## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to\infty}(e^{-2x}\cos x)=0$$
$$A=\lim\limits_{x\to\infty}(e^{-2x}\cos x)$$ *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 $$\lim\limits_{x\to\infty}(e^{-2x}\cos x)=0$$