How do I find the asymptotes of #y=x/sin(x^2)#? Are there even any asymptotes?

1 Answer
Feb 16, 2017

Answer:

Vertical asymptotes : #uarrx=+-sqrt(kpi)darr, k = 0, 1, 2, 3, ...#.

See Socratic graph, for highlights.

Explanation:

#y =(1/x)(x^2/(sinx^2)#

The second factor has an indeterminate form #0/0#, at x = 0.

As # x to 0, y to lim1/x lim(x^2/sin(x^2))=+-oo(1)=+-oo#.

Besides this vertical asymptote x = 0, the others are given by the

zeros #x =+-sqrt(kpi), k = 1. 2. 3, ...#, of #sinx^2#.

Socratic asymptotes-inclusive graph, for #x in [-pi, pi]##, is included.

graph{(y-x/sin(x^2))=0 [-20, 20, -10, 10]}
4
graph{(x+.004y)(x-1.77+.005y)(x+1.77+.005y)(y-x/sin(x^2))(x-2.54-.003y)(x+2.54+.003y)=0 [-3.1416 3.1416 -10, 10]}

Ad hoc scales used to highlight asymptotes relative graph, for #x in (-pi, pi)#.