Question

How do you find the limit of `x/(sqrt(4x^2 - x + 3))` as x approaches negative infinity?

Answer 1

It is

#lim_(x->-oo) x/(sqrt(4x^2 - x + 3))= lim_(x->-oo) x/[absx*sqrt[4-1/x+3/x^2]]= lim_(x->-oo) x/[(-x)*sqrt[4-1/x+3/x^2]]= lim_(x->-oo) -1/[sqrt[4-1/x+3/x^2]]= lim_(x->-oo) -1/[sqrt(4-0+0)]= =-1/2#

Note that as `x->-oo` , `absx=-x` and `1/x->0` , `1/x^2->0`