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Let $t \in \mathbb{R}_+$, $\varepsilon > 0$ and $p_\varepsilon \in [0, 1]$. There is in a book the following relation: $$ \underset{\varepsilon \rightarrow 0}{lim} (1 - p_\varepsilon)^{\lceil t / \varepsilon \rceil} = \underset{\varepsilon \rightarrow 0}{lim} \hspace{3pt} e^{\frac{log(1 - p_\varepsilon)}{\varepsilon}t} \in [0, 1]. $$ I don't understand the above relation. $$e^{\frac{log(1 - p_\varepsilon)}{\varepsilon}t} = e^{\frac{t}{\varepsilon} log(1-p_\varepsilon)} $$ and then?

Thank you!

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    $\begingroup$ It's always true that $a^b=e^{b\ln a}$. And it's always true that $a\ln b=\ln b^a$. Also, you have made a mistake in your calculation, $a^ba^c=a^{b+c}$ not $a^{bc}$. $\endgroup$ Mar 16, 2015 at 12:10
  • $\begingroup$ The problem in the question is not the one expanded in a comment but rather to explain why $$\lim_{\varepsilon\to0+}a(\varepsilon)\cdot\lceil t / \varepsilon \rceil=t\cdot\lim_{\varepsilon\to0+}a(\varepsilon) / \varepsilon.$$ To do so, start from the obvious $$-1+t / \varepsilon\le\cdot\lceil t / \varepsilon \rceil\le t / \varepsilon.$$ $\endgroup$
    – Did
    Mar 16, 2015 at 14:00

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