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 = \deltaabs(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.:)