Proving the Limit of Zero Using Epsilon-Delta Method

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Epsilon-Delta proof of zero??

Homework Statement



Write an epsilon delta proof for the limx[itex]\rightarrow[/itex]2 0 = 0.


The Attempt at a Solution



This is for my discrete math class. I know how to do limit proofs with a variable, like x or x2, but it seems that this is obvious that the limit approaching zero is zero. It is so easy, that it is hard for me to use epsilon and delta to prove. I don't know really where to start for this one. I would appreciate some hints and help!
 
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savtaylor2010 said:

Homework Statement



Write an epsilon delta proof for the limx[itex]\rightarrow[/itex]2 0 = 0.


The Attempt at a Solution



This is for my discrete math class. I know how to do limit proofs with a variable, like x or x2, but it seems that this is obvious that the limit approaching zero is zero. It is so easy, that it is hard for me to use epsilon and delta to prove. I don't know really where to start for this one. I would appreciate some hints and help!

It might help to call f(x) = 0. Write down the definition as you would with f(x) and L and replace f(x) and L by 0.
 


Start by choosing an epsilon maybe let epsilon equal something really close to 0 like 0.2

and assume that x-2< alpha . so that you can so that 0<ε .
 


savtaylor2010 said:

Homework Statement



Write an epsilon delta proof for the limx[itex]\rightarrow[/itex]2 0 = 0.


The Attempt at a Solution



This is for my discrete math class. I know how to do limit proofs with a variable, like x or x2, but it seems that this is obvious that the limit approaching zero is zero. It is so easy, that it is hard for me to use epsilon and delta to prove. I don't know really where to start for this one. I would appreciate some hints and help!

hint: the proof is eactly the same if you replace 0 by c, where c is any constant. that is, if you replace the function f(x) = 0 with g(x) = c, the same argument works for both (the "delta" is really easy to find, for any "epsilon").
 


Thank you guys for all of your help! I know I am really late responding back, but all of the feedback really helped!