# Math Help - How do I find this limit

1. ## How do I find this limit

Limit as x approaches infinity [(x+1)/(x-1)]^x

I divided everything by an x. I see that the numerator would be e; but not too sure what to do or if that is the right process.

Help

2. Originally Posted by Nichelle14
Limit as x approaches infinity [(x+1)/(x-1)]^x

I divided everything by an x. I see that the numerator would be e; but not too sure what to do or if that is the right process.

Help
I am going to find the limit of,
$\lim_{x\to\infty} \left( 1-\frac{1}{x} \right)$. I will assume that it exists, (I will leave that to you to prove).

Begin by saying the limit exists,
$\lim_{x\to\infty}\left(1- \frac{1}{x} \right)=L$
Thus, (using the concept of countinous functions)
$\ln L=\lim_{x\to\infty} \ln \left(1-\frac{1}{x} \right)^x$
Using, the properties of logarithms,
$\ln L=\lim_{x\to\infty} x\ln (1-1/x)$
Since, $|1-1/x|<1$ it is expressable as a Taylor polynomial,
Thus,
$\ln L=\lim_{x\to\infty} x\left( -(1/x)-\frac{(1/x)^2}{2}-\frac{(1/x)^3}{3}-... \right)$
Open parantheses,
$\ln L=\lim_{x\to\infty} -1-\frac{1}{2x}-\frac{1}{3x^2}-...$
Since, each term with and "x" gives a zero you have,
$\ln L=-1-0-0-0...$
Thus,
$\ln L=-1$
Finally,
$L=\frac{1}{e}$
---
Thus, given,
$
\lim_{x\to\infty} \left( \frac{x+1}{x-1} \right)^x$

Divide by "x" as you said,
$\lim_{x\to\infty} \frac{(1+1/x)^x}{(1-1/x)^x}$
Since the numerator and denominator (non-zero) both exist you can conclude that the limit is,
$\frac{e}{\frac{1}{e}}=e^2$

3. Hello, Nichelle14!

You were on the right track . . .

$\lim_{x\to\infty}\left(\frac{x+1}{x-1}\right)^x$

Inside the parentheses, divide top and bottom by $x:\;\;\lim_{x\to\infty} \frac{\left(1 + \frac{1}{x}\right)^x} {\left(1 - \frac{1}{x} \right)^x}$

. . So we have: . $\displaystyle{\frac{\lim_{x\to\infty}\left(1 + \frac{1}{x}\right)^x} {\lim_{x\to\infty}\left(1 - \frac{1}{x}\right)^x} }$

You recognized that the numerator is $e$ . . . good!

But did you know that the denominator is $e^{-1}$ ? **

Therefore, the limit is: . $\frac{e}{e^{-1}} \:=\:\boxed{e^2}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
A very handy theorem: . $\lim_{x\to\infty}\left(1 + \frac{k}{x}\right)^x\;=\;e^k$