Precise Definition of the Limit for f(x)=0

[nietzsche]

Homework Statement

Suppose f(x) = 0. Prove, using the precise definition of the limit, that

<br /> \begin{equation*}<br /> \lim_{x\to a} f(x) = 0<br /> \end{equation*}<br />

Homework Equations

The Attempt at a Solution

I just learned the precise definition of the limit today, and I thought of this example and I couldn't really figure it out. As I understand it, we want to find

<br /> 0&lt;|x-a|&lt;\delta\\<br /> \implies 0&lt;|f(x)-0|&lt;\varepsilon<br />

What I don't get is, does it make sense to say that we're looking for epsilon such that 0 is LESS THAN |f(x)-0| is less than epsilon? If f(x) never deviates from 0 anyway, then f(x) will always be less than epsilon. But isn't |f(x)-L| = |f(x)-0| = |f(x)| always EQUAL to zero?

I'm also still not entirely sure I understand the concept.

 

[Office_Shredder]

I've never seen the 0<|f(x)-0| portion included in the definition of a limit before. Wherever you got the definition from has a type in it, just ignore that part

 

[aPhilosopher]

This is a very simple example but you can still carry the formalism through. If you are just doing this for your own benefit, it might be more instructive to look at the functions f(x) = x, g(x) = 2x and h(x) = x² and prove the limit is 0 at 0. Then go back and look at this one. These ones have enough meat on them to give you some traction.

And what Office Shredder said is true as well. I've seen =< but that is implicit anyways.

 

[nietzsche]

I've never seen the 0<|f(x)-0| portion included in the definition of a limit before. Wherever you got the definition from has a type in it, just ignore that part

Oh okay, I think I may have copied it down wrong then. Thanks for clearing that up.

 

[Mark44]

For this problem, the 0 < |f(x) - 0| part is incorrect, since f(x) is identically zero.