Continuity of f(0): Does It Exist?

In summary, the function provided has some questionable symbols and is not defined properly at \(x=0\). The correct definition may have been intended to be \[f(x)=\begin{cases}1,&\,x=0\\ (x+1)^{1/\sin(x)}-\frac{\cos 2x}{x+1},&\,x<0\\ \frac{\sin^{2}x}{x.3^x-9x^2},&x>0\\ \end{cases}\]
  • #1
Chipset3600
79
0
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
 

Attachments

  • forum.png
    forum.png
    8.9 KB · Views: 50
Physics news on Phys.org
  • #2
Chipset3600 said:
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.
 
  • #3
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).
Sudharaka said:
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.
 
  • #4
Chipset3600 said:
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}\]
 

FAQ: Continuity of f(0): Does It Exist?

What does it mean for "f(0)" to exist?

"f(0)" refers to the value of a function at the input of 0. For a function to exist at a specific input, it means that the function is defined and has a corresponding output at that input value.

How is the continuity of "f(0)" determined?

The continuity of "f(0)" is determined by evaluating the left-hand and right-hand limits of the function at the input of 0. If both limits are equal, then the function is continuous at 0 and "f(0)" exists.

Can "f(0)" exist if the function is not continuous?

No, in order for "f(0)" to exist, the function must be continuous at the input of 0. If the function is not continuous, then "f(0)" does not exist.

What factors can affect the continuity of "f(0)"?

The continuity of "f(0)" can be affected by the presence of a removable or non-removable discontinuity at the input of 0, as well as the behavior of the function near 0. Other factors such as the type of function (e.g. polynomial, rational, exponential) and the existence of a limit at 0 can also impact the continuity of "f(0)".

Is the continuity of "f(0)" necessary for the function to be defined at 0?

No, it is possible for a function to be defined at 0 but not have continuity at that point. For example, a function with a removable discontinuity at 0 is still defined at 0 but is not continuous. However, if "f(0)" does not exist, then the function is not defined at 0.

Similar threads

Back
Top