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#