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

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.) :|

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SammyS
Staff Emeritus
Homework Helper
Gold Member
Welcome to PF .

It looks like a fairly direct δ-ε proof.

What have you tried?

Where are you stuck?

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
Mentor
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.

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