How do you find the vertical, horizontal or slant asymptotes for #f(x)= (x+3)/(x-3)#?

1 Answer
Jun 19, 2016

Answer:

Vertical asymptote is #x=3#
Horizontal asymptote is #y=1#

Explanation:

Tony B

The equation becomes undefined at #x=3# in the denominator as #3-3=0#. Basically this means that mathematically you are not allowed to divide by 0.

#color(blue)("Vertical asymptotes")#

#lim_(xto3^(+))(x+3)/(x-3) -> (x+3)/(0^+)=+oo #

#lim_(Xto3^-)(x+3)/(x-3)-> (x+3)/(0^-)=-oo#

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#color(blue)("Horizontal asymptotes")#

As #x# becomes bigger and bigger then the addition or subtraction of 3 becomes insignificant. Consequently we end up with basically #x/x#

#lim_(xtooo^+) (x+3)/(x-3) ->(+oo)/(+oo) = +1#

#lim_(xtooo^-) (x+3)/(x-3)->(-oo)/(-oo) = +1#

So the horizontal asymptote is +1
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: using polynomial division.

#(x+3)-:(x-3) = 1+6/(x-3)#

#lim_(x->3^(+-) ) = 1+-oo#

#lim_(x->oo^(+-) ) = 1+-0=1#