Proving Limit of x^3sin(1/x) = 0 as x approaches 0

[AimlessWander]

Homework Statement

Prove:

Limit of x³sin(1/x)=0 as x→0

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)?

 

[Mark44]

Homework Statement

Prove:

Limit of x³sin(1/x)=0 as x→0

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)?

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.

 

[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<ε
|x³sin(1/x)|≤x³<ε

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

 

[LCKurtz]

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<ε
|x³sin(1/x)|≤x³<ε

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

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.

 

[AimlessWander]

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.

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