lim [cotx - (1/x)]

x-->0

cotx - 1/x = (x-tanx)/(x tanx)

now apply L'Hospital's Rule, so the limit is also

(1-sec^2 x)/(tanx - xsec^2 x)
and again
-2sec^2 x tanx / (sec^2 x - 2sec^2 x tanx)
= -2sec^2x tanx / (sec^2x(1-2tanx))
= 0/1

so the limit is 0