Limit of sum of wha?!

Relevant wiki: Sum of n, n², or n³

$\begin{aligned} L & = \lim_{n \to \infty} \frac {9n(1^8+2^8+3^8+\cdots+n^8)}{1^9+2^9+3^9+\cdots+n^9} \\ & = \lim_{n \to \infty} \frac {9n \color{#3D99F6} \sum_{k=1}^n k^8}{\color{#3D99F6}\sum_{k=1}^n k^9} & \small \color{#3D99F6} \text{Using Faulhaber's formula} \\ & = \lim_{n \to \infty} \frac {9n\cdot \color{#3D99F6}\frac 1{90}(10n^9+45n^8+60n^7-42n^5+20n^3-3n)}{\color{#3D99F6}\frac 1{20}(2n^{10}+10 n^9+15n^8-14n^6+10n^4-3n^2)} \\ & = \lim_{n \to \infty} \frac {2(10n^{10}+45n^9+60n^8-42n^6+20n^4-3n^2)}{2n^{10}+10 n^9+15n^8-14n^6+10n^4-3n^2} & \small \color{#3D99F6} \text{Divide up and down by }n^{10} \\ & = \lim_{n \to \infty} \frac {2\left(10+\frac {45}n+ \frac {60}{n^2}- \frac {42}{n^4}+\frac {20}{n^6}-\frac 3{n^8}\right)}{2+\frac {10}n+\frac {15}{n^2}-\frac {14}{n^4}+\frac {10}{n^6}-\frac 3{n^8}} \\ & = \boxed{10} \end{aligned}$

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Reference: Faulhaber's formula

Not a solution, but it could be a hint towards one of the methods to solve this: https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula