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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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Thank you! I'm still having trouble seeing how you show that $o(u^2)$ factors don't affect the limit, especially when they occur inside the root. Why do $$\frac{u(1-2\lambda u^2)+u o(u^2)}{\sqrt{(1-(1-u^2)(1-2 \lambda u^2)+(1-u^2)o(u^2)}}$$ and $$\frac{u(1-2 \lambda u^2)}{\sqrt{(1-(1-u^2)(1-2 \lambda u^2)}}$$ have the same limit? –  Haderlump Aug 26 '12 at 16:25
    
Well, the series expansion tells you that $\lim_{u \downarrow 0} =\frac{1-\exp(-2\lambda u^2)}{u^2} = 2 \lambda$. Transform the denominator as $1-(1-u^2)\exp(-2 \lambda u^2) = u^2 \left(\frac{1-\exp(-2\lambda u^2)}{u^2} + \exp(-2 \lambda u^2) \right)$. Does this help to compute the limit? –  Sasha Aug 26 '12 at 16:44
    
@Haderlump I have edited the post to better show evaluation of the limit. –  Sasha Aug 26 '12 at 16:49
    
Now it's clear immediately. Thank you! –  Haderlump Aug 26 '12 at 17:29

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