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?
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?