find the limit without using L'Hospital's rule

Question

lim x->-pi/2 (1+ cos (2x))/(2x+ pi )^2

Answer 1

L = lim {x→ π/2} (1 + cos 2x) / (2x + π)² . . . . . . . . . Note this is NOT of indeterminate form 0/0 at x = π/2

so, we can simply set x = π/2

= (1 - 1) / (2π)²

= 0

Perhaps, the problem should read: lim {x→ π/2} (1 + cos 2x) / (2x - π)² which IS of form 0/0 at x = π/2

Then, by L'Hopital's Rule (notice correct spelling of his name!), we have:

L = lim {x→ π/2} (-2sin 2x) / 4(2x + π)

= 0