Limit of sin(3x)/π-x as x approaches π

[Pithikos]

Homework Statement

lim of sin(3x)/π-x as x->π

Homework Equations

The Attempt at a Solution

I tryied replacing π-x with t so t=π-x and rewrote the whole limit as:
lim of sin(3(π-t))/t as t->0

that gives me(without the lim notation):
sin(3π-3t)/1 * 1/t

then I noticed that if I multiply the dominators and numerators with 3π-3t it doesn't help much because replacing t with 0 in (3π-3t) doesn't give 0 so therefore can't use the standard limit lim sinx/x -> 1 as x -> 0

I have been stuck with this problem 2-3 days now. I made some other mistakes in the first tries not worth mentioning. Now it seems I took the right way into the problem but I just feel stuck :/

 

[micromass]

Well, you should discriminate cases:

1) n=0, then you can use that sin(x)/x --> 1
2) n is not 0, then you will have to do something else... But in this case, we got a "number/0" situation, so this would give positive or negative infinity...

 

[Cygni]

Since for x = pi => sin(3pi) = 0 and pi - pi = 0 one can apply I'Hopital's rule by differentiating the top and the bottom expressions.

 

[Pithikos]

Thing is that we haven't learned the L'Hopital's rule so for the exercise there must be some more "caveman" way to do this. We have only been working with standard limits like the following.

limits of x approaching 0 :

sinx/x -> 1
tanx/x -> 1
ln(1+x)/x -> 1
.. etc.

 

[LCKurtz]

I think a previous poster thought the π symbol was the variable n.

First note that

\sin(3x) = \sin(3x - 3\pi) so

\frac{\sin(3x) }{\pi - x}=\frac{\sin(3x-3\pi) }{\pi - x}

Can you get it from there?

 

[Pithikos]

Thank you! Now I solved it :!)

Though instead of sin(3x-3π) I used sin(3π-3x) so it became:

sin(3π-3x)/3π-3x * 3/1 =

1 * 3


Btw with what tool do you write equations in that mathematical way?

 

[LCKurtz]

Thank you! Now I solved it :!)

Though instead of sin(3x-3π) I used sin(3π-3x) so it became:

sin(3π-3x)/3π-3x * 3/1 =

1 * 3Btw with what tool do you write equations in that mathematical way?

Of course, you need a minus sign. The math is LaTeX using the tex tags. Click on an expression to see how it's written.

See https://www.physicsforums.com/misc/howtolatex.pdf

 

[Strants]

Actually, 3 is correct.

\sin(3x) = -\sin(3x - 3\pi) = \sin(3\pi - 3x)

This makes sense when you think about it: subtracting 3π is essentially the same as rotating the unit circle one half-turn (in either direction), so the sine function would have to change sign (pun not intended).