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

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

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Answer 1 · 2016-03-12T20:21:08

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#

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

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