Can someone explain to me why f(x)=csc(2x) doesn't have any horizontal asymptotes?
What I did was 1/sin2x=1/((sin2x/2x)2x) = 1/1(2x)=1/2x = (1/x)/(2x/x) = 0/2 = 0

but this was deemed wrong...

Question

Can someone explain to me why f(x)=csc(2x) doesn't have any horizontal asymptotes?
What I did was 1/sin2x=1/((sin2x/2x)2x) = 1/1(2x)=1/2x = (1/x)/(2x/x) = 0/2 = 0

but this was deemed wrong...

anonymous · Answer

Also haven't been here in a while, love the upgrades to the site

mathmale · Answer

You might want to graph y=sin x, and then graph y=csc x on the same set of coordinate axes.  Neither sin x nor csc ever "dies out" in the sense of approaching zero as x increases.      sin x simply repeats itself as x increases.  You see the same behavior (repetition of cycles) in the graph of the cosecant function.

How would you define "horizontal asymptote?"  Can you find illustrations of functions with horiz. asymptotes to compare to the graphs of sin x and csc x?