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Find $\lim\limits_{x \to \infty}\frac{2^x}{3^{x^2}}$ I can only reason with this intuitively. since $3^{x^2}$ grows much faster than $2^x$ the limit as $x \to \infty$ of $f$ must be 0.

Is there a more rigorous way to show this?

5 Answers 5

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After so many equivalent examples one more should not be missed because of its form. Here we get the numerator constant and keep the notation in powers of 2 and 3. Denote ß=\frac{\ln(2)}{\ln(3)} Then \frac{2^x}{3^{x^2}}=3^{ßx-x^2}=3^{(ß/2)²-(x-\fracß2)^2} = \frac{2^{ß/4}}{3^{(x-ß/2)^2}} where the numerator is constant. We see also, that the denominator is minimal if x=ß/2 and the whole expression is maximal by the exponent ßx-x^2 of the second term, its first derivative ß-2x (which is zero at $x=ß/2$) and its second derivative $-2$ which is negative and shows, that the whole expression has a maximum there.

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How about comparison to $\displaystyle{\frac{2^x}{3^x}=\left(\frac{2}{3}\right)^x}$?

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    +1 This is the way to go imo. Just notice that $(2/3)^x\to 0$ and notice that $3^{x^2}$ goes to infinity faster than $3^x$.2011-02-09
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Hint: $2^x \lt 3^x$ for sufficiently large $x$.

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    @bobobobo: Note $\frac{3^x}{3^{x^2}} = \frac{1}{3^{x^2-x}}$ and $x^2-x \rightarrow \infty$ and $x \rightarrow \infty$2011-02-09
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How about rewriting $\frac{2^x}{3^{x^2}}$ as $\frac{2^x}{3^x3^{x^2-x}}=(\frac{2}{3})^x\cdot\frac{1}{3^{x^2-x}}$?

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Rewriting the fraction as $\frac{e^{x \ln 2}}{e^{x^2 \ln 3}} = e^{x \ln 2 - x^2 \ln 3}$ may also help you (although perhaps no more than the other answers given already).

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    The$ $direct$ $way$ $to go here is by passing through a 'limit of a composition is the composition of the limit'-type of argument.2011-02-09