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Math Help - Proving infimum

  1. #1
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    Proving infimum

    f(x)=e^{-x}+sinx i need to prove that inf f([0,00))=-1
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  2. #2
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    Re: Proving infimum

    Isn't it because \displaystyle e^{-x} > 0 and \displaystyle -1 \leq \sin{(x)} \leq 1 for all x?
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    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
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    Re: Proving infimum

    Quote Originally Posted by orir View Post
    f(x)=e^{-x}+sinx i need to prove that inf f([0,00))=-1
    Clearly \forall x>0,~f(x)\ge -1. \forall n\in\mathbb{Z}^+ define a_n=\frac{(3n+4)\pi}{2}.

    It should clear to you that {\lim _{n \to \infty }}f({a_n}) =  - 1
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  5. #5
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    Re: Proving infimum

    I need to solve that without sequences.. Havent learnt that yet..
    Is there another way?
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  6. #6
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    Re: Proving infimum

    Quote Originally Posted by orir View Post
    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 \varepsilon  > 0 show that \exists x_{\varepsilon} such that -1<f\left(x_{\varepsilon}\right)<-1+{\varepsilon}.
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