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I have to evaluate $$\lim_{k\to\infty}\frac{k}{3^{k}}$$

At first, I rewrote it as $$\lim_{k\to\infty}\frac{3^{-k}}{\dfrac{1}{k}}$$ and then, applying L'Hopital rule, I got $$\lim_{k\to\infty}\frac{\ln(3)\cdot3^{-k}}{\dfrac{1}{k^{2}}}$$

I feel this is not the way I should follow but I have no ideia how can I evaluate it.

N. F. Taussig
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mvfs314
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    See for example https://math.stackexchange.com/q/173061/42969 or https://math.stackexchange.com/q/1247447/42969 or https://math.stackexchange.com/q/372655/42969 – Martin R Mar 21 '23 at 14:36
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    Don't try to oversmart the problem. Apply L'Hopital without rewriting – Vasili Mar 21 '23 at 14:37

1 Answers1

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Why didn't you apply L'Hôpital's rule in the original form?

$$ \lim_{k\to \infty}\frac{k}{3^k} = \lim_{k\to \infty}\frac{1}{\ln 3 \cdot 3^k} = 0. $$

PierreCarre
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