Show that there is a M>0 such that |f(x)|<M

[mathmari]

Hey!

I have the following exercise:
It is given that $f:R \to R$ continuous and $\lim_{x \to -\infty}f(x)=-5$ & $\lim_{x \to +\infty}f(x)=5$.Show that there is a $M>0 \in R$ such that $|f(x)|<M, \forall x \in R$.Can you find a value for $M$?

Could you tell me if that what I have tried is right or what I could change??

$$\lim_{x \to -\infty}f(x)=-5:$$
$ \forall \epsilon>0 $ there is a $c>0$ such that $\forall x<-c \Rightarrow |f(x)-(-5)|<\epsilon \Rightarrow - \epsilon < f(x)+5< \epsilon \Rightarrow - \epsilon -5< f(x)< \epsilon -5$$ $(1)
$$\lim_{x \to +\infty}f(x)=5:$$
$ \forall \epsilon ' >0$ (also for $\epsilon ' =\epsilon$) there is a $c'>0$ such that $\forall x>c' \Rightarrow |f(x)-5|< \epsilon \Rightarrow -\epsilon < f(x) -5 < \epsilon \Rightarrow 5- \epsilon < f(x)< \epsilon +5 $ $(2)$

From the relations $(1)$ and $(2)$ we get $$- \epsilon -5<f(x)< \epsilon -5 < \epsilon +5$$
$$ - \epsilon +5 <f(x)< \epsilon +5$$
So $|f(x)|< \epsilon +5$.

 

[Evgeny.Makarov]

So $|f(x)|< \epsilon +5$.

This graph (clickable) begs to differ.

[GRAPH]oatb2hhtr1[/GRAPH]

 

[mathmari]

This graph (clickable) begs to differ.

[GRAPH]oatb2hhtr1[/GRAPH]

So is the way I tried to solve it wrong, or just the value of $M$?

 

[Evgeny.Makarov]

\[\lim_{x \to -\infty}f(x)=-5:\]
$ \forall \epsilon>0 $ there is a $c>0$ such that $\forall x<-c \Rightarrow |f(x)-(-5)|<\epsilon \Rightarrow - \epsilon < f(x)+5< \epsilon \Rightarrow - \epsilon -5< f(x)< \epsilon -5$$ $(1)
$$\lim_{x \to +\infty}f(x)=5:$$
$ \forall \epsilon ' >0$ (also for $\epsilon ' =\epsilon$) there is a $c'>0$ such that $\forall x>c' \Rightarrow |f(x)-5|< \epsilon \Rightarrow -\epsilon < f(x) -5 < \epsilon \Rightarrow 5- \epsilon < f(x)< \epsilon +5 $ $(2)$

Up to here everything is correct. I recommend you take the graph as a counterexample and use it to find the error in the last part of your post. Pick a specific value of $\epsilon$, say, $\epsilon=1$, see if (1) and (2) are true and whether the conclusion you derive from them is true.

It is clear that the peak of the graph can be made as tall as possible, so it is not possible to give an upper bound in advance that works for all functions. Another hint is that you should use the fact that a continuous function on a closed segment is bounded (Wikipedia). The challenge is to find that closed segment.

 

[Plato2]

It is given that $f:R \to R$ continuous and $\lim_{x \to -\infty}f(x)=-5$ & $\lim_{x \to +\infty}f(x)=5$.Show that there is a $M>0 \in R$ such that $|f(x)|<M, \forall x \in R$.Can you find a value for $M$?

You do not have to be that abstract. Use the theorem: If f is continuous on [-n,n] then it is bounded there.

From the given:
$\left( {\exists {M_1} \in {\mathbb{Z}^ + }} \right)\left[ {x \geqslant {M_1} \Rightarrow \left| {f(x) - 5} \right| < 1} \right]$ & $\left( {\exists {M_2} \in {\mathbb{Z}^ + }} \right)\left[ {x \le -{M_2} \Rightarrow \left| {f(x) + 5} \right| < 1} \right]$

In each of those cases $|f(x)|<6$.

Now let $N=M_1+M_2$. The continuous function $f$ bounded on $[-N,N]$ by some $B>0$

So $\forall x\in\mathbb{R},~|f(x)|<B+6$

 

[mathmari]

Pick a specific value of $\epsilon$, say, $\epsilon=1$, see if (1) and (2) are true and whether the conclusion you derive from them is true.

For $\epsilon=1$ we get from the relation $(1)$ $-6 <f(x)<-4$, and from the relation $(2)$ we get $4<f(x)<6$.
Does this mean that we cannot say that $|f(x)|<6$?

It is clear that the peak of the graph can be made as tall as possible, so it is not possible to give an upper bound in advance that works for all functions. Another hint is that you should use the fact that a continuous function on a closed segment is bounded (Wikipedia). The challenge is to find that closed segment.

To find that closed segment do we have to use the relations $(1)$ and $(2)$?

 

[mathmari]

Now let $N=M_1+M_2$. The continuous function $f$ bounded on $[-N,N]$ by some $B>0$

So $\forall x\in\mathbb{R},~|f(x)|<B+6$

I got stuck...Could you explain me why |f(x)|<B+6 ?

 

[Plato2]

I got stuck...Could you explain me why |f(x)|<B+6 ?

You have three cases:
1) $x\in (-\infty, N)$ then $|f(x)+5|<1$ implies that $|f(x)|<6$

2) $x\in (N,\infty)$ then $|f(x)-5|<1$ implies that $|f(x)|<6$

3) $x\in [-N, N]$ implies that $|f(x)|<B$

 

[mathmari]

You have three cases:
1) $x\in (-\infty, N)$ then $|f(x)+5|<1$ implies that $|f(x)|<6$

2) $x\in (N,\infty)$ then $|f(x)-5|<1$ implies that $|f(x)|<6$

3) $x\in [-N, N]$ implies that $|f(x)|<B$

Ok, I understand!

And could we also say $|f(x)|< \max \{6, B\}$ instead of $|f(x)|<B+6$ ?

 

[Plato2]

And could we also say $|f(x)|< \max \{6, B\}$ instead of $|f(x)|<B+6$ ?

Well of course you can do do that, but why would you?

 

[mathmari]

Well of course you can do do that, but why would you?

Ok! I just wanted to know...

Thank you very much for your answer!

 

[Deveno]

It appears to me that all we can say is $M$ exists, we have no way of "finding a value for it" (such a function might have an arbitrarily large, but finite maximum, for example).