How to prove the equation

[MathsLearner]

The problem is


If i take the absolute value of x as "0" then the answer [math]f(x) = \frac {\sin(x)} x [/math] becomes undefined as [math] x =0 [/math]. Also [math]\sin(0) = 0 [/math]. The answer should be 0, if [math] \frac 0 0[/math] is defined. Please advise.

 

[MathsLearner]

Yes probably i can use the Taylor series

then i think i am getting the answer of 1.

 

[HallsofIvy]

The problem is
View attachment 19344

If i take the absolute value of x as "0" then the answer [math]f(x) = \frac {\sin(x)} x [/math] becomes undefined as [math] x =0 [/math]. Also [math]\sin(0) = 0 [/math]. The answer should be 0, if [math] \frac 0 0[/math] is defined. Please advise.

Perhaps I do not understand your question. Taking the "absolute value" of x as 0 seems strange- the absolute value of 0 is, of course, 0. Don't you mean just "Take x as 0"?

The problem is to show that xf(x)= sin(x). If x is NOT 0 then f(x)= sin(x)/x so that xf(x)= x(sin(x)/x)= sin(x). If x= 0 then f(x)= 1 so xf(x)= 0(1)= 0. It is also true that sin(0)= 0 so xf(x)= sin(x) in this case also.

In any case the fact that sin(x)/x is undefined at x= 0 is completely irrelevant because f(x)= sin(x)/x only for x NOT equal to 0.

 

[MathsLearner]

Perhaps I do not understand your question. Taking the "absolute value" of x as 0 seems strange- the absolute value of 0 is, of course, 0. Don't you mean just "Take x as 0"?

Yes i mean "Take x as 0".

I am only confused when the question says

x is greater and equal to 0

 

[HallsofIvy]

It's pretty clear: "Show that xf(x)= sin(x) for $0\le x \le \pi$.
Do you understand what "xf(x)" means? Do you understand what "=" means? Do you understand what "sin(x)" means?

There is no "limit" involved here, just reading the definition of f and doing the algebra!

I said before:
If x is NOT 0 then f(x)= sin(x)/x so xf(x)= x(sin(x)/x)= sin(x)

If x= 0 then f(x)= 1 so xf(x)= 0(1)= 0= sin(0).

 

[JeffM]

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.

 

[MathsLearner]

Yes, I understand now. Thank you.