Epsilon-Delta Proof for Limit of x^2sin(1/x) as x Approaches 0

[rman144]

I need to show that:

$\lim_{x \to 0} x^{2}sin(1/x)=0$

Using the epsilon-delta method. I figured delta=sqrt[epsilon] would make the limit hold, but wanted to be sure. Thanks in advance.

 

[╔(σ_σ)╝]

$\lim_{x \to 0} x^{2}sin(1/x)=0$

I believe it can be show that if $$\epsilon$$ = $$\delta$$$abs(x^{2}sin(1/x))\leq x^{2}$ < $$\epsilon$$

 

[Liwuinan]

$$\delta = \sqrt{\varepsilon}$$ is ok too, |sin(...)| will be always less or equal to 1, so $$\varepsilon \left|\sin(1/\sqrt{\varepsilon})\right| \leq \varepsilon$$

 

[╔(σ_σ)╝]

$$\delta = \sqrt{\varepsilon}$$ is ok too, |sin(...)| will be always less or equal to 1, so $$\varepsilon \left|\sin(1/\sqrt{\varepsilon})\right| \leq \varepsilon$$

This is more sufficient.:)