Know your limits

Calculus Level 2

lim ⁡ x → ∞ 1 3 + 2 3 + 3 3 + … + x 3 ( x 2 + 1 x 2 ) 2 = a b \lim_{x\rightarrow \infty} \frac{1^3+2^3 + 3^3+\ldots+x^3}{\left(x^2+\frac{1}{x^2}\right)^2} = \frac ab

The equation above holds true for positive coprime integers a a and b b . Enter the value of a + b a+b


The answer is 5.

Chandrashekar Giridharan by Chandrashekar Giridharan , 18, India

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1 solution

L = lim ⁡ x → ∞ 1 3 + 2 3 + 3 3 + ⋯ + x 3 ( x 2 + 1 x 2 ) 2 = lim ⁡ x → ∞ ( x ( x + 1 ) 2 ) 2 ( x 2 + 1 x 2 ) 2 = lim ⁡ x → ∞ ( x ( x + 1 ) 2 ( x 2 + 1 x 2 ) ) 2 Divide up and down by x 2 = lim ⁡ x → ∞ ( 1 + 1 x 2 2 ( 1 + 1 x 4 ) ) 2 = 1 4 \begin{aligned} L & = \lim_{x \to \infty} \frac {1^3+2^3+3^3+\cdots + x^3}{\left(x^2+\frac 1{x^2}\right)^2} \\ & = \lim_{x \to \infty} \frac {\left(\frac {x(x+1)}2\right)^2}{\left(x^2+\frac 1{x^2}\right)^2} \\ & = \lim_{x \to \infty} \left({\color{#3D99F6}\frac {x(x+1)}{2\left(x^2+\frac 1{x^2}\right)}}\right)^2 & \small \color{#3D99F6} \text{Divide up and down by }x^2 \\ & = \lim_{x \to \infty} \left({\color{#3D99F6} \frac {1+\frac 1{x^2}}{2\left(1+\frac 1{x^4}\right)}}\right)^2 & \small \color{#3D99F6} \\ & = \frac 14 \end{aligned}

Therefore, a + b = 1 + 4 = 5 a+b = 1+4 = \boxed{5} .

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