lim(lnx)/(sin180x) when x goes to 1 (whithout hopital)?

Answer 1

x>1 (lnx)/(sin180x)

let x=p+1 as x >1, p > 0

p>0 log{1+p} / (sin pi(p+1))

p>o log{1+p} / (- sin (pi p)) as sin (pi+t) = - sin t

using log and sine series
p>0 - [p - (1/2)p^2+(1/3)p^3 -...]
.........-----------------------------------
........[pi p - (1/6) pi^3 p^3+(1/120)pi^5 p^5....]
taking p common and deleting

p>0 -[1 - (1/2)p+(1/3)p^2 -....] / [pi - (1/6) pi^2 p^2+...]

put in limit
= - [1 ] / [pi]
answer - 1/pi

Answer 2

i am asuming 180x means 180xdegrees not radians otherwise limit would be zero
let x=u+1 as x->1, u->0
limu->0 ln(1+u)/sin((pi)(u+1))
=ln(1+u)/sin(pi+pi(u))
= - ln(1+u)/sin(pi)u
devide numerator and denominator by u
and put values of limit
you will get -1/(pi)