<s>[INST] Conclude whether the limit $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})+\sin(2x)] $ exists or not . Conclude whether the limit $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})+\sin(2x)] $ exists or not .  
Answer: 
Since $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})]=0 , \ \ and \ \ - 1\leq \sin(2x) \leq 1 $,  the given limit oscillates between $ -1 \ \ to \ \ 1 $. 
So the limit does not exists . 
I need confirmation about my work. Any help is there ?
 [/INST] Let us consider the following two sequences:
$$a_n=n\pi+\frac{\pi}{4}$$
$$b_n=n\pi+\frac{3\pi}{4}$$
and let $f(x)=\ln\left(1+\frac{1}{x}\right)+\sin(2x)$
So, we see that:
$$\begin{align*}f(a_n)=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(2n\pi+\frac{\pi}{2}\right)=\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(\frac{\pi}{2}\right)=\\=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+1\overset{n\to\infty}{\longrightarrow}1
\end{align*}$$
and
$$\begin{align*}f(b_n)=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(2n\pi+\frac{3\pi}{2}\right)=\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(\frac{3\pi}{2}\right)=\\=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)-1\overset{n\to\infty}{\longrightarrow}-1
\end{align*}$$
So, the limit $\lim\limits_{x\to+\infty}f(x)$ does not exist since we have shown that $f$ has at least two accumulation points as $x\to+\infty$.
</s>