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$$\lim_{n\to\infty} \frac{2^{100+5n}}{e^{4n-10}}= \lim_{n\to\infty} \frac{2^{100}2^{5n}}{e^{4n}e^{-10}}= 2^{100}e^{10}\lim_{n\to \infty}\frac{e^{-4n}}{2^{-5n}}= 2^{100}e^{10}\frac{\lim_{n\to \infty} e^{-4n}}{\lim_{n\to \infty}2^{-5n}} = 0$$

I think it's right, both limits exist for the two exponential functions.

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All perfect...until the last step: both limits in the numerator and denominator are zero, so how did you get the whole limit is zero, too?? –  DonAntonio Sep 23 '12 at 3:13
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You'll be better off if you put everything under the same exponent $n$ and see if you get $a^n$ with $|a|<1$ or not. –  Pedro Tamaroff Sep 23 '12 at 3:15
    
The law you broke was attempting to find the limit by pure manipulation, without thinking about the relative sizes of things. –  André Nicolas Sep 23 '12 at 3:29
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up vote 3 down vote accepted

$$\lim_{n\to\infty}\frac{e^{-4n}}{2^{-5n}}=\lim_{n\to\infty}\frac{32^n}{\left(e^4\right)^n}=\lim_{n\to\infty}\left(\frac{32}{e^4}\right)^n=0$$

since

$$\left|\frac{32}{e^4}\right|<1$$

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