# Question about $\lim_{x\to \infty}\frac{\cos(3x)}{e^{8x}}$

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

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)?

• The amplitude of the oscillation shrinks to $0$. Oct 7 '14 at 20:45

Hint: Notice that for all $x \in \mathbb R$, we have that: $$\frac{-1}{e^{8x}} \leq \frac{\cos(3x)}{e^{8x}} \le \frac{1}{e^{8x}}$$
• 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: $$\lim_{x \to \infty} \frac{\cos(3x)}{e^{8x}} = 0$$ Oct 7 '14 at 20:52