# Prove Limit of x^3sin(1/x)

1. Feb 5, 2014

### AimlessWander

1. The problem statement, all variables and given/known data

Prove:

Limit of x3sin(1/x)=0 as x→0

3. The attempt at a solution

Intuitively, I understand that since sin(1/x) always lies between 1 and -1, and since x approaches zero, x3sin(1/x) must also approach zero. When formally expressing proof, however, which do we restrict, x3 or sin(1/x)?

2. Feb 5, 2014

### Staff: Mentor

There are two things I don't understand.
1. How formally do you need to prove this? I.e., do you need to do a δ-ε argument, or can you use a theorem?
2. What do you mean "which do we restrict"? The variable x is approaching zero, so both x3 and sin(1/x) are affected.

3. Feb 5, 2014

### AimlessWander

Preferably delta-epsilon. I know how to prove it with squeeze theorem, but I got stuck at δ-ε proof.
I think I have an idea of how to prove it with δ-ε, but I'm not sure.
EDIT: Oh, and by restrictions, I mean what is taken as δ so we can focus on one function of x, since it's not multiplied by a constant, rather, another function of x.

so if ε>0

-1≤sin(1/x)≤1
x3 |sin(1/x)|≤x3
|x3sin(1/x)|≤x3

since |x3sin(1/x)| = |[x3sin(1/x)]-0| <ε and |x3| <ε
x3
x<cubic√ε
|x|<cubic√ε
|x-0|<cubic√ε

let δ=cubic√ε
0<|x-0|<cubic√ε
0<|x-0|<δ
|[x3sin(1/x)]-0| = |f(x)-0| <ε

4. Feb 5, 2014

### LCKurtz

You definitely have the correct idea and a good argument. But you could write it up a bit nicer. Since you have already solved it, I will suggest how you should write up your argument:

Suppose $\epsilon > 0$. Pick $\delta = \sqrt[3]\epsilon$. Then if $0<|x-0|<\delta$$$\left| x^3\sin(\frac 1 x)\right| =\left| x^3\right|\left| \sin(\frac 1 x)\right| \le |x^3|\cdot 1 < (\sqrt[3]\epsilon)^3=\epsilon$$Note that you can't allow $x=0$ in the argument.

5. Feb 5, 2014

### AimlessWander

Awesome, yeah that's a lot cleaner and less redundant than mine. Thanks so much! Really appreciate it. :)