How do you find the vertical, horizontal and slant asymptotes of: f(x) = x/ sinx?

1 Answer
Aug 6, 2016

f(x) has vertical asymptotes at x=n pi for any non-zero integer n.

It has a hole at x=0

Explanation:

Vertical asymptotes will occur where the denominator is zero and the numerator non-zero.

sin x = 0 if and only if x = n pi for some n in ZZ

Hence f(x) has vertical asymptotes at x = n pi where n in ZZ and n != 0

f(x) has a hole at x=0. The rational expression becomes 0/0, which is undefined, but the right and left limits exist at x=0 and are equal:

lim_(x->0) x/(sin x) = 1

f(x) has no horizontal or slant asymptotes, since we find:

f(2n pi+pi/2) = 2n pi+pi/2 -> +oo as n -> +oo

f(2n pi+pi/2) = 2n pi+pi/2 -> -oo as n -> -oo

f(2n pi+(3pi)/2) = -(2n pi+(3pi)/2) -> -oo as n -> +oo

f(2n pi+(3pi)/2) = -(2n pi+(3pi)/2) -> +oo as n -> -oo