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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

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If $a$ is positive, there is an open interval of positive reals that contains $a$, and $f$ is constantly $1$ on this interval. Now follow the solution suggested by Alex. – Chris Leary Apr 18 '12 at 17:13
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 $1$. – Mikko Korhonen Apr 18 '12 at 17:18
up vote 4 down vote accepted

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.

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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.

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Okay but if $a$ is close to $0$ then in a neighborhood of $a$ I will have negative $x$ values, and $f(x)$ will have a different result. Will it not matter? – yotamoo Apr 18 '12 at 17:21
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. – Adam Rubinson Apr 18 '12 at 18:56

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

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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. – user21436 Apr 18 '12 at 17:19

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