How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x) = (3x)/(Sin2x)#?

1 Answer
Jan 31, 2017

Answer:

Vertical : #uarr x = k/2pi darr, k = +-1, +-2, +-3, ...#

Explanation:

As #x to 0, y to 3/2#.

The denominator is periodic. Yet, the ratio is non-periodic.

As x to #xto k/2pi, f to +-oo, k = +-1, +-2, +-3, ...#,

giving vertical asymptotes

#x= k/2pi, f to +-oo, k = +-1, +-2, +-3, ...#.

graph{(3x)/sin (2x) [-50, 50, -25, 25]}