Question

Find the limits? Are the functions continuous at the points being approached?
lim sin (x-sinx)
x->pi

Answer 1

This is why we studied trig identities until our fingers begain to loose their prints.

Use this: \(\sin(x) = \cos\left(\dfrac{\pi}{2} - x\right)\)

Answer 2

I am having trouble figuring this out?

Answer 3

What do you get after applying the identify I provided?

Answer 4

I am a little confused about solving identity part?

Answer 5

You did take trigonometry, right?

If I say, \(\sin(x)\), you say \(\cos\left(\dfrac{\pi}{2} - x\right)\)

If I say, \(\sin(2x)\), you say \(\cos\left(\dfrac{\pi}{2} - 2x\right)\)

If I say, \(\sin(x-5)\), you say \(\cos\left(\dfrac{\pi}{2} - (x-5)\right)\)

If I say, \(\sin(Banana)\), you say \(\cos\left(\dfrac{\pi}{2} - Banana\right)\)

Here it comes. I say \(\sin(x - sin(x))\). What do you say?

Answer 6

cos(pi/2-(x-sin(x)))*

Answer 7

Let's not forget our algebra. :-)

Okay, substitute this expression into your original problem and see if things work out any better.

Answer 8

so like this;
lim cos(pi/2-(x-sin(x)))
x->pi

Answer 9

Okay, is this any different or any more or less useful than the original expression? So far, we're just playing around with trig expressions. Now for the limit...

1) Did we need the substitution?
2) What is the limit?

Answer 10

I think the substitution was not needed,

Answer 11

But it was fun, wan't it? So, what's the limit?

Answer 12

is it 0?

Answer 13

Check out the cosine version and see if you get the same result.

Answer 14

Thats what I get

Answer 15

Excellent. Trig still works!

Serously, very good work. Sometimes we try things and they teach us, even if it doesn't really help out much.

Answer 16

Thanks for the help.