Limiting Limits

Calculus Level 3

If lim ⁡ x → ∞ x 9 − 8 3 43 x 3 − 256 x 6 − 4 x = A \displaystyle \lim _{ x\to \infty }{ \dfrac { \sqrt [ 3 ]{ { x }^{ 9 }-8 } }{ { 43x }^{ 3 }-\sqrt { 256{ x }^{ 6 }-4x } } =A } , what is the value of 1 A \dfrac { 1 }{ A } ?

25 16 0 27

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Liam P by Liam P , 22, Canada

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1 solution

Chew-Seong Cheong
Sep 27, 2016

A = lim ⁡ x → ∞ x 9 − 8 3 43 x 3 − 256 x 6 − 4 x Divide up and down by x 3 = lim ⁡ x → ∞ 1 − 8 x 9 3 43 − 256 − 4 x 5 = 1 3 43 − 256 = 1 43 − 16 = 1 27 \begin{aligned} A & = \lim_{x \to \infty} \frac {\sqrt[3]{x^9-8}}{43x^3-\sqrt{256x^6-4x}} & \small \color{#3D99F6}{\text{Divide up and down by }x^3} \\ & = \lim_{x \to \infty} \frac {\sqrt[3]{1-\frac 8{x^9}}}{43-\sqrt{256-\frac 4{x^5}}} \\ & = \frac {\sqrt[3]{1}}{43-\sqrt{256}} = \frac 1{43-16} = \frac 1{27} \end{aligned}

⟹ A = 27 \implies A = \boxed{27}

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