Some asymptotics as difference tends to infinity

Question

Let $c>0$ and $\alpha >\frac{c}{2}$. Is it true that, as $q_2-q_1$ goes to infinity, the expression $$ (q_2-q_1)e^{-\alpha(q_2-q_1)} $$ goes faster to $0$ as $$e^{-\frac{c}{2}(q_2-q_1)}?$$

I would think so:

First of all, the expression converges to $0$ indeed:

I set $y:=q_2-q_1$. Then, by L'Hospital, $$ \lim_{y\to\infty}ye^{-\alpha y}=\lim_{y\to\infty}\frac{y}{\frac{1}{e^{-\alpha y}}}=\lim_{y\to\infty}\frac{1}{\alpha e^{\alpha y}}=\lim_{y\to\infty}\frac{1}{\alpha}e^{-\alpha y}=0 $$

But this means that $ye^{-\alpha y}=O(e^{-\alpha y})$ as $y\to\infty$ and $e^{-\alpha y}$ tends to $0$ faster than $e^{-\frac{c}{2}y}$ as $y\to\infty$ since it was assumed that $\alpha > \frac{c}{2}$.

Answer 1

More precisely, we have $\;(q_2-q_1)\mathrm e^{-\alpha(q_2-q_1)}= o\Bigl(\mathrm e^{-\tfrac c2(q_2-q_1)}\Bigr)$ since $$\frac{(q_2-q_1)\mathrm e^{-\alpha(q_2-q_1)}}{e^{-\tfrac c2(q_2-q_1)}}= (q_2-q_1)\mathrm e^{\bigl(-\alpha+\tfrac c2 \bigr)(q_2-q_1)}\xrightarrow[q_2-q_1\to\infty]{} 0$$ because $-\alpha+\frac c2<0$, and $q_2-q_1=o\Bigl(\mathrm e^{r(q_2-q_1)}\Bigr)$ for any $r>0$.

Answer 2

You are right, and it's true that $ye^{-\alpha y}$ goes to $0$ faster than $e^{-\frac{c}{2}y}$ since $ye^{-(a-\frac{c}{2}y)}$ goes to $0$ as $y$ goes to infinity.