# Thread: Limits of Functions in Real Analysis

1. ## Limits of Functions in Real Analysis

Alright, I need some guidance with the following problem. Any help would be appreciated.
Suppose $\displaystyle lim_{x\rightarrow x_{0}}f(x)=b$ and given any $\displaystyle r>0$ the set $\displaystyle f(B'_{r}(a))$ contains both poisitive and negative real numbers,show that $\displaystyle b=0$.

After I talked to some people it sounds like there may be a typo???
Possibly $\displaystyle x_{0}=a$ in this problem? Either way I'm not really sure how to show that b=0.

2. Originally Posted by zebra2147
Alright, I need some guidance with the following problem. Any help would be appreciated.
Suppose $\displaystyle lim_{x\rightarrow x_{0}}f(x)=b$ and given any $\displaystyle r>0$ the set $\displaystyle f(B'_{r}(a))$ contains both poisitive and negative real numbers,show that $\displaystyle b=0$.

After I talked to some people it sounds like there may be a typo???
Possibly $\displaystyle x_{0}=a$ in this problem? Either way I'm not really sure how to show that b=0.

It must be, again, $\displaystyle a=x_0$ : just think that when $\displaystyle x\rightarrow x_0$ , x gets into ANY neighborhood of $\displaystyle x_0$, so if in ANY such

neighborhood there are positive and negative values of f AND the limit exists, this limit must be zero.

Try now to give a proof with $\displaystyle \epsilon\,,\,\delta$ and/or a proof bya contradiction.

Tonio

3. Attempt #1 at proof:

Suppose $\displaystyle D=D_{1}\bigcup D_{2}$ where $\displaystyle D_{1}$ contains only negative real numbers and $\displaystyle D_{2}$ contains only positive real values. Let $\displaystyle a$ be a limit point of $\displaystyle D_{1}$ and $\displaystyle D_{2}$. Then let $\displaystyle f: D\rightarrow R$ be the function $\displaystyle f(B'_{r}(a))$. Thus, the $\displaystyle lim_{x\longrightarrow a}$ exists only if $\displaystyle lim_{x\longrightarrow a}f|_D_{j}$ both exist and are equal.

Then, if both limits exist we have that, for some $\displaystyle \epsilon>0$,when $\displaystyle x_{1}<0$, $\displaystyle |x_{1}-a|<\delta\Rightarrow |f(B'_{r}(a))-b|<\epsilon$. Also, when $\displaystyle x_{2}>0$, $\displaystyle |x_{2}-a|<\delta\Rightarrow |f(B'_{r}(a))-b|<\epsilon$.

The only way that this can happen is when $\displaystyle b=0$?????

4. Originally Posted by zebra2147
Attempt #1 at proof:

Suppose $\displaystyle D=D_{1}\bigcup D_{2}$ where $\displaystyle D_{1}$ contains only negative real numbers and $\displaystyle D_{2}$ contains only positive real values. Let $\displaystyle a$ be a limit point of $\displaystyle D_{1}$ and $\displaystyle D_{2}$. Then let $\displaystyle f: D\rightarrow R$ be the function $\displaystyle f(B'_{r}(a))$. Thus, the $\displaystyle lim_{x\longrightarrow a}$ exists only if $\displaystyle lim_{x\longrightarrow a}f|_D_{j}$ both exist and are equal.

Then, if both limits exist we have that, for some $\displaystyle \epsilon>0$,when $\displaystyle x_{1}<0$, $\displaystyle |x_{1}-a|<\delta\Rightarrow |f(B'_{r}(a))-b|<\epsilon$. Also, when $\displaystyle x_{2}>0$, $\displaystyle |x_{2}-a|<\delta\Rightarrow |f(B'_{r}(a))-b|<\epsilon$.

The only way that this can happen is when $\displaystyle b=0$?????

I think you confused stuff here big time: the positive and negative numbers are in the image of f , NOT in its domain!

We know that $\displaystyle \forall\epsilon>0\,\exists\delta_\epsilon>0\,\,s.t .\,\,x\in B_{\delta_\epsilon}(x_0)\Longrightarrow f(x)\in B_\epsilon(b)$.

Suppose $\displaystyle b>0$ , and let us choose $\displaystyle \epsilon=b/2$ . By the above, we get that $\displaystyle f(x)\in B_\epsilon(b)$ for any

point $\displaystyle x\in B_{\delta_\epsilon}(x_0)$ , for some $\displaystyle \delta_\epsilon>0$ ,or in other words: $\displaystyle f\left(B_{\delta_\epsilon}(x_0)\right)\subset B_\epsilon(b)$ ...but this can't be since

in $\displaystyle f\left(B_{\delta_\epsilon}(x_0)\right)$ there's some negative number, whereas in $\displaystyle B_\epsilon(b)$ all the numbers are positive...

Read the above slowly and with a pencil and a piece of paper by your side...draw diagrams if that helps you.

Tonio

5. Oh ok. That makes more sense. Then I'm assuming that you showed how b>0 does not work, so if we show that b<0 does not work by a similar proof then b=0 must be true?

6. Originally Posted by zebra2147
Oh ok. That makes more sense. Then I'm assuming that you showed how b>0 does not work, so if we show that b<0 does not work by a similar proof then b=0 must be true?

Of course. I think you don't even need to show for $\displaystyle b<0$ since it is plain that this case is completely similar to the one already proven.

Tonio