Proving lim x→a f(x) = 0 for All Nonzero x and f(0)=c

[dmatador]

If f(x) = 0 for all nonzero x and f(0) = some constant c, how can you show that lim = 0 for any x as x approaches a?

I tried using the definition of limit, but this is going nowhere.

 

[HallsofIvy]

The definition of limit should work nicely. Show what you did and we may be able to give tips.

 

[dmatador]

|f(x) - L| = |f(x) - 0| = |f(x)| < ϵ.

Given the definition of the function, the only way this holds is if x is nonzero.

At this point I tried to consider 0 < |x - a| < δ. I'm trying to show that you can pick a δ such that i dunno...

 

[HallsofIvy]

If a is 0, any \delta will work. If a is not 0, any \delta&lt; |a| will work.

The definition of 'limit' only requires that there exist such a \delta it doesn't say it is unique- in fact, it can't be given an \delta that works, any positive value less than that will also work.

 

[dmatador]

I know that I need only to find a sing delta given an epsilon, but without knowing a, what good is knowing that |x - a| < 1?, for instance. What about considering the open interval (-∞, 0) U (0, ∞)? With this, I can show that for any x in I, the limit goes to zero, correct?

Could you elaborate on the second inequality please?

 

[Maxter]

Actually, taking a limit as you approach x₀=a requires that you look at all points close to x₀=a but not equal to it.

That is, you are looking for all values x where 0<|x-a|<\delta, which means that \delta can be any positive number - the zero part of the number says that you won't be looking "at a". For fun, take delta to be pi

In summary, you never have to worry about x being equal to a, no matter what delta is, since, by definition, that can't happen when taking a limit. It's when you are looking at continuity that you have to worry about that.

 

[dmatador]

I thank for your replies. I was definitely making it much harder than it should have been.