Non- zero real value

Calculus Level 3

f ( x ) = x p ( x + 1 3 + x − 1 3 − 2 x 3 ) \large f(x) = x^p\left(\sqrt[3]{x+1} + \sqrt[3]{x-1}- 2\sqrt[3]{x}\right)

Find the value of p p so that limit of the function above is some non-zero real number as x → ∞ x \to \infty . If the value of p p can be expressed as a b \dfrac{a}{b} , where a a and b b are coprime integers, submit your answer as a + b a+b .

9 11 10 6 5 8 7

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Brian Moehring
Jul 31, 2018

We use x + h 3 − x + k 3 = ( x + h 3 − x + k 3 ) ( ( x + h 3 ) 2 + x + h 3 x + k 3 + ( x + k 3 ) 2 ) ( x + h 3 ) 2 + x + h 3 x + k 3 + ( x + k 3 ) 2 = h − k ( x + h 3 ) 2 + x + h 3 x + k 3 + ( x + k 3 ) 2 \sqrt[3]{x+h} - \sqrt[3]{x+k} = \frac{(\sqrt[3]{x+h} - \sqrt[3]{x+k})\left(\left(\sqrt[3]{x+h}\right)^2 + \sqrt[3]{x+h}\sqrt[3]{x+k} + \left(\sqrt[3]{x+k}\right)^2\right)}{\left(\sqrt[3]{x+h}\right)^2 + \sqrt[3]{x+h}\sqrt[3]{x+k} + \left(\sqrt[3]{x+k}\right)^2} = \frac{h-k}{\left(\sqrt[3]{x+h}\right)^2 + \sqrt[3]{x+h}\sqrt[3]{x+k} + \left(\sqrt[3]{x+k}\right)^2}

a few times to write

[Note: The following can be made simpler by selectively using asymptotics. In fact, that was my original method, but the rules for doing so are often misunderstood, so I've written out all the steps exactly here, even if it's quite messy.]

x + 1 3 + x − 1 3 − 2 x 3 = ( x + 1 3 − x 3 ) − ( x 3 − x − 1 3 ) = 1 ( x + 1 3 ) 2 + x + 1 3 x 3 + ( x 3 ) 2 − 1 ( x 3 ) 2 + x 3 x − 1 3 + ( x − 1 3 ) 2 = ( ( x 3 ) 2 + x 3 x − 1 3 + ( x − 1 3 ) 2 ) − ( ( x + 1 3 ) 2 + x + 1 3 x 3 + ( x 3 ) 2 ) ( ( x + 1 3 ) 2 + x + 1 3 x 3 + ( x 3 ) 2 ) ( ( x 3 ) 2 + x 3 x − 1 3 + ( x − 1 3 ) 2 ) = ( x − 1 3 − x + 1 3 ) x + 1 3 + x 3 + x − 1 3 ( ( x + 1 3 ) 2 + x + 1 3 x 3 + ( x 3 ) 2 ) ( ( x 3 ) 2 + x 3 x − 1 3 + ( x − 1 3 ) 2 ) = − 2 ( x + 1 3 + x 3 + x − 1 3 ) ( ( x − 1 3 ) 2 + x − 1 3 x + 1 3 + ( x + 1 3 ) 2 ) ( ( x + 1 3 ) 2 + x + 1 3 x 3 + ( x 3 ) 2 ) ( ( x 3 ) 2 + x 3 x − 1 3 + ( x − 1 3 ) 2 ) = x − 5 / 3 ⋅ − 2 ( 1 + 1 x 3 + 1 + 1 − 1 x 3 ) ( ( 1 − 1 x 3 ) 2 + 1 − 1 x 3 1 + 1 x 3 + ( 1 + 1 x 3 ) 2 ) ( ( 1 + 1 x 3 ) 2 + 1 + 1 x 3 + 1 ) ( 1 + 1 − 1 x 3 + ( 1 − 1 x 3 ) 2 ) \begin{aligned} \sqrt[3]{x+1} &+ \sqrt[3]{x-1} - 2\sqrt[3]{x} = \left(\sqrt[3]{x+1} - \sqrt[3]{x}\right) - \left(\sqrt[3]{x} - \sqrt[3]{x-1}\right) \\ &= \frac{1}{\left(\sqrt[3]{x+1}\right)^2 + \sqrt[3]{x+1}\sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2} - \frac{1}{\left(\sqrt[3]{x}\right)^2 + \sqrt[3]{x}\sqrt[3]{x-1} + \left(\sqrt[3]{x-1}\right)^2} \\ &= \frac{\left(\left(\sqrt[3]{x}\right)^2 + \sqrt[3]{x}\sqrt[3]{x-1} + \left(\sqrt[3]{x-1}\right)^2\right) - \left(\left(\sqrt[3]{x+1}\right)^2 + \sqrt[3]{x+1}\sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2\right)}{\left(\left(\sqrt[3]{x+1}\right)^2 + \sqrt[3]{x+1}\sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2\right)\left(\left(\sqrt[3]{x}\right)^2 + \sqrt[3]{x}\sqrt[3]{x-1} + \left(\sqrt[3]{x-1}\right)^2\right)} \\ &= \left(\sqrt[3]{x-1} - \sqrt[3]{x+1}\right)\frac{\sqrt[3]{x+1} + \sqrt[3]{x} + \sqrt[3]{x-1}}{\left(\left(\sqrt[3]{x+1}\right)^2 + \sqrt[3]{x+1}\sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2\right)\left(\left(\sqrt[3]{x}\right)^2 + \sqrt[3]{x}\sqrt[3]{x-1} + \left(\sqrt[3]{x-1}\right)^2\right)} \\ &= \frac{-2\left(\sqrt[3]{x+1} + \sqrt[3]{x} + \sqrt[3]{x-1}\right)}{\left(\left(\sqrt[3]{x-1}\right)^2 + \sqrt[3]{x-1}\sqrt[3]{x+1} + \left(\sqrt[3]{x+1}\right)^2\right)\left(\left(\sqrt[3]{x+1}\right)^2 + \sqrt[3]{x+1}\sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2\right)\left(\left(\sqrt[3]{x}\right)^2 + \sqrt[3]{x}\sqrt[3]{x-1} + \left(\sqrt[3]{x-1}\right)^2\right)} \\ &= x^{-5/3} \cdot \frac{-2\left(\sqrt[3]{1+\frac{1}{x}} + 1 + \sqrt[3]{1-\frac{1}{x}}\right)}{\left(\left(\sqrt[3]{1-\frac{1}{x}}\right)^2 + \sqrt[3]{1-\frac{1}{x}}\sqrt[3]{1+\frac{1}{x}} + \left(\sqrt[3]{1+\frac{1}{x}}\right)^2\right)\left(\left(\sqrt[3]{1+\frac{1}{x}}\right)^2 + \sqrt[3]{1+\frac{1}{x}} + 1\right)\left(1 + \sqrt[3]{1-\frac{1}{x}} + \left(\sqrt[3]{1-\frac{1}{x}}\right)^2\right)} \\ \end{aligned}

