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$\lim_{x\to \infty}\dfrac{\cos(3x)}{e^{8x}}$

The answer is $0$. Why is the answer $0$? The top oscillates between $-1$ and $1$ and the bottom becomes huge, but since the top is oscillating, shouldn't the answer be DNE (does not exist)?

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    $\begingroup$ The amplitude of the oscillation shrinks to $0$. $\endgroup$ Oct 7 '14 at 20:45
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Hint: Notice that for all $x \in \mathbb R$, we have that: $$ \frac{-1}{e^{8x}} \leq \frac{\cos(3x)}{e^{8x}} \le \frac{1}{e^{8x}} $$

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  • $\begingroup$ Is this the squeeze theorem? could you show some steps to how you would use the Squeeze theorem on this type of problem, thanks +1 $\endgroup$
    – Emi Matro
    Oct 7 '14 at 20:49
  • $\begingroup$ Notice that: $$ \lim_{x \to \infty} \frac{-1}{e^{8x}} = \lim_{x \to \infty} \frac{1}{e^{8x}} = 0 $$ Combining this with the above inequality, it follows by the Squeeze Theorem that: $$ \lim_{x \to \infty} \frac{\cos(3x)}{e^{8x}} = 0 $$ $\endgroup$
    – Adriano
    Oct 7 '14 at 20:52

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