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?
Answer
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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