求极限：当 n 趋于无穷时 1/2((n+1)^2H + (n-1)^2H - 2n^2H) / (H(2H-1)n^(2H-2)) 的极限值

首先，将分子展开：/n/n//begin{aligned} &/frac{1/2((n+1)^{2H}+(n-1)^{2H}-2n^{2H})}{H(2H-1)n^{2H-2}}// =&/frac{1}{2H(2H-1)n^{2H-2}}((n+1)^{2H}+(n-1)^{2H}-2n^{2H})// =&/frac{1}{2H(2H-1)n^{2H-2}}((n^{2H}+2n^{2H-1}+O(n^{2H-2}))+(n^{2H}-2n^{2H-1}+O(n^{2H-2}))-2n^{2H})// =&/frac{1}{2H(2H-1)n^{2H-2}}(4n^{2H-1}+O(n^{2H-2}))// =&/frac{2n}{(2H-1)n^{2H-1}}+O(n^{-2}) /end{aligned}/n/n因此，当 $n$ 趋于无穷时，极限为 $0$。