proving that $\sup \limits_{x > 0} \frac{x\sin x}{x+1}=1$ I am having trouble proving that $$\sup ~ \Big\{  \frac{x\sin x}{x+1} \,:\, x>0 \Big\}=1$$ for my homework assignment. I have managed to prove that there is no $x$ so that $f(x) >1$ but cant seem to manage to prove there is no smaller number then $1$ for which that is true.
Can someone please help me out? Thanks.
 A: Consider the sequence $(x_n)$ defined by $x_n = \frac{\pi}{2} + 2\pi n$. We have
$$
\begin{align*}
\frac{x_n\sin(x_n)}{x_n+1} &= \frac{\frac{\pi}{2} + 2\pi n}{\frac{\pi}{2} + 2\pi n + 1}
\end{align*}
$$
which can be as close to $1$ as you please by taking $n$ sufficiently large.
A: Obviously, when $x>0$ we have $|\frac{x}{x+1}\sin x| \leq |\sin x| \leq 1$, therefore we have the inequality $\frac{x}{x+1}\sin x \leq 1$.
Now, to prove that the supremum is indeed 1, we need to find a sequence $x_n$ such that the expression evaluated at $x_n$ tends to $1$. This can be done using the sequence defined in Austin Mohr's answer, but I think it is important to understand why that is the sequence you need. when you look at the expression $\frac{x}{x+1}\sin x$ you should notice that $\lim_{x \to \infty}\frac{x}{x+1}=1$. The problem is that $\lim_{x \to \infty} \sin x$ does not exist. ( In fact, for every $\alpha \in [-1,1]$ there exists a sequence $x_n \to \infty$ such that $\sin x_n \to \alpha$.)
Now, since $\sin $ is periodic, there exists a sequence $x_n \to \infty$ such that $\sin x_n=1$, which is exactly the sequence chosen in the reffered answer. 
