Lim[x->infty][(x+a)/(x-a)]^x = e for what value of a?

[theschaef]

For what value of "a" is the following equation true?

lim [x -> infty] [(x + a) / (x - a)]^x = e

 

[galactus]

Where are you stuck?.

Evaluate the limit. What value of 'a' makes your solution equal to e?.

I'll get you started:

\(\displaystyle \L\\\lim_{x\to\infty}\left(\frac{x+a}{x-a}\right)^{x}\)

=\(\displaystyle \L\\\lim_{x\to\infty}e^{xln\left(\frac{x+a}{x-a}\right)}\)

Rewrite using L'Hopital's rule:

=\(\displaystyle \L\\e^{{-}2a\left(\lim_{x\to\infty}\frac{x^{2}}{(x+a)(x-a)}\right)}\)

=\(\displaystyle \L\\e^{{-}2a\left(\lim_{x\to\infty}\frac{1}{-1+\frac{a^{2}}{x^{2}}}\right)}\)

Can you finish?. Almost there.

 

[pka]

Here is another way: \(\displaystyle \L\lim _{x \to \infty } \left( {\frac{{x + a}}{{x - a}}} \right)^x = \lim _{x \to \infty } \left( {1 + \frac{{2a}}{{x - a}}} \right)^x = ?\)