How would you do this calculus problem involving limits?

Question

limit as x->0 of (x csc2x) / csc5x ??

Answer 1

I don't like csc, so rewrite as sin. You get

x sin5x / sin2x

As x -> 0, sin5x -> 5x, sin2x -> 2x. And 5x / 2x -> 5/2.
So what you get is limit of x(5/2). And as x goes to zero, this goes to zero too.

Answer 2

id rewrite this first so we can use lh

so um
xsin5x/sin2x

lim as x->0=0/0
lh rule this time
lim as x-> of the deriviative of the top divided by the deriv of the bottom.
(sin5x+5xcos5x*)/(cos2x)2
plug in 0 and you get 0/1=0

Answer 3

=x/sin2x *sin5x
x/sin2x==>1/2 and sin 5x ==>0so the limit is 0