Real Analysis Limits of Functions

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October 18th 2010, 06:23 AM
zebra2147

Real Analysis Limits of Functions

Suppose $lim_{x\longrightarrow x_{0}}f(x)=b$ and there is a $r>0$ such that $f(B'_{r}(a)) \subseteq [0,\infty)$ . Show that $b\geq 0$ .

I understand the definition of limit but I'm having trouble getting this proof started. Any help would be appreciated.
October 18th 2010, 07:15 AM
tonio

Quote:

Originally Posted by zebra2147

Suppose $lim_{x\longrightarrow x_{0}}f(x)=b$ and there is a $r>0$ such that $f(B'_{r}(a)) \subseteq [0,\infty)$ . Show that $b\geq 0$ .

I understand the definition of limit but I'm having trouble getting this proof started. Any help would be appreciated.

What is the relation between $x_0\,\,and\,\,a$ ? What is $B'_r(a)$ ? ....

Tonio
October 18th 2010, 07:36 AM
zebra2147

well... $x_{0}$ is a limit point of $f$ . And $a$ is a point that the ball $B'_{r}(a)$ is centered around but $a$ is not contained in $B'_{r}(a)$ .
October 18th 2010, 08:05 PM
tonio

Quote:

Originally Posted by zebra2147

well... $x_{0}$ is a limit point of $f$ . And $a$ is a point that the ball $B'_{r}(a)$ is centered around but $a$ is not contained in $B'_{r}(a)$ .

Oh, so a punctured ball...but then the claim is false: $f(x)=\left\{\begin{array}{rl}-1&,\,if\,\,\,x\leq 0\\2&,\,if\,\,\,x>0\end{array}\right.$ is such that

$f(x)\xrightarrow [x\to -1]{}-1\,,\,\,and\,\,\,f((1,2))\subset [0,\infty)$ , and nevertheless $b=-1<0$ ...

Note that $(1,2)=B_{1/2}(3/2)$ , and I don't care if you want to puncture it and take the middle point $3/2$ out or not.

It's not hard to come up with a counterexample where f is continuous, so I think something must be missing in your question...or I misunderstood, of course.

Tonio
October 19th 2010, 05:32 AM
zebra2147

Well, I looked over his exercise and I typed it in right, and typically he doesn't try and trick us. So are you saying that your example of $f(x)$ is continuous? Isn't that a jump discontinuity? Or am I confused?
October 19th 2010, 08:00 AM
tonio

Quote:

Originally Posted by zebra2147

Well, I looked over his exercise and I typed it in right, and typically he doesn't try and trick us. So are you saying that your example of $f(x)$ is continuous? Isn't that a jump discontinuity? Or am I confused?

No, it is not continuous. What I said is that's easy to give a continuous counterexample to your claim, and that's why I thought some data must be missing.

Tonio
October 19th 2010, 09:47 AM
zebra2147

Oh ok. I apologize for me misunderstanding. As far as I can see there is no information missing so maybe he left something out in his notes.
October 19th 2010, 10:29 AM
Plato

I rather suspect that in the OP the $a$ should be $x_0$ . Or visa versa.
October 19th 2010, 11:58 AM
tonio

Quote:

Originally Posted by Plato

I rather suspect that in the OP the $a$ should be $x_0$ . Or visa versa.

That's exactly what I thought, but when I asked I was told all is fine...if $a=x_0$ then the claim is straightforward.

Tonio