已知数列\(\left\{x_n\right\}\)满足\(\lim\limits_{n\to \infty}\left(x_n-x_{n-2}\right)=0\)，求证：\(\lim\limits_{n\to\infty}\dfrac{x_n-x_{n-1}}{n}=0\)．
证明：设\(y_n=x_n-x_{n-1}\)，\(n\in \mathcal N^*\)，则原问题转化为已知\(\lim\limits_{n\to \infty}\left(y_n+y_{n-1}\right)=0\)，求证：\(\lim\limits_{n\to\infty}\dfrac{y_n}{n}=0\)．
根据已知，有\(\forall \varepsilon>0\)，\(\exists N(\varepsilon)\in\mathcal N^*\)，使得当\(n>N\)时，均有\[\left|y_n+y_{n-1}\right|<\dfrac 12\varepsilon,\]此时有\[y_n=\left(y_n+y_{n-1}\right)+(-1)\left(y_{n-1}+y_{n-2}\right)+(-1)^2\left(y_{n-2}+y_{n-3}\right)+\cdots+(-1)^{n-N-1}\left(y_{N+1}+y_{N}\right)+(-1)^{n-N}y_N,\]从而\[\left|y_n\right|\leqslant \dfrac 12(n-N)\varepsilon+\left|y_N\right|,\]从而\[\dfrac{\left|y_n\right|}{n}\leqslant \dfrac 12\varepsilon -\dfrac{N}{2n}\varepsilon+\dfrac{\left|y_N\right|}{n}<\dfrac 12\varepsilon +\dfrac{\left|y_N\right|}{n},\]此时只需要取\(N_0=\max\left\{N(\varepsilon,\left[\dfrac{2\left|y_N\right|}{\varepsilon}\right]+1\right\}\)即可使得\[\dfrac{\left|y_n\right|}{n}<\varepsilon,\]因此\[\lim_{n\to\infty}\dfrac{\left|y_n\right|}{n}=0,\]进而原命题得证．