How do you find the horizontal asymptote for #y=(13x)/(x+34)#?

1 Answer
Dec 7, 2015

Answer:

Find #lim_(x->+-oo) (13x)/(x+34) = 13# hence asymptote #y=13#

Explanation:

One way is to divide both numerator and denominator by #x# and look at the limit as #x->+-oo#

#f(x) = (13x)/(x+34) = 13/(1+34/x)#

So #lim_(x->oo) f(x) = 13/(1+0) = 13#

and #lim_(x->-oo) f(x) = 13/(1+0) = 13#

So the horizontal asymptote is #y = 13#

In general if #f(x) = (p(x))/(q(x))# for some polynomials #p(x)# and #q(x)# of the same degree, then #f(x)# will have a horizontal asymptote equal to the quotient of the leading coefficients.

For example:

#lim_(x->oo) (3x^2+4x-5)/(2x^2-11x+1)=lim_(x->oo) (3+4/x-5/(x^2))/(2-11/x+1/(x^2)) = 3/2#