Answer:
1
Explanation:
Evaluate the following
Lim (xcotx)
x->0
Note that cotx = cosx/sinx
The expression becomes
Lim (xcotx) = Lim (xcosx/sinx)
x->0 x->0
= 0cos0/sin0
= 0/0 )(ind)
Apply l'hospital rule
= Lim d/dx(xcosx/sinx)
x->0
= lim (-xsinx + cosx)/cosx
x->0
= -0sin0+cos0/cos0
= 0+1/1
= 1/1
= 1
Hence the limit of xcotx as x tends to zero is 1