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

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  • #1
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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!
 

Answers and Replies

  • #2
Char. Limit
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I don't know if this would work, but have you tried multiplying both sides by x and then taking the limit?
 
  • #3
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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!
 
  • #4
Char. Limit
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Well, what do the two outer sides approach as x approaches zero?
 
  • #5
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You need to prove that

[tex]\forall \epsilon>0: \exists \delta>0: \forall x: |x|<\delta~\Rightarrow~|f(x)|<\epsilon[/tex]

You have given that [tex]|f(x)|< B|x|[/tex]. It seems you need to take [tex]\delta=\epsilon/B[/tex].
 
  • #6
Char. Limit
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Why would you do all that, when you can just use the squeeze theorem?
 
  • #7
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I just like using the definition instead of theorems :smile:
 
  • #8
Char. Limit
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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.
 
  • #9
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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!
 
  • #10
Char. Limit
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That seems correct, yeah.
 
  • #11
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Awesome! Thank you very much! It makes a lot more sense now! :D
 
  • #12
Char. Limit
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No problem at all.
 

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