I'm trying to find an exponentially decaying upper bound for the function $$ f(\mu)=\frac{\mu(\rho-1)e^{-\mu(\rho-1)a}}{\rho-e^{-\mu(\rho-1)a}}, $$ as $\mu\rightarrow\infty$ where $\rho>1$ and all variables are nonnegative, i.e., in the form $$ f(\mu)\leq C_{1}e^{-C_{2}\mu}. $$ I can bound $f(\mu)$ as follows: $$ f(\mu)=\frac{\mu(\rho-1)e^{-\mu(\rho-1)a}}{\rho-e^{-\mu(\rho-1)a}}\leq\frac{\mu(\rho-1)e^{-\mu(\rho-1)a}}{\rho-1}=\mu e^{-\mu(\rho-1)a}. $$ I know that $\mu(\rho-1)e^{-\mu(\rho-1)a}$ is quasi-concave with a maximum of $1/(ea(\rho-1))$ but I haven't been able to find a bound with $\mu$ in the exponent as above.
Showing a function goes to zero exponentially fast
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calculus
real-analysis
limits
asymptotics
1 Answers
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Substituting $x = \mu (\rho-1)a$, you're looking for $C_1$ and $C_2$ such that \begin{equation} \frac{x e^{-x}}{\rho - e^{-x}} \le C_1 e^{-C_2 x} \end{equation} Taking the natural log of both sides and rearranging gives \begin{equation} \ln x - \ln\left[C_1(\rho - e^{-x})\right] \le (1 - C_2) x \end{equation} So as long as $C_2 < 1$, your function will be bounded above by $C_1 e^{-C_2 x}$ for sufficiently large $x$.