Answer
$$\lim\limits_{x\to\infty}(e^{-2x}\cos x)=0$$
Work Step by Step
$$A=\lim\limits_{x\to\infty}(e^{-2x}\cos x)$$
*Strategy: Apply the Squeeze Theorem
1) We see that $$-1\le\cos x\le1$$
Since $e^{2x}\gt0$ for $x\in R$, so $\frac{1}{e^{2x}}\gt0$ for $x\in R$
Which means, $e^{-2x}\gt0$ for $x\in R$
Therefore, $$-e^{-2x}\le e^{2x}\cos x\le e^{-2x}\hspace{.5cm}(1)$$ (the inequality direction does not change since $e^{-2x}\gt0$
2) We calculate
$\lim\limits_{x\to\infty}(-e^{-2x})=-\lim\limits_{x\to\infty}(\frac{1}{e^{2x}})$
As $x\to\infty$, $e^{2x}\to\infty$. So, $\frac{1}{e^{2x}}\to0$
Therefore, $\lim\limits_{x\to\infty}(-e^{-2x})=0$
Similarly, $\lim\limits_{x\to\infty}(e^{-2x})=\lim\limits_{x\to\infty}(\frac{1}{e^2x})=0$
So, $\lim\limits_{x\to\infty}(-e^{-2x})=\lim\limits_{x\to\infty}(e^{2x})=0\hspace{.5cm}(2)$
3) From $(1)$ and $(2)$, according to the Squeeze Theore, we conclude $$\lim\limits_{x\to\infty}(e^{-2x}\cos x)=0$$
