Prove if abs|f(x)| <= B for all x#0, then lim as x->0 of xf(x) = 0,

[anthonyg]

1. if there is a number B such that abs|f(x)| <= B for all x#0, then lim as x->0 of xf(x) = 0,2. Would I be able to get some help on how to prove this?3. Given Epsilon > 0 such that delta = epsilon/B will prove the limit equals 0. (I know that may not make much sense but how am I supposed to make a solution when I don't know where to start... but I was warned so here is my attempt.) :|

 

[SammyS]

Welcome to PF .

It looks like a fairly direct δ-ε proof.

What have you tried?

Where are you stuck?

 

[anthonyg]

Thanks, the thing is I don't know where to start. I mean my teacher always starts out with given ε > 0 such that δ... I honestly just don't know where to go either.

 

[Mark44]

Thanks, the thing is I don't know where to start. I mean my teacher always starts out with given ε > 0 such that δ... I honestly just don't know where to go either.

I think your teacher starts out with a given ε > 0, and then figures out what δ needs to be.

Look at your text or class notes for some examples.

 

[anthonyg]

Oh, yea, thanks! I didn't think of that!