Limits and radicals?

$\begin{aligned} k & = \lim_{n \to \infty} \frac {(n^3-2n^2+1)^\frac 12 + (n^4+1)^\frac 13}{(n^6+6n^5+2)^\frac 14-(n^7+3n^3+1)^\frac 15} \\ & = \lim_{n \to \infty} \frac {n^\frac 32(1-2n^{-1}+n^{-3})^\frac 12 + n^\frac 43(1+n^{-4})^\frac 13}{n^\frac 32(1+6n^{-1}+2n^{-6})^\frac 14 - n^\frac 75(1+3n^{-4}+n^{-7})^\frac 15} & \small \color{#3D99F6}{\text{Divide up and down by }n^\frac 32} \\ & = \lim_{n \to \infty} \frac {(1-2n^{-1}+n^{-3})^\frac 12 + n^{-\frac 16}(1+n^{-4})^\frac 13}{(1+6n^{-1}+2n^{-6})^\frac 14 - n^{-\frac 1{10}}(1+3n^{-4}+n^{-7})^\frac 15} \\ & = \frac {(1-2\cancel{n^{-1}}^0+\cancel{n^{-3}}^0)^\frac 12 + \cancel{n^{-\frac 16}}^0(1+\cancel{n^{-4}}^0)^\frac 13}{(1+\cancel{6n^{-1}}^0+2\cancel{n^{-6}}^0)^\frac 14 - \cancel{n^{-\frac 1{10}}}^0(1+3\cancel{n^{-4}}^0+\cancel{n^{-7}}^0)^\frac 15} \\ & = \boxed{1} \end{aligned}$