Answer
$\lim\limits_{x\to\infty}(e^{-x}+2\cos{3x})$ does not exist.
Work Step by Step
$$\lim\limits_{x\to\infty}(e^{-x}+2\cos3x)$$$$=\lim\limits_{x\to\infty}e^{-x}+2\lim\limits_{x\to\infty}\cos3x$$$$=\lim\limits_{x\to\infty}(\frac{1}{e^x})+2\lim\limits_{x\to\infty}\cos3x$$$$=\frac{1}{\lim\limits_{x\to\infty}e^x}+2\lim\limits_{x\to\infty}\cos3x$$$$=A+B$$
- Consider A:
As $x\to\infty$, $e^x$ approaches $\infty$. So, $\lim\limits_{x\to\infty}e^x=\infty$
Therefore, $A=\frac{1}{\lim\limits_{x\to\infty}e^x}=0$
- Consider B:
We know that $\cos3x$ oscillates constantly between $1$ and $-1$ as $x$ increases, so it does not reach any definite number.
In other words, $\lim\limits_{x\to\infty}\cos{3x}$ does not exist.
Therefore, $A+B$ does not exist.
Overall, $\lim\limits_{x\to\infty}(e^{-x}+2\cos{3x})$ does not exist.
