Continuity of f(0): Does It Exist?

[Chipset3600]

Hello MHB, the f(0) of this function doesn't exist, so I am i wrong or this question don't hv solution?

View attachment 406

 

[Sudharaka]

Hello MHB, the f(0) of this function doesn't exist, so I am i wrong or this question don't 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]

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, 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}\]