Have you seen this limit before?

**remusmp** · June 1st 2011, 12:39 AM

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.

**mr fantastic** · June 1st 2011, 02:33 AM

Originally Posted by remusmp

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 $e^{\frac{T(t)}{2}}$ ....

**Prove It** · June 1st 2011, 02:33 AM

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?

**remusmp** · June 1st 2011, 05:24 AM

That was easy. Thanks!

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