# Thread: Have you seen this limit before?

1. ## 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.

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

3. 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?

4. That was easy. Thanks!