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Math Help - Have you seen this limit before?

  1. #1
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    Have you seen this limit before?

    Hi,

    I don't know how to solve this limit I am aware of the approximation of the exponential function around 0 but it is not that helpful since T(t)/2 is non-zero (T is a temperature of about 200 degrees). If you have any ideas please let me know.

    \lim_{\Delta t\rightarrow 0}\frac{e^{\frac{T(t)}{2}}-e^{\frac{T(t-\Delta t)}{2}}}{\Delta t}

    Thanks a lot!

    Regards,
    Remus.
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  2. #2
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    Quote Originally Posted by remusmp View Post
    Hi,

    I don't know how to solve this limit I am aware of the approximation of the exponential function around 0 but it is not that helpful since T(t)/2 is non-zero (T is a temperature of about 200 degrees). If you have any ideas please let me know.

    \lim_{\Delta t\rightarrow 0}\frac{e^{\frac{T(t)}{2}}-e^{\frac{T(t-\Delta t)}{2}}}{\Delta t}

    Thanks a lot!

    Regards,
    Remus.
    Surely it's simply the limit definition of the derivative of the function e^{\frac{T(t)}{2}} ....
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  3. #3
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    Why not define \displaystyle E(t) = e^{\frac{T(t - \Delta t)}{2}}, which means \displaystyle E(t + \Delta t) = e^{\frac{T(t)}{2}}? Then

    \displaystyle \begin{align*} \lim_{\Delta t \to 0}\frac{e^{\frac{T(t)}{2}} - e^{\frac{T(t - \Delta t)}{2}}}{\Delta t} &= \lim_{\Delta t \to 0}\frac{E(t + \Delta t) - E(t)}{\Delta t} \\ &= E'(t) \end{align*}

    What does \displaystyle E'(t) equal?
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  4. #4
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    That was easy. Thanks!
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