# Does the lim x→0 of (x^4)(sin1/x) exist?

1. Dec 18, 2011

### smeiste

1. The problem statement, all variables and given/known data
I need to solve this limit.
lim (x^4)(sin1/x)
x→0

2. Relevant equations

3. The attempt at a solution

Since as 1/x → 0+, the limit is +∞ but when 1/x → 0-, the limit is -∞. Since these two-sided limits don't match I thought the limit could not be calculated. apparently this is incorrect. any thoughts?

2. Dec 18, 2011

### USN2ENG

Yes it does exist. Basically you have to rearrange the function (I know I'm not using great math vocab here).

You take x^4*sin(1/x) and make it x^4 / (1/sin(1/x)). you then have 0/1...which the limit is 0.

Correct me if I am way off but that is how I would do it.

Last edited: Dec 18, 2011
3. Dec 18, 2011

### awkward

The limit as x approaches 0 of sin(1/x) does not exist, so you can't use the usual theorems about products (or quotients) of limits.

But since $|\sin x| \leq 1$,
$$|x^4 \sin(1/x)| = |x^4| \cdot |\sin(1/x)| \leq |x^4| \cdot 1$$
so the limit does exist and is zero.

4. Dec 18, 2011

### USN2ENG

Yep, thats right.

5. Dec 18, 2011

### smeiste

awesome. thanks guys!

6. Dec 18, 2011

### D H

Staff Emeritus
The problem isn't asking about 1/x. It's asking about sin(1/x), and in particular, the it is asking about the product of x4 with the sine of 1/x. You are correct in the sense that sin(1/x) is undefined as x→0. It is however bounded: $-1\le \sin(1/x) \le 1$ for all x. Thus $-x^4\le x^4\sin(1/x)≤x^4$. What does the squeeze theorem say about this as x→0?

7. Dec 18, 2011

### smeiste

that since the limits on both of those sides equals 0 then the limit must also equal 0. that is extremely helpful. Thank you!