Prove a function is continuous at a point (2)

[Hernaner28]

As in my previous thread we had:
"Let f a function which satisfies $$|f(x)|\leq|x| \forall{x\in{\mathbb{R}}}$$

Proof that is continuous at 0.
We concluded that since f(0)=0 then we found a delta equal epsilon so $$|f(x)|≤|x|<ϵ$$.

But now I have:

$$\textrm{g continuous at 0 and g(0)=0}\Longleftrightarrow{\displaystyle\lim_{x \to{0}}{g(x)}=g(0)=0} $$
$$ \left . \begin{matrix}{g(0)=0}\\{|f(x)|\leq |g(x)|}\end{matrix}\right \} \Longrightarrow{}f(0)=0$$

And I have to prove that f is cont. at 0.

So I have to fid a delta such that:
$$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \varepsilon}$$

And I'm in the same problem as before, right? I just choose δ equal ε . If this is OK then I don't understand why it tells me that g is cont. at 0 because I never used that.

Thank you!

 

[Fredrik]

Consider the g defined by
$$
g(x)=
\begin{cases}
0 &\text{if }x=0\\
1 &\text{if }x\neq 0
\end{cases}
$$ and the f defined by
$$
f(x)=
\begin{cases}
0 &\text{if }x=0\\
\frac 1 2 &\text{if }x\neq 0.
\end{cases}
$$

 

[Bearded Man]

You do use g is cont at 0 by writing
I. if |x| < delta, then |(g(x) - g(0)| = |g(x)| < epsilon
Because |f(x)| <= |g(x)|,
if |x| < delta, then modify I. by substituting |f(x)| for |g(x)| to get
if |x| < delta, then |f(x)| < epsilon

In this case, delta = delta. In the last case, you substituted in the |x-a| < delta inequality, in this one you substituted in the epsilon's inequality.

 

[Hernaner28]

Ahmm... so what I did was right? Thanks

 

[Fredrik]

No, because the choice δ=ε doesn't work. To "choose δ=ε" is to say that for all x such that |x|<ε, we have |f(x)|<ε. But how would you know that the latter inequality is implied by the former? You only know that |f(x)|≤|g(x)|. So for the choice δ=ε to work, you need something that proves that for all x such that |x|<ε, we have|g(x)|<ε.

 

[Hernaner28]

From ALL what I know (from the hypothesys, including g) I can say that:
I have to a delta such that this condition is OK:

$$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \varepsilon}$$

So if I choose epsilon equal delta:
$$|x|< \varepsilon$$ and this implies:
$$|f(x)|< \varepsilon$$

EDIT: Sorry I got confused, I cannot deduce the latter inequality. But now I look at it... as Bearded Man said I have that $$|g(x)|<\varepsilon$$... so that may be useful..

 

[Hernaner28]

I think I got something, please anybody verify if this is right:

This is what I have to proove:
$$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \varepsilon}$$

And I know that g is cont. at 0 so:
$$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|g(x)-0|< \varepsilon}$$
And
$$|f(x)|\leq |g(x)|$$

So having that two conditions I can deduce that
$$|f(x)|<\varepsilon$$ and therefore I can conclude that f is continuous at 0... am I missing sth?

 

[Fredrik]

This is the right way to do it. Just make sure to mention that the delta that you use for f is the one you got from the stuff you know about g.

 

[Hernaner28]

OK, yes, I am supposing that epsilon and delta are the same numbers for g and f. Thank you very much!