How to find the asymptotes of #f(x) = (x+6)/(2x+1)# ?

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Answer 1 · 2016-01-21T16:54:07

This function has vertical asymptote #x=-1/2# and horizontal asymptote #y=1/2#

To check if a rational function has a vertical asymtote(s) you have to look for zeroes of the denominator.

In this case there is one zero #x_0=-1/2#. So #x=-1/2# is a vertical asymptote.

To look for the horizontal asymptotes you have to calculate

#lim_{x->-oo}f(x)# and #lim_{x->+oo}f(x)#. If the limits are finite and

equal to #l#, then the line #y=l# is the asymptote.

In this example we have:

#lim_{x->-oo}(x+6)/(2x-1)=lim_{x->-oo}(1+6/x)/(2-1/x)=1/2#

#lim_{x->+oo}(x+6)/(2x-1)=lim_{x->+oo}(1+6/x)/(2-1/x)=1/2#

The limits are finite and equal, so #y=1/2# is a horizontal asymptote.