What is the Limit as x Approaches Negative Infinity of 2x - sqrt(4x^2 + x + 1)?

[Suy]

Homework Statement

lim x->-inf 2*x-sqrt(4*x^2+x+1)

Homework Equations

The Attempt at a Solution

For positive inf, i got -inf(not sure)
but negative inf, i got -1/4
http://www.wolframalpha.com/input/?i=lim+2*x-sqrt%284*x^2%2Bx%2B1%29
but wolfram have switch the answer, which is right, so i don't know which part i am wrong
lim X->-inf
2*x-sqrt(4*x^2+x+1)
2*x-sqrt(4*x^2+x+1)*((2*x+sqrt(4*x^2+x+1))/(2*x+sqrt(4*x^2+x+1))
(4x^2+4x^2-x-1)/(2*x-sqrt(4*x^2+x+1))
-x-1/(2*x-sqrt(4*x^2+x+1))
-x/(2x-sqrt(4x^2) X<0 sqrt(4x^2) =|2x|=-2x
-x/(2x-(-2x))=-1/4

lim x->+inf
-x/(2x-sqrt(4x^2) X>0 sqrt(4x^2) =|2x|=2x
-x/0= -inf?
Thanks

 

[Mentallic]

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

Yes, good.

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

No, you should have \frac{4x^2-(4x^2+x+1)}{2x+\sqrt{4x^2+x+1}}
You got the numerator wrong - specifically it should be -4x² not +4x². And in the denominator, you changed it to a minus when you showed a plus in the previous line.