supremum proof question

**transgalactic** · August 4th 2011, 04:18 AM

prove that $sup\{\frac{xsinx}{x+1}|x>0\}=1$
here is the solution:
for $x>0$ $x+1>x$ so $0<\frac{x}{x+1}<1$ so 1 is upper bound so
there is a supremum suppose 1 is not the smallest upper bound
suppose $sup\{f(R)\}=1-\epsilon$ . $lim_{x->\infty}\frac{x}{x+1}=1$
so by the limit definition for every epsilon there is M>0 so for every
x for which x>M $|\frac{x}{x+1}-1|<\epsilon$
so $\frac{x}{x+1}>1-\epsilon$
and it works for every x for which x>M
we take $x_{0}=2\pi([M]+45)+\frac{\pi}{2}$ $sin(x_{0})=1$
for this $x_{0}$ it works too
$\frac{x_{0}}{x_{0}+1}=\frac{x_{0}sin(x_{0})}{x_{0} +1}>1-\epsilon$
where is the contradiction here??
we were talking about the supremum of $\frac{x}{x+1}$ not
$\frac{xsin(x)}{x+1}$

??

**girdav** · August 4th 2011, 04:43 AM

No, the solution shows that if the supremum is $1-\varepsilon$ then using the fact that $\lim_{x\to+\infty} \frac x{x+1}=1$ we find $x_0$ such that $\frac{x_0\sin x_0}{x_0+1}>1-\varepsilon$ which is a contradiction.
But here, we don't have to prove the result by contradiction. Since $\frac{x\sin x}{x+1}\leq 1$ we know that the supremum is $\leq 1$ . Then consider for $n\in\mathbb{N}$ : $x_n :=2n\pi+\frac{\pi}2$ (the idea is quite the same as the solution gives, but I used a "direct way").

**transgalactic** · August 4th 2011, 05:06 AM

why its a contradiction
what it contradicts
?

**girdav** · August 4th 2011, 05:14 AM

In the solution, we assumed that $\sup \left\{\frac{x\sin x}{x+1}\right\}\neq 1$ and since we know that this supremum is $\leq 1$ we should have $\sup \left\{\frac{x\sin x}{x+1}\right\}< 1$ . Hence we can find $\varepsilon>0$ such that $\sup \left\{\frac{x\sin x}{x+1}\right\}= 1-\varepsilon$ . Do you agree that we contradict that if we can find $x_0$ such that $\frac{x_0\sin x_0}{1+x_0}$ ?

**transgalactic** · August 4th 2011, 05:37 AM

ok
we assumed that the supremum is 1-epsilon

but we fund a certain x in which the function is bigger
so its not a supremum

regarding your solution
you picked a certain x
how it proves that 1 is a supremum
?

**girdav** · August 4th 2011, 05:43 AM

We define $f(x):=\frac{x\sin x}{x+1}$ . We have $f(x_n) =\frac{2\pi n+\frac{\pi}2}{2\pi n+\frac{\pi}2+1}$ , and if you take the limit when $n\to \infty$ you will find $1$ . Now apply the definition of supremum and you are done.

**transgalactic** · August 4th 2011, 05:58 AM

supremum is the smallest upper bound
i dont how this definition will help here

?
and you found a limit to a certain group of x
not all of them

**girdav** · August 4th 2011, 07:11 AM

To conclude, use $\sup\left\{f(x);x>0\right\}\geq \sup \left\{f(x_n),n\in\mathbb{N}\right\}$ and compute $\sup \left\{f(x_n),n\in\mathbb{N}\right\}$ .

**transgalactic** · August 4th 2011, 10:27 AM

why the supremum of this subgroup is >= of the bigger group

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