One way of doing this is to express #(x^2+7x+11)/(x+5)# into partial fractions.
Like this : #f(x)=(x^2+7x+11)/(x+5) color(red)= (x^2+7x +10-10+11)/(x+5) color(red)= ((x+5)(x+2)+1)/(x+5) color(red)= (cancel((x+5))(x+2))/cancel((x+5))+1/(x+5) color(red)= color(blue)((x+2)+1/(x+5))#
Hence #f(x)# can be written as : #x+2+1/(x+5)#
From here we can see that the oblique asymptote is the line #y=x+2#
Why can we conclude so?
Because as #x# approaches #+-oo# , the function #f# tends to behave as the line #y=x+2#
Look at this : #lim_(xrarroo)f(x)=lim_(xrarroo)(x+2+1/(x+5))#
And we see that as #x# becomes larger and larger, #1/(x+5) " tends to " 0#
So #f(x)# tends to #x+2# , which is like saying that the function #f(x)# is trying to behave as the line #y=x+2#.
