# question in sup

• Mar 27th 2013, 01:00 AM
orir
question in sup
i need to prove that
sup $\displaystyle {(xsin(x))/(x+1) |x>0}}=1$

i succeded proving that 1 is an upper bound of $\displaystyle f(x)$
now i need to prove that 1 is the lowest upper bound. i tried to assume that there's another number (let called $\displaystyle c$), that is also an upper bound of $\displaystyle f(x)$ but is lower than 1.
$\displaystyle c<1$
but i don't really know how to go forward.. (Worried)
what can i do? i'm so suck in proving things...
• Mar 27th 2013, 01:02 AM
Prove It
Re: question in sup
Hang on, I can barely read this, are you asking to prove that the supremum of $\displaystyle \displaystyle \frac{\sin{(x)}}{x} + 1$ is 0 if x > 0?
• Mar 27th 2013, 01:06 AM
orir
Re: question in sup
sorry, i don't know how to add math to a post..
i ment the supremum of (xsinx)/(x+1)=1 (at x>0)
• Mar 27th 2013, 04:56 AM
HallsofIvy
Re: question in sup
That can be written $\displaystyle \frac{x}{x+1}sin(x)$. Now look at $\displaystyle x= N\pi/2$ for N a very large odd number.
• Mar 27th 2013, 05:21 AM
orir
Re: question in sup
Quote:

Originally Posted by HallsofIvy
That can be written $\displaystyle \frac{x}{x+1}sin(x)$. Now look at $\displaystyle x= N\pi/2$ for N a very large odd number.

yeah, in this condition (N approaches to infinity) it will all approach 1. but is it just enough - does it necessarly tell me that there's no other upper bound of $\displaystyle f(x)$ which is smaller than 1??
• Mar 27th 2013, 05:35 AM
orir
Re: question in sup
(i accidently posted another post, and i can't delete it..)
• Mar 27th 2013, 06:32 AM
Plato
Re: question in sup
Quote:

Originally Posted by orir
yeah, in this condition (N approaches to infinity) it will all approach 1. but is it just enough - does it necessarly tell me that there's no other upper bound of $\displaystyle f(x)$ which is smaller than 1??

Note that $\displaystyle {\lim _{x \to \infty }}\frac{x}{{x + 1}} = 1$

Also if $\displaystyle n\in\mathbb{Z}^+$ then $\displaystyle \sin \left( {\frac{{(1 + 4n)\pi }}{2}} \right) = 1$.

If you put those two facts together, then it easy to see why $\displaystyle 1\text{ is the }\sup$.
• Mar 27th 2013, 06:48 AM
orir
Re: question in sup
i'm sorry, but it's not easy to me to see from that that 1 is the sup. can you explain more? (and why 1+4n?)
• Mar 27th 2013, 07:13 AM
Plato
Re: question in sup
Quote:

Originally Posted by orir
i'm sorry, but it's not easy to me to see from that that 1 is the sup. can you explain more? (and why 1+4n?)

No I cannot teach you basic mathematics.
• Mar 27th 2013, 08:20 AM
orir
Re: question in sup
Quote:

Originally Posted by Plato
No I cannot teach you basic mathematics.

i didn't ask you to teach me basic math, so you don't need to be so supercilious.. just wanted a bit more explanation, that's it.
• Mar 27th 2013, 03:19 PM
Prove It
Re: question in sup
Plato is pointing out that the sine function can never be any greater than 1. Since the two functions are always bounded above by 1, then so must be their product...