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Assuming that the following is trivial:

  1. $\lim_{x \to +\infty} \frac{2^x}{2.1^x}=0$

  2. $\lim_{x \to +\infty} \frac{2^x}{2^{x^2}}=0$

What is the most simple, intuitive way to show that:

$\lim_{x \to +\infty} \frac{2^{x^2}2.1^x}{2.1^{x^2}}=0$

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    $\begingroup$ I would just put everything over a single base. $2^{x^2}2.1^{x}/2.1^{x^2}=2^{(1 - \log_2 2.1)x^2 + x \log_2 2.1}\rightarrow 0$ because the exponent goes to $-\infty$. $\endgroup$
    – mjqxxxx
    Oct 29, 2013 at 19:27

1 Answer 1

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I think I got an explanation:

$2^{x^2}=({2^2})^{\frac{1}{2}x^2}=4^{\frac{1}{2}x^2}$

when $x \to +\infty$

$\frac{1}{2}x^2>x $

$\rightarrow\lim_{x \to +\infty} \frac{2.1^x}{4^{\frac{1}{2}x^2}}=0\rightarrow 2.1^x=o(2^{x^2})$

From this point it's really trivial (using $\lim_{x \to +\infty} \frac{2^{x^2}}{2.1^{x^2}}=0$)

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