# Find$\lim\limits_{n\rightarrow\infty}$ $\frac{n+\sin\left(n^{2}\right)}{n+\cos\left(n\right)}$

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

• 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 $1$. – samjoe Jan 4 '18 at 11:56
• @SamjoeThanks i am gonna try it. Brother there is one more question i posted it yesterday math.stackexchange.com/questions/2590230/… – Mohan Sharma Jan 4 '18 at 11:59
• @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 ? – Mohan Sharma Jan 4 '18 at 12:03
• 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 $1$ – samjoe Jan 4 '18 at 12:09

$$\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}-1=\lim_{n\to\infty}\frac{\sin^2(n)-\cos(n)}{n+\cos(n)}=0,$$since the numerator is bounded and the denominator tends to $+\infty$. Therefore$$\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}=1.$$
• $\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}$-1 Did you use any theorem .Why did you used -1 ? – Mohan Sharma Jan 4 '18 at 12:07
• @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 $\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}-1=0$. – José Carlos Santos Jan 4 '18 at 12:14
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 $$\frac{\sin(n^{2})}{n} \to0~~~and ~~~\frac{\cos(n)}{n} \to0$$
• 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 ? – Mohan Sharma Jan 4 '18 at 12:08