How do you find the end behavior of #y = (2x+3) /( x+2)#?

1 Answer
May 10, 2015

The dominant terms in the numerator and denominator as #x -> +-oo#, will be the ones in #2x# and #x# respectively.

The limit of #(2x)/x# as #x->oo or -oo# is 2.

More formally, we have

#y = (2x+3)/(x+2) = (2x+4-1)/(x+2) = (2(x+2)-1)/(x+2) #

#= 2(x+2)/(x+2)-1/(x+2) = 2-1/(x+2)#

The term #1/(x+2)->0# as #x->+-oo#.