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

1 Answer
Aug 6, 2016

Answer:

#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#