# Proving infimum

• Jun 16th 2013, 05:08 AM
orir
Proving infimum
$\displaystyle f(x)=e^{-x}+sinx$i need to prove that inf f([0,00))=-1
• Jun 16th 2013, 05:19 AM
Prove It
Re: Proving infimum
Isn't it because $\displaystyle \displaystyle e^{-x} > 0$ and $\displaystyle \displaystyle -1 \leq \sin{(x)} \leq 1$ for all x?
• Jun 16th 2013, 05:52 AM
orir
Re: Proving infimum
That is saying (-1) is a lowe bound of the function.
It is something but it does not necesseraly say there is no bigger lower bound (inf).
I need to prove that -1 is the biggest lower bound, and there is no else
• Jun 16th 2013, 06:03 AM
Plato
Re: Proving infimum
Quote:

Originally Posted by orir
$\displaystyle f(x)=e^{-x}+sinx$i need to prove that inf f([0,00))=-1

Clearly $\displaystyle \forall x>0,~f(x)\ge -1$. $\displaystyle \forall n\in\mathbb{Z}^+$ define $\displaystyle a_n=\frac{(3n+4)\pi}{2}$.

It should clear to you that $\displaystyle {\lim _{n \to \infty }}f({a_n}) = - 1$
• Jun 16th 2013, 08:00 AM
orir
Re: Proving infimum
I need to solve that without sequences.. Havent learnt that yet..
Is there another way?
• Jun 16th 2013, 08:34 AM
Plato
Re: Proving infimum
Quote:

Originally Posted by orir
I need to solve that without sequences.. Havent learnt that yet..
Is there another way?

How are we expected to know what you have studied?

Show that if $\displaystyle \varepsilon > 0$ show that $\displaystyle \exists x_{\varepsilon}$ such that $\displaystyle -1<f\left(x_{\varepsilon}\right)<-1+{\varepsilon}$.