Question 363622
a. {{{ lim(x->0, cotx(1-cosx) )  = lim(x->0, cosx(1-cosx)/sinx )}}}
={{{lim(x->0, cosx)}}}*{{{lim(x->0, (1-cosx)/sinx )}}} = 1*0 = 0.  The second limit was obtained by applying LHopital's rule.

b.  Let {{{y= (sinx)^tanx}}}. then {{{lny = tanx*lnsinx = lnsinx/cotx}}}.  The limit as x approaches 0 has the form {{{infinity/infinity}}}, so apply the LH rule.  The new expression becomes {{{(cosx/sinx)/(-csc^2x)}}}.  Simplifying this, we get {{{cosx/(-cscx) = -sinxcosx}}}.  As x goes to 0, -sinx*cosx approaches 0.  Hence, lny approaches 0 as x approaches 0, and thus y approaches 1.