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I've recently learned the Limit Laws and came across the following problem. $$\lim_{x\to \infty}(e^{-2x}\cos x)$$

I know the answer is 0 but I do not know how to solve it algebraically. When I use the rules that I know, I get $\lim_{x\to \infty}\cos x$ in the numerator, which is undefined. In addition to checking it on CAS, since the denominator approaches numbers much larger than the domain of $\cos x$, I concluded the limit was 0; however, how could I solve this algebraically?

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Use the squeeze theorem, and note that $-e^{-2x} \le e^{-2x}\cos x \le e^{-2x}$. You have that $\lim -e^{-2x} = 0$ and $\lim e^{-2x} = 0$. You can conclude that $\lim e^{-2x}\cos x = 0$.

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Apply the squeeze theorem to $-e^{-2x} \le e^{-2x} \cos x \le e^{-2x}$.

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$$\lim _{x \rightarrow \infty} |e^{-2x}cosx| \leq \lim _{x \rightarrow \infty} |e^{-2x}|=\lim _{x \rightarrow \infty} e^{-2x}=0$$

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