Math Help - Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity

1. Limit of e^(x)-1)/(e^(x)+1) as X -> Infinity

Hey all,

I been playing around with limits, when i solve for the following function:

$\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$
But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks

2. Originally Posted by Oiler
Hey all,

I been playing around with limits, when i solve for the following function:

$\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$
But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations? also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks
????

If you graph that, you obtain a line y=1

3. Originally Posted by dwsmith
????

If you graph that, you obtain a line y=1
dwsmith, sorry just made some changes to the formula.

4. Originally Posted by Oiler
dwsmith, sorry just made some changes to the formula.
It is -1 when you go to negative infinity.

5. Originally Posted by Oiler
also how does $\{e^{x}-1}=1+e^{x}$ ?. Thanks
Let x = 0

0 = 2 is that true?

6. ofcourse, how dim of me. Thanks dwsmith.

7. $\displaystyle\lim_{x\to +\infty}\frac{e^x-1}{e^x+1}=\frac{\infty}{\infty}$

Applying L'Hopitals Rule
$\displaystyle\Rightarrow\lim_{x\to +\infty}\frac{e^x}{e^x}=1$

8. Originally Posted by dwsmith
$\displaystyle\lim_{x\to +\infty}\frac{e^x-1}{e^x+1}=\frac{\infty}{\infty}$

Applying L'Hopitals Rule
$\displaystyle\Rightarrow\lim_{x\to +\infty}\frac{e^x}{e^x}=1$
L'Hôpital's Rule is overkill for this. XD

$\lim\limits_{x\to\infty}\dfrac{e^x-1}{e^x+1} = \lim\limits_{x\to \infty}\dfrac{1-e^{-x}}{1+e^{-x}} = 1$ since $e^{-x}\rightarrow0$ as $x\rightarrow\infty$.

9. Originally Posted by Chris L T521
L'Hôpital's Rule is overkill for this. XD

$\lim\limits_{x\to\infty}\dfrac{e^x-1}{e^x+1} = \lim\limits_{x\to \infty}\dfrac{1-e^{-x}}{1+e^{-x}} = 1$ since $e^{-x}\rightarrow0$ as $x\rightarrow\infty$.
Sometimes you just have to kill it.

10. $\displaystyle \lim_{x \to \infty}\frac{e^x - 1}{e^x + 1} = \lim_{x \to \infty}1 - \frac{2}{e^x + 1}$

$\displaystyle = 1 - 0$

$\displaystyle = 1$.

11. Originally Posted by dwsmith
Sometimes you just have to kill it.
But you never have to "over-kill". When you are showing someone how to do something, the simplest way is always best.

12. Originally Posted by Oiler
Hey all,

I been playing around with limits, when i solve for the following function:

$\displaystyle\lim_{x\rightarrow\infty }\frac{e^{x}-1}{e^{x}+1}=1$

But when i graph it, i can see that -1 is also a limit. Am i doing something wrong in my calculations?

also how does $\{e^{x}-1\}=\{1+e^{x}\}$ ?.

Thanks
For this fraction, the denominator is always 2 greater than the numerator.

However, the "ratio" gets closer and closer to 1, as we increase x above 0.

$x\rightarrow\infty$ means x increases without bound above 0.

Hence, as x "approaches infinity", you need infinitely many decimal places
to express the value of the fraction as a value "other than 1".

That's part of the concept of limits.

viz-a-viz

$\displaystyle\frac{e^1-1}{e^1+1}=0.46211715726$

$\displaystyle\frac{e^2-1}{e^2+1}=0.76159415596$

$\displaystyle\frac{e^{10}-1}{e^{10}+1}=0.99990920426$

....onward.

If $x<0$ and decreases without bound below 0.

$x=-y$

$x\rightarrow\ -\infty\Rightarrow\ y\rightarrow\infty$

$\displaystyle\lim_{x \to -\infty}\frac{e^x-1}{e^x+1}=\lim_{y \to \infty}\frac{e^{-y}-1}{e^{-y}+1}=\lim_{y \to \infty}\frac{\frac{1}{e^y}-1}{\frac{1}{e^y}+1}$

$\displaystyle\lim_{y \to \infty}\left[\frac{\frac{1}{e^y}}{\frac{1}{e^y}}\right]\;\frac{1-e^y}{1+e^y}\right]=-\lim_{y \to \infty}\frac{e^y-1}{e^y+1}=-1$

When you have the concept, you can apply the fast methods as you please later.

13. Originally Posted by HallsofIvy
But you never have to "over-kill". When you are showing someone how to do something, the simplest way is always best.
To me, that was mighty simply. The constant disappear and the exponentials are their same derivatives.