For what values of x, if any, does #f(x) = 1/((12x+4)sin(pi+(6pi)/x) # have vertical asymptotes?

1 Answer
Jan 26, 2017

Answer:

Vertical asymptotes : #x = uarr-1/3;# # uarr +-6, +-3, +-2, +-1.5, ...,+-6/n, ...darr#

Horizontal asymptote : #larry = 0rarr#

Explanation:

The asymptotes are given by

#(3x+1)sin(6pi/x)=0#. So,

#x= -1/3 ; x = 6/n, n=+-1, +-2, +-3, ...# give asymptotes.

The horizontal asymptote is revealed by

#x to +-oo#, as y to 0#.

The horizontal space between consecutive vertical asymptotes

diminishes from #3 to 0#, as #x to 0_(+-)#. The vertical end-

asymptotes #x = +-6# are marked, in the first graph

You can study the second graph, for shape near the exclusive

asymptote #x = -1/3#.

I have used ad hoc ( for the purpose ) scales, for clarity.

graph{(4y(3x+1)sin(6pi/x)+1)(x-6-.01y)(x+6+.01y)=0 [-16, 16, -.5, .5]}

graph{(4y(3x+1)sin(6pi/x)+1)(x+.333-.00001y)=0 [-.4 -.0,-10, 10]}