{x \rightarrow \infty}\left(x \sin\left(\frac{1}{x}\right) \right) = 1[/LATEX]

Wouldn't seem an "obvious" limit to me.

LeixlipRed

Well no, as x goes to infinity that will blow up. I assume you mean as x goes to zero

**Timbuk2**

But as x tends to infinity, doesn't 1/x tend to 0, and thus sin(1/x) tends to 0. As sin(1/x) tends to 0, xsin(1/x) gets closer and closer to 0 (as anything times 0 = 0), so shouldn't the limit be 0? But looking at the graph on Wolfram alpha, it is very different to 0. although 1 looks like a likely candidate!

These functions containing sin, like xsinx, xsin(x^2) and xsin(1/x) confuse me

LeixlipRed

Ahh, I was wrong actually, I didn't work it out. It is 1!

LeixlipRed

I should add that you can calculate it using L'Hoptals Rule.

Michael Collins

**Timbuk2** wrote: »

But as x tends to infinity, doesn't 1/x tend to 0, and thus sin(1/x) tends to 0. As sin(1/x) tends to 0, xsin(1/x) gets closer and closer to 0 (as anything times 0 = 0), so shouldn't the limit be 0? But looking at the graph on Wolfram alpha, it is very different to 0. although 1 looks like a likely candidate!

These functions containing sin, like xsinx, xsin(x^2) and xsin(1/x) confuse me

Some heuristic arguments:

[latex] \displaystyle \lim_{x\to\infty}x \sin\left(\frac{1}{x}\right) = 1 [/latex]

since as [latex] x \to \infty [/latex], [latex] \frac{1}{x} \to 0 [/latex] and we can write [latex] \sin\left(\frac{1}{x}\right) \approx \frac{1}{x}\right [/latex] so:

[latex] \displaystyle \lim_{x\to\infty}x \sin\left(\frac{1}{x}\right) = \lim_{x\to\infty} x \cdot \frac{1}{x} = 1[/latex]

As for

[latex] \displaystyle \lim_{x \to 0}x \sin\left(\frac{1}{x}\right) [/latex]

we know that [latex] \sin(\theta) [/latex] is bounded for real arguments, so max it can be is 1, so multiplying a finite number by zero gives zero (we're assuming the limit actually exists, of course), hence:


[latex] \displaystyle \lim_{x \to 0}x \sin\left(\frac{1}{x}\right) = 0[/latex]

COUCH WARRIOR

don't forget the lower limit so -1<=Sin(1/x)<=1 so you have

lim -x <= lim xSin(1/x) <= lim x giving (lim for limit as x tends to 0)

0 <= lim xSin(1/x) <= 0 or lim xSin(1/x) = 0