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Question Find $$\lim_{n\rightarrow\infty}\frac{n+\sin\left(n^{2}\right)}{n+\cos\left(n\right)}$$

My Approach $$\lim_{n\rightarrow\infty}\frac{n+\sin\left(n^{2}\right)}{n+cos\left(n\right)}=\lim_{n\rightarrow\infty}\left[\frac{n}{n+\cos\left(n\right)}+\frac{\sin\left(n^{2}\right)}{n+\cos n}\right] =\lim_{n\rightarrow\infty}\left[\frac{1}{1+\frac{\cos\left(n\right)}{n}}+\frac{\sin\left(n^{2}\right)}{n+\cos n}\right]$$

Applying L ' Hospital is not working here

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    $\begingroup$ Because its not $0/0$ form! Simply divide by $n$ on numerator and denominator and split the limit on numerator and denominator. You have the limit as $1$. $\endgroup$ – samjoe Jan 4 '18 at 11:56
  • $\begingroup$ @SamjoeThanks i am gonna try it. Brother there is one more question i posted it yesterday math.stackexchange.com/questions/2590230/… $\endgroup$ – Mohan Sharma Jan 4 '18 at 11:59
  • $\begingroup$ @samjoe lim$_{n\rightarrow\infty}$$\frac{sin\left(n^{2}\right)}{n}$ is it Intermediate Form or lim$_{n\rightarrow\infty}$$\frac{sin\left(n^{2}\right)}{n}$= 0 ? $\endgroup$ – Mohan Sharma Jan 4 '18 at 12:03
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    $\begingroup$ I think you mean Indeterminate. Well this is not and yes indeed $\lim_{n\to \infty}\frac{sin\left(n^{2}\right)}{n} = 0$. Note that numerator always lies between $-1$ and $1$ $\endgroup$ – samjoe Jan 4 '18 at 12:09
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$$\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}-1=\lim_{n\to\infty}\frac{\sin^2(n)-\cos(n)}{n+\cos(n)}=0,$$since the numerator is bounded and the denominator tends to $+\infty$. Therefore$$\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}=1.$$

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  • $\begingroup$ Thanks . there is one more question i posted it yesterday math.stackexchange.com/questions/2590230/… $\endgroup$ – Mohan Sharma Jan 4 '18 at 12:05
  • $\begingroup$ $\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}$-1 Did you use any theorem .Why did you used -1 ? $\endgroup$ – Mohan Sharma Jan 4 '18 at 12:07
  • $\begingroup$ @MohanSharma None. SInce both the numerator and the denominator behaved as $n$ when $n\gg1$, the limit had to be $1$. I justified that by proving that $\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}-1=0$. $\endgroup$ – José Carlos Santos Jan 4 '18 at 12:14
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Observes that $$\frac{n+\sin\left(n^{2}\right)}{n+\cos\left(n\right)}=\frac{1+\frac{\sin(n^{2})}{n}}{1+\frac{\cos(n )}{n})} \to 1$$

Since $$\frac{\sin(n^{2})}{n} \to0~~~and ~~~\frac{\cos(n)}{n} \to0$$

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  • $\begingroup$ lim$_{n\rightarrow\infty}$$\frac{sin\left(n^{2}\right)}{n}$ is it Intermediate Form or lim$_{n\rightarrow\infty}$$\frac{sin\left(n^{2}\right)}{n}$= 0 ? $\endgroup$ – Mohan Sharma Jan 4 '18 at 12:08
  • $\begingroup$ @MohanSharma Yes of course see the edit $\endgroup$ – user518372 Jan 4 '18 at 12:09

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