Find $\lim_{x\to\infty} \frac{e^{2x}-1}{e^{2x}+1}$ and $\lim_{x\to-\infty} \frac{e^{2x}-1}{e^{2x}+1}$

Question

How do I calculate $\displaystyle \lim_{x\to\infty} \frac{e^{2x}-1}{e^{2x}+1} \ , \ \lim_{x\to-\infty}\frac{e^{2x}-1}{e^{2x}+1}$.

Please help me.

Thanks!

Answer 1

For the limit as $x\to\infty$, divide top and bottom by $e^{2x}$.

For the limit as $x\to-\infty$, it is enough to look.

Answer 2

In the second limit you can just use the fact that $e^{-\infty}=0$ and get the limit.

In the first limit we can write $\displaystyle \frac{e^{2x}-1}{e^{2x}+1}=\frac{1-\frac{1}{e^{2x}}}{1+\frac{1}{e^{2x}}}$, hence $$\lim_{x\to\infty}\frac{e^{2x}-1}{e^{2x}+1}=\lim_{x\to\infty}\frac{1-\frac{1}{e^{2x}}}{1+\frac{1}{e^{2x}}}=\frac{1}{1}=1$$