Answer
$$\lim\limits_{x\to\infty}\frac{1-e^x}{1+2e^x}=\frac{-1}{2}$$
Work Step by Step
$$A=\lim\limits_{x\to\infty}\frac{1-e^x}{1+2e^{x}}$$$$=\lim\limits_{x\to\infty}\frac{X}{Y}$$
Divide both numerator and denominator by $e^x$, we have
- In the numerator:
$X=\frac{1-e^x}{e^x}=\frac{1}{e^x}-1$
- In the denominator:
$Y=\frac{1+2e^x}{e^x}=\frac{1}{e^x}+2$
Therefore, $$A=\lim\limits_{x\to\infty}\frac{\frac{1}{e^x}-1}{\frac{1}{e^x}+2}$$
$$A=\frac{\lim\limits_{x\to\infty}(\frac{1}{e^x})-1}{\lim\limits_{x\to\infty}(\frac{1}{e^x})+2}$$
As $x\to\infty$, $e^x\to\infty$. So, $\frac{1}{e^x}\to0$. Which means, $\lim\limits_{x\to\infty}(\frac{1}{e^x})=0$
$$A=\frac{0-1}{0+2}=\frac{-1}{2}$$
