How do you find all the asymptotes for #f(x) = x+1+sqrt(x^2+4x)#?

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Answer 1 · 2015-05-05T16:18:23

There is one horizontal asymptote #y= -1#

It is to be investigated here, how f(x) behaves if x#->oo or -oo#

If x#->oo#, f(x) #->oo# is quite obvious. For determining the limit if #x->-oo#, find the limit of f(-x) for x#->oo#.

f(-x) = -x+1 +#sqrt(x^2-4x)# (Now multiply by the conjugate as follows)

= #(-x+1 +sqrt(x^2 -4x))* (-x+1-sqrt(x^2 -4x))/(-x+1-sqrt(x^2 -4x))#

=#(x^2 -2x +1 -x^2+4x)/(-x+1-sqrt(x^2-4x))# =#(2x+1)/(-x+1-xsqrt(1-4/x)#

=#(2+1/x)/(-1+1/x -sqrt(1-4/x))# Now apply the limit x#->oo#

=#2/-2# = -1. Hence there is an asymptote y= -1