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For $0 \le s < 1$, $t \ge 0$ let $G(s,t) := \frac{e^{-t} s}{\sqrt{1-(1-e^{-2t})s}}$ For $\lambda > 0$ compute the limit of $G(e^{-2\lambda e^{-2t}},t)$ as $t \rightarrow \infty$.

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Well, $ G\left( \mathrm{e}^{-2 \lambda \exp(-2t)}, t\right) = \frac{ \exp(-t) \cdot \exp\left( -2 \lambda \exp(-2t) \right)}{\sqrt{1-(1-\exp(-2t) \exp\left( -2 \lambda \exp(-2t) \right) }} $ Let $u = \exp(-t)$. Then $ \lim_{t \to \infty} G\left( \mathrm{e}^{-2 \lambda \exp(-2t)}, t\right) = \lim_{u \downarrow 0} \frac{u \exp(-2 \lambda u^2)}{\sqrt{1-\left(1-u^2\right) \exp(-2 \lambda u^2) }} $ Use l'Hospital's rule, or Taylor series expansion of the exponential $\exp(-2 \lambda u^2) = 1 - 2 \lambda u^2 + \mathcal{o}(u^2)$ to get $ \lim_{t \to \infty} G\left( \mathrm{e}^{-2 \lambda \exp(-2t)}, t\right) = \lim_{u \downarrow 0} \frac{u \exp(-2\lambda u^2)}{\sqrt{1-(1-u^2)\exp(-2\lambda u^2)}} = \lim_{u \downarrow 0} \frac{\color\green{\exp\left(- 2 \lambda u^2\right)}}{\sqrt{\color\red{\frac{1-\exp\left(- 2 \lambda u^2\right)}{u^2}} + \color\green{\exp\left(- 2 \lambda u^2\right)} }} = \frac{\color\green{1}}{\sqrt{\color\red{2 \lambda} + \color\green{1}}} = \frac{1}{\sqrt{1+2\lambda}} $ where the Taylor series can be used to see that $\lim_{u \downarrow 0} \frac{1-\exp\left(- 2 \lambda u^2\right)}{u^2} = 2 \lambda$.

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    Now it's clear immediately. Thank you!2012-08-26