lim x>1 (lnx/sinpix)

8 years agoSince since sinpi is 0 (sinpi/x = 0 b/c x goes to 1), we have an indeterminate form (0/0). We can use L'Hopital's rule: take the derivative of the top and bottom. Get... (1/x)/cos(pi/x)(-pi/x^2) Plug in x=1 and get 1/(-1)(-pi) = 1/pi The answer is 1/pi

8 years agoAnother method is by series. Realizing lim x>1 (lnx/sinpix) is the same thing as lim x>0 [ ln(1-x) / sin(pi(1-x)) ], and that sin(pi(1-x)) = sin(pi(x)) we get the new limit: lim x>0 [ ln(1-x) / sin(pi*x) ] we may then use taylor expansions of both functions: ln(1-x) = x + (1/2)x^2 + (1/3)x^3 + ... sin(pi*x) = pi*x - (1/3!)(pi*x)^3 + (1/5!)(pi*x)^5 - ... when placed in fractional form we may divide out an x, leaving 1 and a series on top, pi and some series on the bottom, as x -> 0 the series left on top and bottom obviously approach 0, leaving 1 and pi, the solution being 1/pi

8 years ago