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**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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