Why $lim_(x->oo) (sqrt(4x^2+x-1)-sqrt(x^2-7x+3)) = lim_(x->oo) (3x^2+8x-4)/(2x+...+x+...)=oo$ ?

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Answer 1 · 2018-02-28T07:29:19

$"See explanation"$

$"Multiply by "$

$1 = (sqrt(4 x^2 + x - 1) + sqrt(x^2 - 7 x + 3))/(sqrt(4 x^2 + x - 1) + sqrt(x^2 - 7 x + 3))$

$"Then you get"$

$lim_{x->oo} (3 x^2 + 8 x - 4)/(sqrt(4 x^2 + x - 1) + sqrt(x^2 - 7 x + 3))$

$"(because "(a-b)(a+b) = a^2-b^2")"$

$=lim_{x->oo} (3 x^2 + 8 x - 4)/(sqrt(4 x^2 (1 + 1/(4x) - 1/(4x^2))) + sqrt(x^2(1 - 7/x + 3/x^2))$

$=lim{x->oo} (3 x^2 + 8 x - 4)/(2x sqrt(1 + 0 - 0) + x sqrt(1 - 0 + 0))$

$"(because "lim_{x->oo} 1/x = 0")"$

$=lim{x->oo} (3 x^2 + 8 x - 4)/(3 x)$

$=lim{x->oo} (x + (8/3) - (4/3)/x)$

$=oo + 8/3 - 0$

$=oo$

Why lim_(x->oo) (sqrt(4x^2+x-1)-sqrt(x^2-7x+3)) = lim_(x->oo) (3x^2+8x-4)/(2x+...+x+...)=oolim_(x->oo) (sqrt(4x^2+x-1)-sqrt(x^2-7x+3)) = lim_(x->oo) (3x^2+8x-4)/(2x+...+x+...)=oo?