Proove that a function is cointinuous at one point

[emailanmol]

I've done this so far and I think it's the same situation as above (same demonstration):

We 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.

Hey.
g(x) is continuos is reauired.
Your proof uptil now hasn't painted the whole picture.

Think it this way.
g(0) is 0.

Now if g(x) wasn't continuos g(0.000001) [taking v small value)could hve had value 50000.

In that case f(0.000001) could have been anywhere between -50,000 and 50,000 which would make f(x) non-continuos like g(x).

The basic method is to realize the f(x) gets sandwiched as it reaches 0.

You can prove this by the graphical method and also the algebraic method (which is more important as you camnot always draw graphs) of your previous question.

The method should be same as what you wrote for your last term.
Just replace x with g(x) everywhere.

Also make sure the algebra method makes sense to you.It's important you realize how you got the answer and ask yourself if the answer has loopholes :-)