$\lim _{x\to -\infty }\left(\frac{\left(e^x-1\right)}{\left(e^{2x}+1\right)}\right)$

Question

How can I calculate the following limit?

$\lim _{x\to -\infty }\left(\frac{\left(e^x-1\right)}{\left(e^{2x}+1\right)}\right)$

If the limit is

$\lim _{x\to +\infty }\left(\frac{\left(e^x-1\right)}{\left(e^{2x}+1\right)}\right)$

then it is quiet easy, as I just need to make something like

$\lim _{x\to +\infty }\left(\frac{e^x\left(1-\frac{1}{e^x}\right)}{e^{2x}\left(1+\frac{1}{e^{2x}}\right)}\right)$

and it evident it is 0.

But with limit to negative infinity, I cannot do the same, as a I go back to an undetermined form, like

$\frac{0*\infty}{0*\infty}$

So don't know what I should do. Any suggestion?

Answer 1

It is $$\lim_{x\to -\infty}e^{2x}=\lim_{x\to -\infty}e^x=0$$

Answer 2

Your idea is fine, if negative infinite makes confusion to you, using $y=-x \to \infty$ we have

$$\lim _{x\to -\infty }\frac{e^x-1}{e^{2x}+1}=\lim _{y\to \infty }\frac{e^{-y}-1}{e^{-2y}+1}=\frac{0-1}{0+1}=-1$$

Answer 3

Note that $\lim_{x\rightarrow-\infty}e^{x}=\lim_{u\rightarrow\infty}e^{-u}=\lim_{u\rightarrow\infty}\dfrac{1}{e^{u}}=\dfrac{1}{\lim_{u\rightarrow\infty}e^{u}}=0$, similarly, $\lim_{x\rightarrow-\infty}e^{2x}=0$, so the whole limit is $-1$.

Answer 4

As $x$ tends to $-\infty$, both $e^x$ and $e^{2x}$ tend to $0$, therefore the given limit is $\frac{-1}{1}=-1$