Answer
$$\lim\limits_{x\to\infty}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}=1$$
Work Step by Step
$$A=\lim\limits_{x\to\infty}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$$$$=\lim\limits_{x\to\infty}\frac{X}{Y}$$
Divide both numerator and denominator by $e^{3x}$, we have
- In the numerator:
$X=\frac{e^{3x}-e^{-3x}}{e^{3x}}=1-e^{-6x}$
- In the denominator:
$Y=\frac{e^{3x}+e^{-3x}}{e^{3x}}=1+e^{-6x}$
Therefore, $$A=\lim\limits_{x\to\infty}\frac{1-e^{-6x}}{1+e^{-6x}}$$
$$A=\frac{1-\lim\limits_{x\to\infty}(e^{-6x})}{1+\lim\limits_{x\to\infty}(e^{-6x})}$$
$$A=\frac{1-\lim\limits_{x\to\infty}(\frac{1}{e^{6x}})}{1+\lim\limits_{x\to\infty}(\frac{1}{e^{6x}})}$$
As $x\to\infty$, $e^{6x}$ approaches $\infty$. So, $\frac{1}{e^{6x}}$ approaches $0$.
Therefore, $\lim\limits_{x\to\infty}(\frac{1}{e^{6x}})=0$
So, $$A=\frac{1-0}{1+0}=1$$
