# Homework Help: Find the limit

1. Apr 22, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
$\lim_{x->0} x\sin\frac{1}{x}$

3. 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.

2. Apr 22, 2014

### LCKurtz

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

3. Apr 22, 2014

### utkarshakash

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

4. Apr 22, 2014

### LCKurtz

So....?

5. Apr 22, 2014

### 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.

6. Apr 22, 2014

### LCKurtz

7. Apr 22, 2014

### Yanick

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

8. Apr 22, 2014

### utkarshakash

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

9. Apr 22, 2014

### LCKurtz

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

10. Apr 22, 2014

### ChrisVer

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

11. Apr 23, 2014

### utkarshakash

When x approaches 0 sin(1/x) approaches 1. So the limit should be 0, right?

12. Apr 23, 2014

### Saitama

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. Apr 23, 2014

### utkarshakash

But the correct answer is 1 according to the solution.

14. Apr 23, 2014

### Saitama

15. Apr 23, 2014

### Curious3141

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. Apr 23, 2014

### utkarshakash

I too think the given answer is incorrect.

17. Apr 23, 2014

### LCKurtz

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

Last edited: Apr 23, 2014
18. Apr 23, 2014

### vthem

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

19. Apr 23, 2014

### utkarshakash

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

$-1 \leq \sin \frac{1}{x} \leq 1 \\ -x \leq \sin \frac{1}{x} \leq x \\ lim_{x \to 0} -x = lim_{x \to 0} x = 0$

The limit should therefore be 0.

Is my reasoning correct?

20. Apr 23, 2014

### LCKurtz

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 $\left| x\sin(\frac 1 x)\right|$.

 Note lurflurf's following comment about the missing $x$ which I just added.

Last edited: Apr 23, 2014
21. Apr 23, 2014

### 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

Last edited: Apr 23, 2014
22. Apr 24, 2014

### utkarshakash

$0 \leq \left|\sin(\frac 1 x)\right| \leq 1 \\ 0 \leq \left|x\sin(\frac 1 x)\right| \leq |x|$

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

23. Apr 24, 2014

### vthem

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 $\lim_{x \to α} |f(x)|$

Last edited: Apr 24, 2014
24. Apr 24, 2014

### utkarshakash

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

25. Apr 24, 2014

### vthem

I just edited it!