Hello
Can anybody help me to find the lim zero value of an exponential function where, as you can imagine, at t=0 you have a division by zero situation.
I've put everything on the attached image with the exact values of the constants.
Thanks
Hello
Can anybody help me to find the lim zero value of an exponential function where, as you can imagine, at t=0 you have a division by zero situation.
I've put everything on the attached image with the exact values of the constants.
Thanks
Thanks Haven
In fact, I found, by playing around with small values of a and b, that lim zero (exp(bx) - 1)/(exp(ax)-exp(bx)) was quite simply b/(a-b).
Could you guide me through to the mathematical demonstration of this result. I have heard of hôpital's rule, but either my teachers deemed we didn't need to know it, or we weren't up to that level. In case you're wondering, this lim zero is the second boundary condition for solving a second order differential equation.
Thanks
L'Hopital's rule says that if $\displaystyle \lim_{x\to a} f(x)= 0$ and $\displaystyle \lim_{x\to a}g(x)= 0$ then
$\displaystyle \lim_{t\to a}\frac{f(t)}{g(t)}= \lim_{t\to a}\frac{f'(t)}{g'(t}}$ if that last limit exists.
Here, $\displaystyle f(t)= e^{\beta t}- 1$ and $\displaystyle g(t)= e^{\alpha t}- e^{\beta t}$, both of which go to 0 as t goes to 0.
So $\displaystyle f'(t)= \beta e^{\beta t}$ and $\displaystyle g'(t)= \alpha e^{\alpha t}- \beta e^{\beta t}$
So the problem becomes to find $\displaystyle \lim_{t\to 0}\frac{\beta e^{\beta t}}{\alpha e^{\alpha t}- \beta e^{\beta t}}$'
which is fairly easy.
Perfect, and Ive just done a Wiki on Hôpital, all that was mastered 400 years ago, makes me feel rather small....
(I posted another query earlier on in the day but it was straightforward so I've edited it away)