- Thread starter
- #1

- Apr 14, 2013

- 4,601

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