Question

find the
limx→1(sinΠx)/(x−1)
t=x-1
*without using l'hopital rule
answer should be -Π but I don't know how to do without l'hopital rule

Answer 1

is that a pi symbol?

Answer 2

yup

Answer 3

i will post something in just a sec k?

Answer 4

okay, no biggie, and dont use l'hopital rule, make it as clear as you think I can catch up

Answer 5

check that attachment out

Answer 6

let me know if you have any questions

Answer 7

remember as g->0, sing/g->1

Answer 8

hey polpak can you prove the sin(x+y)=sinxcosy+sinycosx off the top of your head

Answer 9

I can, but I usually do it with euler's stuff..

Answer 10

hey heart, so you like totally got it?

Answer 11

oh don't worry about showing me i will look up later

Answer 12

i just forgot how to prove it

Answer 13

\[sin(x+y) = Im( e^{(x+y)i})\]\[= Im (e^{xi} \cdot e^{yi})\]\[e^{xi} \cdot e^{yi} = (cos(x) + isin(x))(cos(y) + isin(y))\]\[= cos(x)cos(y) - sin(x)sin(y) \]\[+ i(cos(x)sin(y) + sin(x)cos(y)) \]

\[\implies Im (e^{xi} \cdot e^{yi}) =cos(x)sin(y) + sin(x)cos(y) = sin(x+y)\]

Answer 14

Where Im(a) is the imaginary part of a.

Answer 15

thats real pretty

Answer 16

A lot of the trig identities become really pretty when you use eulier's formula: \[e^{\theta i} = cos(\theta) + isin(\theta)\]

Answer 17

And for free you get the identity for cos(x+y) too.. It's just the real part of the same thing so you get

\[cos(x+y) = cos(x)cos(y) - sin(x)sin(y)\]

Answer 18

lol

Answer 19

What's funny?

Answer 20

for free

Answer 21

Well sorta free since you had to find it to get the imaginary part ;p

Answer 22

i get a discount

Answer 23

thanks polpak

Answer 24

Of course!

/tips lego hat