evaluation of Trigonometric limit

\begin{align*} \lim_{k\rightarrow\infty}\dfrac{1}{2k[\cos^{2}(2k\pi/2)+2k\sin^{2}(2k\pi/2)]}=\lim_{k\rightarrow\infty}\dfrac{1}{2k}=0, \end{align*} while \begin{align*} &\lim_{k\rightarrow\infty}\dfrac{1}{(2k+1)[\cos^{2}((2k+1)\pi/2)+(2k+1)\sin^{2}((2k+1)\pi/2)]}\\ &=\lim_{k\rightarrow\infty}\dfrac{1}{(2k+1)(2k+1)}\\ &=0, \end{align*} so \begin{align*} \lim_{n\rightarrow\infty}\dfrac{1}{n[\cos^{2}(n\pi/2)+n\sin^{2}(n\pi/2)]}=0. \end{align*}

Hint: $$\frac{1}{n^2}=\frac{1}{n\left(\color{red}n\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+\color{red}{\require{cancel}\cancel{n}}\sin^2 \frac{n\pi}{2}\right)}=\frac1n$$

$$\lim_{n\rightarrow \infty}\dfrac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}=\lim_{n\rightarrow \infty}\dfrac{1}{n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2}}$$ As $n\to\infty$, since $\sin^2u$ and $\cos^2u$ both oscillate in $[0,1]$ and $n,n^2\to\infty$, So, $$(n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2})\to\infty\implies\lim_{n\rightarrow \infty}\dfrac{1}{n(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2})}=0$$