Therefore lim ⁡ x → ∞ x p ( x + 1 3 + x − 1 3 − 2 x 3 ) = lim ⁡ x → ∞ x p − 5 3 ⋅ lim ⁡ x → ∞ − 2 ( 1 + 1 x 3 + 1 + 1 − 1 x 3 ) ( ( 1 − 1 x 3 ) 2 + 1 − 1 x 3 1 + 1 x 3 + ( 1 + 1 x 3 ) 2 ) ( ( 1 + 1 x 3 ) 2 + 1 + 1 x 3 + 1 ) ( 1 + 1 − 1 x 3 + ( 1 − 1 x 3 ) 2 ) = lim ⁡ x → ∞ x p − 5 3 ⋅ − 2 ( 3 ) 3 3 = { 0 p < 5 3 − 2 9 p = 5 3 − ∞ p > 5 3 \begin{aligned} \lim_{x\to\infty} & x^p\left(\sqrt[3]{x+1} + \sqrt[3]{x-1} - 2\sqrt[3]{x}\right) \\ &= \lim_{x\to\infty} x^{p-\frac{5}{3}} \cdot \lim_{x\to\infty} \frac{-2\left(\sqrt[3]{1+\frac{1}{x}} + 1 + \sqrt[3]{1-\frac{1}{x}}\right)}{\left(\left(\sqrt[3]{1-\frac{1}{x}}\right)^2 + \sqrt[3]{1-\frac{1}{x}}\sqrt[3]{1+\frac{1}{x}} + \left(\sqrt[3]{1+\frac{1}{x}}\right)^2\right)\left(\left(\sqrt[3]{1+\frac{1}{x}}\right)^2 + \sqrt[3]{1+\frac{1}{x}} + 1\right)\left(1 + \sqrt[3]{1-\frac{1}{x}} + \left(\sqrt[3]{1-\frac{1}{x}}\right)^2\right)} \\ &= \lim_{x\to\infty} x^{p-\frac{5}{3}} \cdot \frac{-2(3)}{3^3} \\ &= \begin{cases} 0 & p < \frac{5}{3} \\ -\frac{2}{9} & p = \frac{5}{3} \\ -\infty & p > \frac{5}{3}\end{cases} \end{aligned}

Therefore the answer is a + b = 5 + 3 = 8 a+b = 5+3 = \boxed{8}

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