# Real Analysis Limits of Functions

• Oct 18th 2010, 06:23 AM
zebra2147
Real Analysis Limits of Functions
Suppose $\displaystyle lim_{x\longrightarrow x_{0}}f(x)=b$ and there is a $\displaystyle r>0$ such that $\displaystyle f(B'_{r}(a)) \subseteq [0,\infty)$. Show that $\displaystyle b\geq 0$.

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

Originally Posted by zebra2147
Suppose $\displaystyle lim_{x\longrightarrow x_{0}}f(x)=b$ and there is a $\displaystyle r>0$ such that $\displaystyle f(B'_{r}(a)) \subseteq [0,\infty)$. Show that $\displaystyle b\geq 0$.

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 $\displaystyle x_0\,\,and\,\,a$ ? What is $\displaystyle B'_r(a)$ ? ....

Tonio
• Oct 18th 2010, 07:36 AM
zebra2147
well...$\displaystyle x_{0}$ is a limit point of $\displaystyle f$. And $\displaystyle a$ is a point that the ball $\displaystyle B'_{r}(a)$ is centered around but $\displaystyle a$ is not contained in $\displaystyle B'_{r}(a)$ .
• Oct 18th 2010, 08:05 PM
tonio
Quote:

Originally Posted by zebra2147
well...$\displaystyle x_{0}$ is a limit point of $\displaystyle f$. And $\displaystyle a$ is a point that the ball $\displaystyle B'_{r}(a)$ is centered around but $\displaystyle a$ is not contained in $\displaystyle B'_{r}(a)$ .

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

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

Note that $\displaystyle (1,2)=B_{1/2}(3/2)$ , and I don't care if you want to puncture it and take the middle point $\displaystyle 3/2$ 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
• Oct 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 $\displaystyle f(x)$ is continuous? Isn't that a jump discontinuity? Or am I confused?
• Oct 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 $\displaystyle f(x)$ 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
• Oct 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.
• Oct 19th 2010, 10:29 AM
Plato
I rather suspect that in the OP the $\displaystyle a$ should be $\displaystyle x_0$. Or visa versa.
• Oct 19th 2010, 11:58 AM
tonio
Quote:

Originally Posted by Plato
I rather suspect that in the OP the $\displaystyle a$ should be $\displaystyle x_0$. Or visa versa.

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

Tonio