# Limit of xsin(1/x) as x→0

1. Dec 17, 2011

### PirateFan308

1. The problem statement, all variables and given/known data
$\displaystyle\lim_{x\rightarrow 0}~~ {xsin(1/x)}$

3. The attempt at a solution
I've attempted to solve this limit two different ways and get different answers.

Attempt #1:
If $(x_n)$ is a sequence and xn→0 then because sin(1/xn) is bounded xsin(1/x)→0. So $\displaystyle\lim_{x\rightarrow 0} ~~{xsin(1/x)}$=0

Attempt #2:
$\displaystyle\lim_{x\rightarrow 0} ~~{xsin(1/x)}$ = $\displaystyle\lim_{x\rightarrow 0}~~ {\frac{(1/x)(x)(sin(1/x))}{(1/x)}}$ = $(1/x)(x) ~~\displaystyle\lim_{x\rightarrow 0} ~~ {\frac{sin(1/x)}{(1/x)}}$= (1)(1) = 1

Can you guys tell me which is correct, and what is my error in the incorrect attempt? Thanks!

2. Dec 17, 2011

### Frogeyedpeas

Would you happen to be familiar with what a taylor series is? Because if you know the series function for sine then the answer would be very clear,

nevertheless here is another way to look at it:

What happens to expression 1/Xn as Xn approaches zero...

3. Dec 17, 2011

### Frogeyedpeas

Follow up to that what happens then to sine (1/Xn) as Xn approaches zero

4. Dec 17, 2011

### PirateFan308

as Xn→0, 1/Xn approaches ±∞. The limit as it approaches from the left does not equal the limit as it approaches from the right, so the limit of 1/Xn as Xn approaches 0 does not exist. So the limit of sin(1/xn) also doesn't exist?

Can you pinpoint exactly where I went wrong in my attempts or is this question simply a special case?

5. Dec 17, 2011

### Frogeyedpeas

don't think it does not exist JUST YET...

What is the range of sin(x)?

What happens to the actual graph as sin(1/x) approaches zero from -1 to 0 and from 1 to zero?

6. Dec 17, 2011

### Frogeyedpeas

The limit as x approaches zero of x * sin(1/x)

is taking the limit as x approaches zero and multiplying it with the limit as sin(1/x) approaches zero...

So we're trying to find out what happens to the behavior as it gets closer to zero...

Keep in mind that sin of anything is restricted to a range of [-1, 1]

One additional clue, The function sin(1/x) oscillates increasingly faster as x approaches zero so at zero itself its undefined because it could be any number in [-1, 1]...

But what is the value of x as it approaches zero? and what happens if you multiply the value of sin(1/x) with x? Does it converge back to an answer from being undefined or is it still undefined?

Lastly:

Recalculate the Limit as x approaches 0 for sin(1/x)/(1/x) and tell me what answer you get

7. Dec 17, 2011

### PirateFan308

The range of sin x is [-1,1], so the range of sin (1/x) is also [-1,1]. Because the limit of x as x→0 = 0, multiplying this by sin(1/x) will give us 0 (because range of sin(1/x) is bounded). So I would think that the limit of (x)(sin1/x) as x→0 would equal 0.

So the limit of (sin1/x)/(1/x) is not defined, because sin (1/x) stays between [-1,1] but (1/x) will increase/decrease to ±∞ as x approaches 0. Because the bottom is increasing/decreasing to ±∞, it the limit of (sin 1/x)/(1/x) is also 0. Is this now correct?

I'm curious as to why I can't use the property that the limit of (sin x)/x as x→0 = 1 for this example, plugging in (1/x) for x?

8. Dec 17, 2011

### kru_

Why can't you?

9. Dec 17, 2011

### vela

Staff Emeritus
Use the sandwich or squeeze theorem.

First you say the limit isn't defined, and then you say it's also 0. Which is it?

It's because 1/x goes to ∞ as x goes to 0, so you have
$$\lim_{x \to 0} \frac{\sin (1/x)}{1/x} = \lim_{u \to \infty} \frac{\sin u}{u}$$The second limit would equal 1 if u→0, but that's not what you have.

10. Dec 17, 2011

### PirateFan308

Could I also use the domination principle? Because sin(1/x) is between [-1,1] then |x*sin(1/x)| < |x| where x→0 so |x|→0 and so |x||sin(1/x)| also converges to 0?

Thank you for the explanation!!

11. Dec 17, 2011

### vela

Staff Emeritus
Another mistake in your second attempt is that you can't simply pull the factors containing x out of the limit since you're taking the limit as x→0. You can only pull constant factors out.

12. Dec 17, 2011

### SammyS

Staff Emeritus
As vela has already suggested, use the squeeze theorem.

-|x| ≤ x sin(1/x) ≤ |x|

13. Dec 17, 2011

### vela

Staff Emeritus
This is essentially an application of the squeeze theorem. You have $0 \le |x||\sin(1/x)| \le |x|$ (note it should be ≤, not <, because sin(1/x) can equal 1), and because both the left and right go to 0, the middle must also go to 0.

So you've shown that
$$\lim_{x \to 0}\ |x\sin(1/x)| = 0$$You still need to show why this implies
$$\lim_{x \to 0}\ x\sin(1/x) = 0$$

14. Dec 17, 2011

### PirateFan308

Is this simply because |x| = 0 then x=0? So $\lim_{x \to 0}\ x\sin(1/x)$ must equal 0 because its absolute value limit is 0?

15. Dec 17, 2011

### Karamata

$$\lim_{x\to 0}{\left|x \sin{\frac{1}{x}}\right|}=0$$
Or
$$\forall \varepsilon>0\hskip10pt \exists \delta>0 \hskip10pt\ 0<|x-0|<\delta \hskip10pt\ \left|\left|x\sin{\frac{1}{x}}\right|-0\right|<\varepsilon$$
But, $$\left|x\sin{\frac{1}{x}}\right|=\left|\left|x\sin{\frac{1}{x}}\right|\right|<\varepsilon$$
so we can say
$$\lim_{x\to 0}{x\sin{\frac{1}{x}}}=0$$
Is this OK?