计算frac12n+1^2H+n-1^2H-2n^2HH2H-1n^2H-2的极限

首先，我们可以将分子化简： \begin{aligned} &\frac{1}{2}\left(\frac{(n+1)^{2H}}{n^{2H-2}}+\frac{(n-1)^{2H}}{n^{2H-2}}-\frac{2n^{2H}}{n^{2H-2}}\right)\ =&\frac{1}{2}\left(\left(\frac{n+1}{n}\right)^{2H}+\left(\frac{n-1}{n}\right)^{2H}-2\right)\ =&\frac{1}{2}\left[\left(1+\frac{1}{n}\right)^{2H}+\left(1-\frac{1}{n}\right)^{2H}-2\right]\end{aligned} 接下来，我们将分母中的$n^{2H-2}$移到分式的前面： \begin{aligned} &\frac{1}{H(2H-1)n^{2H-2}}\ =&\frac{\frac{1}{n^{2H}}}{H(2H-1)n^{-2}}\ =&\frac{1}{H(2H-1)n^{-2}}\cdot\frac{1}{n^{2H}}\ =&\frac{1}{H(2H-1)}\left(\frac{1}{n^2}\right)^{H-1}\end{aligned} 现在，我们可以将分式的分子和分母带入极限公式： \begin{aligned} \lim_{n\to\infty}\frac{\frac{1}{2}\left[\left(1+\frac{1}{n}\right)^{2H}+\left(1-\frac{1}{n}\right)^{2H}-2\right]}{\frac{1}{H(2H-1)}\left(\frac{1}{n^2}\right)^{H-1}} &=\lim_{n\to\infty}\frac{\frac{1}{2}\left[\left(1+\frac{1}{n}\right)^{2H}+\left(1-\frac{1}{n}\right)^{2H}-2\right]}{\frac{1}{H(2H-1)}\cdot\frac{1}{n^{2H-2H+1}}}\ &=\lim_{n\to\infty}\frac{\frac{1}{2}\left[\left(1+\frac{1}{n}\right)^{2H}+\left(1-\frac{1}{n}\right)^{2H}-2\right]\cdot n^{2H-1}}{\frac{1}{H(2H-1)}}\ &=\frac{1}{2}\cdot H(2H-1)\cdot 2\ &=\boxed{H(2H-1)}\end{aligned}