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
Answer
$$\lim_{n\to\infty}\frac{n+\sin^2(n)}{n+\cos(n)}-1=\lim_{n\to\infty}\frac{\sin^2(n)-\cos(n)}{n+\cos(n)}=0,$$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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