What is the limit?

Question

lim #(2x+sqrt(4x^2+3x-2))/(3x+2)#
for x -> neg(infinite)

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Answer 1 · 2018-02-27T10:18:31

#lim_(x->-oo) (2x+sqrt(4x^2+3x+2))/(3x+2) = 0#

Note that:

#(2x+sqrt(4x^2+3x+2))/(3x+2) = (2x+sqrt((4x^2)(1+3/(4x)+1/(2x^2))))/(3x+2) #

#(2x+sqrt(4x^2+3x+2))/(3x+2) = (2x+2absxsqrt((1+3/(4x)+1/(2x^2))))/(3x+2) #

For #x < 0#, #absx = -x#, so:

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

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

Then:

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

graph{(2x+sqrt(4x^2+3x+2))/(3x+2) [-10, 10, -5, 5]}