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

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Hey!!! :eek:

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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
\[\lim_{x \to -\infty}f(x)=-5:\]
$ \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.
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
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 ?
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
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 $|f(x)|<B+6$ ?
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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-known member
MHB Site Helper
Apr 14, 2013
4,047
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!!! :eek:
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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).