Yes i mean "Take x as 0".
I am only confused when the question says
View attachment 19352
x is greater and
equal to 0
NO NO NO NO
[MATH]x \ge 0[/MATH] does
not mean
x is greater than 0 and x is also equal to 0.
That is obvious nonsense. It means
either x is equal to 0 or else x is greater than 0.
It is the same as saying
x is not less than zero,
which is perfectly straight forward.
Now going back to your problem, the issue of
[MATH]\dfrac{sin(0)}{0}[/MATH] simply does not arise because, BY DEFINITION, we have
[MATH]f(0) \equiv 1.[/MATH]
So you have to do your proof by cases.
[MATH]0 \le x \le \pi \implies 0 = x \text { or } 0 < x \le \pi.\\
\text {Case I: } 0 = x.\\
\therefore sin(x) = 0.\\
\text {And } f(0) \equiv 1 \implies 0 * f(0) = 0 * 1 = 0.[/MATH]Can you finish the proof for that case? Now how about the other case? Can you prove that one?
Taylor series have zilch to do with the proof. They may help justify why f(x) was defined as it was, but all you need to worry about is the definition provided.