for what value of a is the following equation true?

lim ((x+a)/(x-a))^x = e

x->inf

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- Nov 6th 2012, 10:12 AMpnfullerlimits / equations?
for what value of a is the following equation true?

lim ((x+a)/(x-a))^x = e

x->inf - Nov 6th 2012, 10:57 AMMarkFLRe: limits / equations?
Take the natural log of both sides, so that you wind up with:

$\displaystyle \lim_{x\to\infty}\frac{\ln\left(\frac{x+a}{x-a} \right)}{\frac{1}{x}}=1$

Now use L'Hôpital's rule to find $\displaystyle a$. - Nov 6th 2012, 12:03 PMpnfullerRe: limits / equations?
after applying L'Hôpital's rule i got the limit as x approaches infinity ((x-a)(-2a))/(-x^-2) = 1

and now what do i do from there? - Nov 6th 2012, 12:13 PMMarkFLRe: limits / equations?
You have differentiated the numerator incorrectly. You want:

$\displaystyle \frac{d}{dx}\left(\ln\left(\frac{x+a}{x-a} \right) \right)=\frac{1}{\frac{x+a}{x-a}}\cdot\frac{-2a}{(x-a)^2}=\frac{2a}{a^2-x^2}$

Now, putting this together with the derivative of the denominator, we have:

$\displaystyle \lim_{x\to\infty}\frac{2ax^2}{x^2-a^2}=1$

which we can write as:

$\displaystyle 2a\lim_{x\to\infty}\frac{1}{1-\frac{a^2}{x^2}}=1$

From here, it is a piece of cake. - Nov 6th 2012, 12:20 PMpnfullerRe: limits / equations?
im not seeing how this is a piece of cake?

- Nov 6th 2012, 12:36 PMPlatoRe: limits / equations?
- Nov 6th 2012, 12:37 PMMarkFLRe: limits / equations?
Sorry, bad choice of words. :)

What is $\displaystyle \lim_{x\to\infty}\frac{a^2}{x^2}$ ? - Nov 6th 2012, 12:42 PMpnfullerRe: limits / equations?
the limit as a^2/x^2 approaches infinity equals 0? is that what you mean?

- Nov 6th 2012, 01:09 PMMarkFLRe: limits / equations?
Yes, so what does that tell you about the limit at the end of post #4?

- Nov 6th 2012, 03:06 PMpnfullerRe: limits / equations?
that the limit as x approaches infinity is 1/1 and it equals 1 but what does a equal?

- Nov 6th 2012, 03:10 PMProve ItRe: limits / equations?
Reread that equation carefully. You were told $\displaystyle \displaystyle \begin{align*} 2a \cdot \lim_{x \to \infty}\frac{1}{1 - \frac{a^2}{x^2}} = 1 \end{align*}$

- Nov 6th 2012, 03:11 PMMarkFLRe: limits / equations?
Okay, since the limit is 1, you simply have:

$\displaystyle 2a=1$

Now solve for $\displaystyle a$. - Nov 6th 2012, 03:19 PMpnfullerRe: limits / equations?
so a = 1/2

- Nov 6th 2012, 03:21 PMPlatoRe: limits / equations?
- Nov 6th 2012, 03:26 PMpnfullerRe: limits / equations?
why is it frustrating for you?