Prove lim x->0 f(x) = 0, if there is a number B?

[LilTaru]

Prove lim x-->0 f(x) = 0, if there is a number B?

Homework Statement

Prove that if there is a number B such that |f(x)/x| <= B for all x # 0, then lim x-->0 f(x) = 0.

Homework Equations

The Attempt at a Solution

I don't understand how B comes into play in this question, or how I use it to prove the limit equals 0? I thought of using it in the epsilon-delta proof of a limit, but I don't know how that would work with it being |f(x)/x| <= B... We learned the pinching theorem earlier in the chapter and I thought that could help, but again I don't know where to begin! Help will be most appreciated!

 

[Char. Limit]

I don't know if this would work, but have you tried multiplying both sides by x and then taking the limit?

 

[LilTaru]

So, it would be -Bx < f(x) < Bx? I got that far in one of my many side attempts of solving... let's see if I can take it all the way to a full solution!

 

[Char. Limit]

Well, what do the two outer sides approach as x approaches zero?

 

[micromass]

You need to prove that

$$\forall \epsilon>0: \exists \delta>0: \forall x: |x|<\delta~\Rightarrow~|f(x)|<\epsilon$$

You have given that $$|f(x)|< B|x|$$. It seems you need to take $$\delta=\epsilon/B$$.

 

[Char. Limit]

Why would you do all that, when you can just use the squeeze theorem?

 

[micromass]

I just like using the definition instead of theorems

 

[Char. Limit]

Meh. I like using theorems because they make it easier. I don't like having to start from scratch every time I do a problem.

 

[LilTaru]

I think a light bulb went off in my head... hopefully! So using the squeeze theorem: since -Bx and Bx both approach zero as x approaches zero then f(x) also approaches zero! Making the lim x-->0 f(x) = 0? Hopefully this is correct because then it all actually makes sense :D!

 

[Char. Limit]

That seems correct, yeah.

 

[LilTaru]

Awesome! Thank you very much! It makes a lot more sense now! :D

 

[Char. Limit]

No problem at all.