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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

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    $\begingroup$ 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 $\endgroup$ – fGDu94 Dec 8 '19 at 7:34
  • $\begingroup$ seems to agree with what i said $\endgroup$ – fGDu94 Dec 8 '19 at 7:39
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    $\begingroup$ Another thing you could do is write the denominator as $(n-1)\sin^2(\frac{n \pi}{2})+1$ and bound below by 1 $\endgroup$ – fGDu94 Dec 8 '19 at 7:40
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$$\lim_{n\rightarrow \infty}\dfrac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}=\lim_{n\rightarrow \infty}\dfrac{1}{n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2}}$$ As $n\to\infty$, since $\sin^2u$ and $\cos^2u$ both oscillate in $[0,1]$ and $n,n^2\to\infty$, So, $$(n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2})\to\infty\implies\lim_{n\rightarrow \infty}\dfrac{1}{n(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2})}=0$$

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\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*}

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  • $\begingroup$ 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 "( )". $\endgroup$ – Kevin. S Dec 8 '19 at 8:26
  • $\begingroup$ I have the parenthesis. $\endgroup$ – user284331 Dec 8 '19 at 8:28
  • $\begingroup$ 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? $\endgroup$ – Kevin. S Dec 8 '19 at 8:30
  • $\begingroup$ If you see the George comment, what I wrote is pretty similar to his claim. $\endgroup$ – user284331 Dec 8 '19 at 8:42
  • $\begingroup$ The point is that, those $\sin$ and $\cos$ are zero immediately, even without limit. $\endgroup$ – user284331 Dec 8 '19 at 8:42
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Hint: $$\frac{1}{n^2}=\frac{1}{n\left(\color{red}n\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+n\sin^2 \frac{n\pi}{2}\right)}<\frac{1}{n\left(\cos^2 \frac{n\pi}{2}+\color{red}{\require{cancel}\cancel{n}}\sin^2 \frac{n\pi}{2}\right)}=\frac1n$$

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