# evaluation of Trigonometric limit

Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}$$

What i try

Put $$\displaystyle \frac{n\pi}{2}=x,$$ when $$n\rightarrow\infty,$$ Then $$x\rightarrow \infty$$

$$\frac{\pi^2}{2}\lim_{x\rightarrow \infty}\bigg(\frac{ x^{-1}}{\pi\cos^2(x)+2x\sin^2(x)}\bigg)$$

How do i solve it help me please

• For $n$ odd the limit is $\frac{1}{n(0+n)} \to 0$. For $n$ even the limit is $\frac{1}{n(1+0)} \to 0$. Hence the limit is zero – fGDu94 Dec 8 '19 at 7:34
• seems to agree with what i said – fGDu94 Dec 8 '19 at 7:39
• Another thing you could do is write the denominator as $(n-1)\sin^2(\frac{n \pi}{2})+1$ and bound below by 1 – fGDu94 Dec 8 '19 at 7:40

$$\lim_{n\rightarrow \infty}\dfrac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}=\lim_{n\rightarrow \infty}\dfrac{1}{n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2}}$$ As $$n\to\infty$$, since $$\sin^2u$$ and $$\cos^2u$$ both oscillate in $$[0,1]$$ and $$n,n^2\to\infty$$, So, $$(n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2})\to\infty\implies\lim_{n\rightarrow \infty}\dfrac{1}{n(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2})}=0$$
\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*}
• I think you mistyped the equation, It is $n^2\sin^2(n\pi/2)$ in OP's expression which should be $(2k)^2\sin^2(2k\pi/2)$ in your expression. Please correct it. Note that there exists a pair of "( )". – Kevin. S Dec 8 '19 at 8:26
• I mean... in OP's equation it's $n^2\sin^2(n\pi/2)$ Notice $n^2$, but in yours it is just $n$. Did you see the difference? – Kevin. S Dec 8 '19 at 8:30
• The point is that, those $\sin$ and $\cos$ are zero immediately, even without limit. – user284331 Dec 8 '19 at 8:42
Hint: $$\frac{1}{n^2}=\frac{1}{n\left(\color{red}n\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+\color{red}{\require{cancel}\cancel{n}}\sin^2 \frac{n\pi}{2}\right)}=\frac1n$$