# 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.

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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.

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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.

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hmmmm.. Cool to formally probe this do I have to prove that the limit of that at x approaching infinity is 1? – Jason Nov 25 '11 at 20:58
@Jason Prove the limit of what is 1? If you mean my expression, then this is what I mean by "can be as close to 1 as you please by taking $n$ sufficiently large" (i.e. the limit of that expression as $n$ goes to infinity is 1). If you think your instructor wants to see the details, then you should include them. If you mean the original expression $\frac{x\sin x}{x+1}$, then no, because that limit is not equal to $1$ (and you wouldn't need it to be, anyway). – Austin Mohr Nov 25 '11 at 21:02
Yep I meant the limit of what expression. Thanks :) - BTW Just out of curiosity why is the limit of $xsinx/x+1$ not 1? – Jason Nov 25 '11 at 21:04
@Jason Consider the sequence $(x_n)$ defined by $x_n = -\frac{\pi}{2} + 2\pi n$. Plug it in and see that you can get as close as you like to $-1$, as well. Since we can get close to both $1$ and $-1$ infinitely often, the limit does not exist. – Austin Mohr Nov 25 '11 at 21:07