# Continuity in point

#### Chipset3600

##### Member
Hello MHB, the f(0) of this function doesn't exist, so im i wrong or this question dont hv solution?

#### Sudharaka

##### Well-known member
MHB Math Helper
Hello MHB, the f(0) of this function doesn't exist, so im i wrong or this question dont hv solution?

View attachment 406
Hi Chipset3600,

Your function seem to have some strange symbols which I don't understand. Is it,

$f(x)=\begin{cases}\frac{2}{\pi}\mbox{arctan}\left(\frac{x+1}{x^2}\right),&\,x=0\\ (x+1)^{1/\sec(x)}-\frac{\cos 2x}{x+1},&\,x<0\\ \frac{\sec^{2}x}{x.3^x-9x^2},&x>0\\ \end{cases}$

In that case the definition seem to be problematic since $$f(0)$$ is not defined properly. $$f(0)$$ should be a constant value whereas in the definition it's not.

Kind Regards,
Sudharaka.

#### Chipset3600

##### Member
Well, my language is portuguese, i forgot the translate of symbols: sen^2(x) = sin^(x), arctg(x)=arctan(x)..
and in the exercise is sin^2(x) and not sec^2(x).
Hi Chipset3600,

Your function seem to have some strange symbols which I don't understand. Is it,

$f(x)=\begin{cases}\frac{2}{\pi}\mbox{arctan}\left(\frac{x+1}{x^2}\right),&\,x=0\\ (x+1)^{1/\sec(x)}-\frac{\cos 2x}{x+1},&\,x<0\\ \frac{\sec^{2}x}{x.3^x-9x^2},&x>0\\ \end{cases}$

In that case the definition seem to be problematic since $$f(0)$$ is not defined properly. $$f(0)$$ should be a constant value whereas in the definition it's not.

Kind Regards,
Sudharaka.

#### Sudharaka

##### Well-known member
MHB Math Helper
Well, my language is portuguese, i forgot the translate of symbols: sen^2(x) = sin^(x), arctg(x)=arctan(x)..
and in the exercise is sin^2(x) and not sec^2(x).
Good. But the problem here is that the function is not defined at $$x=0$$ properly. We know that,

$\lim_{x\rightarrow 0}\left[\frac{2}{\pi}\mbox{arctan}\left( \frac{x+1}{x^2}\right)\right]=1$

So maybe the author would have meant ,

$f(x)=\begin{cases}1,&\,x=0\\ (x+1)^{1/\sin(x)}-\frac{\cos 2x}{x+1},&\,x<0\\ \frac{\sin^{2}x}{x.3^x-9x^2},&x>0\\ \end{cases}$