prove that $\displaystyle sup\{\frac{xsinx}{x+1}|x>0\}=1$

here is the solution:

for $\displaystyle x>0$ $\displaystyle x+1>x$ so $\displaystyle 0<\frac{x}{x+1}<1$ so 1 is upper bound so

there is a supremum suppose 1 is not the smallest upper bound

suppose $\displaystyle sup\{f(R)\}=1-\epsilon$. $\displaystyle lim_{x->\infty}\frac{x}{x+1}=1$

so by the limit definition for every epsilon there is M>0 so for every

x for which x>M $\displaystyle |\frac{x}{x+1}-1|<\epsilon$

so $\displaystyle \frac{x}{x+1}>1-\epsilon$

and it works for every x for which x>M

we take $\displaystyle x_{0}=2\pi([M]+45)+\frac{\pi}{2}$ $\displaystyle sin(x_{0})=1$

for this $\displaystyle x_{0}$ it works too

$\displaystyle \frac{x_{0}}{x_{0}+1}=\frac{x_{0}sin(x_{0})}{x_{0} +1}>1-\epsilon$

where is the contradiction here??

we were talking about the supremum of $\displaystyle \frac{x}{x+1}$ not

$\displaystyle \frac{xsin(x)}{x+1}$

??