Continuous extension

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joeyjon

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This is a multiple choice question from my homework. The problem is:

f(x) =Sin x / absolute value x

Graph the following function to see whether it appears to have a continuous extension to the origin.
Can it be extended to be continuous at the origin from the right or left?

(This I can do without a problem) but...

The CORRECT ANSWER according to the book is...

The function cannot be extended to be continuous at x=0. It can be extended to be continuous at the origin from the right if f(0)=1, or from the left if f(0)=-1.


So is the book wrong? The function never intersects the origin.

Are they saying the origin isn't necessarily (0,0)?

Are they saying if f(0)=1, then the line is continuous at the origin because both points

(0,1) (from the right)
(3,0)

or

(0,-1) (from the left)
(-3,0)

both the x axis and the y axis intersects are continuous, although not at the origin, but this constitutes the line being continuous at the origin?
 
Thinking about it more, maybe trigonometric functions have an origin that is different than (0,0) depending when it starts on the unit circle?
 
It is not continuous at x=0. The origin is at (0,0).

\(\displaystyle f(x)=\left\{\begin{array}{cc} \frac{sin(x)}{|x|}, \;\ x\neq 0\\ 1, \;\ x=0 \end{array}\right\)

\(\displaystyle \displaystyle\lim_{x\to 0^{-}}f(x)=\displaystyle \lim_{x\to 0^{-}}\frac{sin(x)}{-x}=\displaystyle -\lim_{x\to 0^{-}}\frac{sin(x)}{x}=-1\)

Thus, f is not continuous at x=0.

Looking at the graph, one can see the discontinuity at x=0. f(x) is 1 when approaching from the right and f(x)=-1 when approaching from the left.
 
At the origin

But my question is, why is it extended at the origin when f(0)=1 or f(0)=-1?
 
Multiple choices

Here are the 4 multiple choice answers they let you answer...

A. The function cannot be extended to be continuous at x=0. It can be extended to be continuous at the origin only from the right, if f(0)=1.

B. The function cannot be extended to be continuous at x=0. It can be extended to be continuous at the origin from the right, if f(0)=1, or from the left if f(0)=-1.

C. The function cannot be extended to be continuous at x=0. It can be extended to be continuous at the origin only from the left, if f(0)=-1.

D. The function can be extended.


The book says B is correct, but I think the answer isn't any of them because of the origin.
 
origin

Basically I am wondering if there is an advanced definition of origin I am unaware of. Maybe a trigonometric function's starting point is considered its origin?

I am pretty sure the book made a mistake but I am just a student.
 
No, not that I am aware of. The origin is the origin (0,0)

In this case, the origin is the same as any other origin.

Look at what I posted and then compare that to B.
 
your answer/ book answer

What you say is that the function can be extended at (0,1) from the right by defining f(0)=1
or the function can be extended at (0,-1) from the left by defining f(0)=-1.

The answer b is saying...

What you say is that the function can be extended at (0,0) from the right by defining f(0)=1
or the function can be extended at (0,0) from the left by defining f(0)=-1.

That was the source of my confusion...
 
the origin must be continuous

B. The function cannot be extended to be continuous at x=0. It can be extended to be continuous at the origin from the right, if f(0)=1, or from the left if f(0)=-1.

Or rather...

The book is saying if f(0)=1 or if f(0)=-1 then (0,0) is continuous.
 
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