Integral Inside a Limit

Calculus Level 3

lim ⁡ n → ∞ n 2 ∫ 0 1 n x 2018 x + 1 d x = ? \large\lim_{n\to\infty}n^2\int_0^{\frac 1n}x^{2018x+1}\, dx=\, ?


The answer is 0.5.

Carwaniwer Qee by Carwaniwer Qee , 25, China

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1 solution

Brian Lie
Jun 12, 2018

L = lim ⁡ n → ∞ n 2 ∫ 0 1 n x 2018 x + 1 d x Let t = 1 n = lim ⁡ t → 0 + ∫ 0 t x 2018 x + 1 d x t 2 Using L’H o ˆ pital’s rule = lim ⁡ t → 0 + t 2018 t + 1 2 t = 1 2 ( lim ⁡ t → 0 + t t ) 2018 = 1 2 = 0.5 \begin{aligned} L&=\lim_{n\to\infty}n^2\int_0^{\tfrac 1n}x^{2018x+1}\, dx&\small\color{#3D99F6}\text{Let }t=\frac 1n \\&=\lim_{t\to 0^+}\frac{\displaystyle\int_0^tx^{2018x+1}\, dx}{t^2}&\small\color{#3D99F6}\text{Using L'Hôpital's rule} \\&=\lim_{t\to 0^+}\frac{t^{2018t+1}}{2t} \\&=\frac 12\left(\lim_{t\to\ 0^+}t^{t}\right)^{2018} \\&=\frac 12=\boxed{0.5} \end{aligned}

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