How do you find the horizontal asymptote for #fR(x)=(3x+5) /(x-6)#?

1 Answer
Mar 16, 2018

Answer:

The horizontal asymptote is #y=3#

Explanation:

Calculate the limit as the function #f(x)# tend to #+oo# and #-oo#

The function is

#f(x)=(3x+5)/(x-6)#

The domain of #f(x)# is #RR-{6}#

#f(x)=(3x+5)/(x-6)=(cancelx(3+5/x))/(cancelx(1-6/x))=(3+5/x)/(1-6/x)#

The limits are

#lim_(x->+oo)(5/x)=0#

#lim_(x->+oo)(6/x)=0#

#lim_(x->-oo)(5/x)=0#

#lim_(x->-oo)(6/x)=0#

And finally

#lim_(x->+oo)f(x)=lim_(x->+oo)(3+5/x)/(1-6/x)=(3-0)/(1-0)=3#

The horizontal asymptote is #y=3#

graph{(y-(3x+5)/(x-6))(y-3)=0 [-34.77, 47.43, -16.94, 24.16]}