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,Originally Posted by Nichelle14
$\displaystyle \lim_{x\to\infty} \left( 1-\frac{1}{x} \right)$. I will assume that it exists, (I will leave that to you to prove).
Begin by saying the limit exists,
$\displaystyle \lim_{x\to\infty}\left(1- \frac{1}{x} \right)=L$
Thus, (using the concept of countinous functions)
$\displaystyle \ln L=\lim_{x\to\infty} \ln \left(1-\frac{1}{x} \right)^x$
Using, the properties of logarithms,
$\displaystyle \ln L=\lim_{x\to\infty} x\ln (1-1/x) $
Since, $\displaystyle |1-1/x|<1$ it is expressable as a Taylor polynomial,
Thus,
$\displaystyle \ln L=\lim_{x\to\infty} x\left( -(1/x)-\frac{(1/x)^2}{2}-\frac{(1/x)^3}{3}-... \right) $
Open parantheses,
$\displaystyle \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,
$\displaystyle \ln L=-1-0-0-0...$
Thus,
$\displaystyle \ln L=-1$
Finally,
$\displaystyle L=\frac{1}{e}$
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Thus, given,
$\displaystyle
\lim_{x\to\infty} \left( \frac{x+1}{x-1} \right)^x $
Divide by "x" as you said,
$\displaystyle \lim_{x\to\infty} \frac{(1+1/x)^x}{(1-1/x)^x}$
Since the numerator and denominator (non-zero) both exist you can conclude that the limit is,
$\displaystyle \frac{e}{\frac{1}{e}}=e^2$
Hello, Nichelle14!
You were on the right track . . .
$\displaystyle \lim_{x\to\infty}\left(\frac{x+1}{x-1}\right)^x$
Inside the parentheses, divide top and bottom by $\displaystyle x:\;\;\lim_{x\to\infty} \frac{\left(1 + \frac{1}{x}\right)^x} {\left(1 - \frac{1}{x} \right)^x}$
. . So we have: .$\displaystyle \displaystyle{\frac{\lim_{x\to\infty}\left(1 + \frac{1}{x}\right)^x} {\lim_{x\to\infty}\left(1 - \frac{1}{x}\right)^x} }$
You recognized that the numerator is $\displaystyle e$ . . . good!
But did you know that the denominator is $\displaystyle e^{-1}$ ? **
Therefore, the limit is: .$\displaystyle \frac{e}{e^{-1}} \:=\:\boxed{e^2}$
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**
A very handy theorem: .$\displaystyle \lim_{x\to\infty}\left(1 + \frac{k}{x}\right)^x\;=\;e^k$