Removable Discontinuity Trig

[johnjones]

I really have no clue what this question is asking me.

Is the discontinuity of f(x) = [sin(pi x)]/3x at x=0 removable? If not, eplain why, and if yes, give a value for f(0) that will make f(x) continuous.
Can someone clarify and guide me thro this, thx?

I know sin x/x = 1.

 

[stapel]

I know sin x/x = 1.

If you mean "I know that the limit, as x goes to zero, of sin(x)/x is 1", then you are correct. And this will be what you'll use to solve this exercise.

. . . . .[sin(pi x)] / [3x]

. . . . .= [pi/pi] [sin(pi x)] / [3x]

. . . . .= [pi/3] [sin(pi x)] / [pi x]

Now apply that limit.

Eliz.

 

[johnjones]

I know sin x/x = 1.

If you mean "I know that the limit, as x goes to zero, of sin(x)/x is 1", then you are correct. And this will be what you'll use to solve this exercise.

. . . . .[sin(pi x)] / [3x]

. . . . .= [pi/pi] [sin(pi x)] / [3x]

. . . . .= [pi/3] [sin(pi x)] / [pi x]

Now apply that limit.

Eliz.

I got pi/3 for a final answer.... is that correct?

I simply said [sin pi x]/pi x = 1

 

[stapel]

Yes.

Eliz.

 

[johnjones]

I really have no clue what this question is asking me.

Is the discontinuity of f(x) = [sin(pi x)]/3x at x=0 removable? If not, eplain why, and if yes, give a value for f(0) that will make f(x) continuous.
Can someone clarify and guide me thro this, thx?

I know sin x/x = 1.

I grahped the function, and I can't seem to find it to be discontinuous anywhere.
Y1 = sin(pi x)/3x

Did I graph it right on my TI calculator?

 

[stapel]

Since we have no idea what you may have entered into your calculator, I don't think there's any way for us to answer that question. But if your calculator's TABLE function showed y1 to be defined at x = 0, then I'd suspect you entered something incorrectly.

Eliz.