Find the limit and use the second squeeze principle to prove that answer. lim (3

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Take the limit: lim_(x->infinity) (3 x^2+7 x-4)/(x^3-4) Divide numerator and denominator by the highest power of x: = lim_(x->infinity) (-4/x^3+7/x^2+3/x)/(1-4/x^3) = (lim_(x->infinity) (7/x^2-4/x^3+3/x))/(lim_(x->infinity) (1-4/x^3)) = (lim_(x->infinity) (7/x^2-4/x^3+3/x))/(1-4 (lim_(x->infinity) 1/x^3)) = (-4 (lim_(x->infinity) 1/x^3)+7 (lim_(x->infinity) 1/x^2)+3 (lim_(x->infinity) 1/x))/(1-4 (lim_(x->infinity) 1/x^3)) = (-4/(lim_(x->infinity) x^3)+7 (lim_(x->infinity) 1/x^2)+3 (lim_(x->infinity) 1/x))/(1-4/(lim_(x->infinity) x^3) = (7 (lim_(x->infinity) 1/x^2)+3 (lim_(x->infinity) 1/x)-4/(lim_(x->infinity) x)^3)/(1-4/(lim_(x->infinity) x)^3) = 7 (lim_(x->infinity) 1/x^2)+3 (lim_(x->infinity) 1/x) = 7/(lim_(x->infinity) x^2)+3 (lim_(x->infinity) 1/x) = 7/(lim_(x->infinity) x^2)+3/(lim_(x->infinity) x) The limit of x as x approaches infinity is infinity: = 7/(lim_(x->infinity) x^2) Using the power law, write lim_(x->infinity) x^2 as (lim_(x->infinity) x)^2: = 7/(lim_(x->infinity) x)^2 The limit of x as x approaches infinity is infinity: = 0place n place of x thank you n please rate it

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