1
$\begingroup$

Given the function $f(x)= \left\{\begin{matrix} 1 & x \gt 0 \\ 0 & x =0 \\ -1 & x \lt 0 \end{matrix}\right\}$

What is $\lim_{a}f$ for all $a \in \mathbb{R},a \gt 0$?

It seems easy enough to guess that the limit is $1$, but how do I take into account the fact that $f(x)=-1$ when $x \lt 0$? Thanks

  • 0
    With limits, the only thing that matters is how $f$ behaves near $a$. Since a > 0, if we get close enough to $a$, all the numbers are positive and $f(x) = 1$ for any $x$ close enough to $a$. Thus the limit equals $1$.2012-04-18

3 Answers 3

4

For $a>0$, let $\delta=a$. Then for any $\epsilon>0$, we have that |x-a|<\delta\implies x>0\implies f(x)=1\implies |f(x)-1|<\epsilon hence $\lim\limits_{x\to a}f(x)=1$, as we have satisfied the definition of limit.

0

Well for a>0, a/2 is between 0 and a, so all x in [a/2,b] with b>a have the property f(x) = 1. Moreover, the question asks you to find the limit for all a>0. You don't need to worry what the value of f is for x =< 0, only what it is for x>0.

  • 0
    No. If$a$> 0, then no matter how close$a$is to 0, a/2 will ALWAYS be between$0$and a. This is a property of the real numbers and is the main reason as to why the real numbers is different from the integers... because between ANY two real numbers b and c there is a real number (namely b+c)/2, whereas for the integers, this is not always true, e.g. there is no integer between 5 and 6.2012-04-18
-3

Here $\lim_{x\to 0^+}f(x)=1$ & $\lim_{x\to 0^-}f(x)=-1$ both limit are not equal.

therefore limit is not exists

  • 0
    What you note is right; but does not answer the question. Please read the question before you'd answer. (I have been a victim of this disease myself. :)) BTW, I did not downvote.2012-04-18