Finding supremum of $\frac{x}{x+1} \cdot \sin x$ I have to find the supremum of the following function: $$f(x)=\frac{x}{x+1} \cdot \sin x$$, where $x \in (0,\infty)$
I think I know it is equal to $1$ but I can't prove it.
Where I'm stuck proving that $\sup f =1$:

Let $\sup f = y$
Let $\varepsilon>0$
We'll prove that for every $\varepsilon>0$ there is a value of $x$ such that $\frac{x}{x+1} \cdot \sin x$ > $y - \varepsilon$ - That is the formal definition in my text book.

Thanks!
 A: Hint : $\sin(x)=1$ has arbitary large solutions and we have $$\lim_{x\rightarrow \infty} \frac{x}{x+1}=1$$
Moreover, $\frac{x}{x+1}$ is strictly increasing for $x>0$
A: The sup of your function is $+\infty$; in fact, you have that $\lim _{x\to-1}f(x)=+\infty$ if $x$ tends to $-1$ from the right. With the domain restriction, we can observe that $f(x)< \frac x {x+1}$ that is $<1$ and tends to $1$ as $x\to+\infty$.
A: Let $\epsilon>0$ be given.  Choose $N$ so big that $1-\frac{1}{N+1} = \frac{N}{N+1} > 1 -\epsilon.$  Then for any integer $M$ such that $\alpha = \frac{(2M+1)\pi}{2} >N$, we have  
$$1 - \frac{\alpha}{1+\alpha} \sin(\alpha) = 1-  \frac{\alpha}{1+\alpha} > 1-\epsilon. $$
So the supremum is at least $1.$
A: (1) First we argue that $f(x)\le1$ for all $x\ge0$.
Indeed, $\sin x\le1$ implies $x\sin x\le x$ implies $x\sin x\le x+1$ implies $f(x)\le1$.
(2) Next we argue that $f$ comes arbitrarily close to $1$.
For this, consider $x_k=\frac{\pi}{2}+k\pi$ for an integer $k\ge0$. Since $\sin(x_k)=1$, we get
$$f(x_k) = \frac{x_k}{(x_k+1)} 
= \frac{\frac{\pi}{2}+k\pi}{\frac{\pi}{2}+k\pi+1} 
\ge \frac{\frac{\pi}{2}+k\pi}{\frac{\pi}{2}+k\pi+ \frac{\pi}{2}}
= \frac{2k+1}{2k+2}.$$
Now if we pick $k>(1-2\varepsilon)/\varepsilon$, we have
$$ \frac{2k+1}{2k+2}>1-\varepsilon. $$
Hence for every $\varepsilon>0$, there indeed exists the desired 
value $x$ for which $f(x)>1-\varepsilon$.
(3) This means that the supremum is $1$.
