Question

Find lim as x-> pi ((pi-x)^2)/(sin^2x) ?

Answer 1

\[
\lim_{x\to \pi}\frac{(\pi-x)^2}{\sin^2(x)}
\]Is this correct?

Answer 2

correct

Answer 3

Do you know l'Hospital's rule?

Answer 4

as far as I remember, I should take derivative of the top and bottom, right?

Answer 5

yes

Answer 6

so I will get 2(pi-x) / cos^2x ?

Answer 7

sorry, I was distracted...

Answer 8

it's OK

Answer 9

\[
d((\pi-x)^2) = 2(\pi-x)d(\pi-x) = -2(\pi-x)dx
\]So the top would be: \[
2x-2\pi
\]

Answer 10

\[
d(\sin^2(x)) = 2\sin(x)d(\sin(x)) = 2\sin(x)\cos(x)dx =\sin(2x)dx
\]So the bottom is: \[
\sin(2x)
\]

Answer 11

This still leads to the indeterminate form of \(0/0\), so we have to do l'Hospital a second time.

Answer 12

Does it make sense how I got the derivatives?

Answer 13

yes, I just checked online, thank you

Answer 14

I took derivative one more time and got -1/cos 2x which is -1

Answer 15

....

Answer 16

\[
d(2x-2\pi) = 2dx
\]Top derivative is \(2\).

Answer 17

\[
d(\sin(2x)) = \cos(2x)d(2x) = 2\cos(2x)dx
\]Bottom derivative is \(2\cos(2x)\).

Answer 18

right.. I mistakenly put 2pi- 2x

Answer 19

So final answer?

Answer 20

1

Answer 21

thank you very much! Wish you all the best!