Limit of fraction with exponential functions

BJU

New member
Joined
Jul 15, 2016
Messages
21
I was here a while ago asking a question about the limit of a function. I was kindly pointed to L'Hospital's Rule and it has served me well. But now I have a fraction I can't solve.

Here is my fraction


\(\displaystyle \frac{e^{a w} \left(e^{a v} (a (v-w)-1)+e^{a w}\right)}{\left(e^{a v}-e^{a w}\right)^2}\)

with v > w and a>0.

I want to know what happens as a goes against 0, w against v and a or v go to infinity. I (think I) know that for a->0 and w->v, the fraction goes to 1/2, and for a or v->infinity, the fraction goes to 0, but can't proof it.

Take a-> 0, for example.

The fraction becomes

\(\displaystyle \frac{1 \left(1 (0 (v-w)-1)+1\right)}{\left(1-1\right)^2}=\frac{0}{0}\)


so I use
L'Hospital's Rule and I get

\(\displaystyle \frac{(v-w) e^{a (v+w)} \left(e^{a v} (a (v-w)-2)+e^{a w} (a (v-w)+2)\right)}{\left(e^{a w}-e^{a v}\right)^3}=\frac{0}{0}\)

and so on and so on. I assume it will always remain 0/0, but at least the
denominator will remain 0 since the exponent just goes up by 1 with each derivative. So no 1/2 in sight.

I have no idea what to do. Can anybody help?
 
Ok, I'm clearly stupid. Of course it works, you just have to do it right:

L'Hospital's Rule gives

\(\displaystyle \frac{w e^{a w} \left(e^{a v} (a (v-w)-1)+e^{a w}\right)+e^{a w} \left(v e^{a v} (a (v-w)-1)+e^{a v} (v-w)+w e^{a w}\right)}{2 \left(e^{a v}-e^{a w}\right) \left(v e^{a v}-w e^{a w}\right)}\)

which for a->0 becomes

\(\displaystyle \frac{w \left(0 (v-w) - 1 + 1\right)+ \left(v (0 (v-w) - 1)+ (v - w)+w \right)}{2 \left(e^{a v}-e^{a w}\right) \left(v e^{a v}-w e^{a w}\right)}\)

which is still 0/0, but the next step gives

\(\displaystyle \frac{e^{a w} \left(e^{a v} (v+w) \left(a v^2-w (a w+3)+v\right)+4 w^2 e^{a w}\right)}{4 v^2 e^{2 a v}-2 (v+w)^2 e^{a (v+w)}+4 w^2 e^{2 a w}}\)

which for a->0 becomes

\(\displaystyle \frac{(v+w) \left(-3w + v \right)+4 w^2}{4 v^2 -2 (v+w)^2 +4 w^2 }=\frac{(v-w)^2}{ 2(v-w)^2 }=\frac{1}{2}\)

So the limit is indeed 1/2.

Sorry for wasting everybody's time with this.
 
Ok, I'm clearly stupid. Of course it works, you just have to do it right:

L'Hospital's Rule gives

\(\displaystyle \frac{w e^{a w} \left(e^{a v} (a (v-w)-1)+e^{a w}\right)+e^{a w} \left(v e^{a v} (a (v-w)-1)+e^{a v} (v-w)+w e^{a w}\right)}{2 \left(e^{a v}-e^{a w}\right) \left(v e^{a v}-w e^{a w}\right)}\)

which for a->0 becomes

\(\displaystyle \frac{w \left(0 (v-w) - 1 + 1\right)+ \left(v (0 (v-w) - 1)+ (v - w)+w \right)}{2 \left(e^{a v}-e^{a w}\right) \left(v e^{a v}-w e^{a w}\right)}\)

which is still 0/0, but the next step gives

\(\displaystyle \frac{e^{a w} \left(e^{a v} (v+w) \left(a v^2-w (a w+3)+v\right)+4 w^2 e^{a w}\right)}{4 v^2 e^{2 a v}-2 (v+w)^2 e^{a (v+w)}+4 w^2 e^{2 a w}}\)

which for a->0 becomes

\(\displaystyle \frac{(v+w) \left(-3w + v \right)+4 w^2}{4 v^2 -2 (v+w)^2 +4 w^2 }=\frac{(v-w)^2}{ 2(v-w)^2 }=\frac{1}{2}\)

So the limit is indeed 1/2.

Sorry for wasting everybody's time with this.
See your other thread.
\(\displaystyle \dfrac{e^t\, (t-1)\, +\, 1}{(e^t\, -\, 1)^2}\)
might be easier to consider.
 
Top