Find lim x-> inf [(3x-2)/(3x+2)]^x

[confused_07]

Find lim x-> inf [(3x-2)/(3x+2)]^x

Here's what I got so far

= lim {ln[(3x-2)/(3x+2)]^x}
= lim {x * ln[(3x-2)/(3x+2)]}
= lim {[ln(3x-2)/(3x+2)] / (1/x)}

l'Hopitals':

To find the derivatives, would you set it up like this using the laws of logarithms:

lim {[ln(3x-2) - ln(3x+2)] / (1/x)}

Then use Dx= u'/u ?

 

[soroban]

Hello, confused_07!

Find \(\displaystyle \L\,\lim_{x\to\infty}\left(\frac{3x\,-\,2}{3x\,+\,2}\right)^x\)

Here's what I got so far:

\(\displaystyle \L= \:\lim_{x\to\infty} \left[\ln\left(\frac{3x-2}{3x+2}\right)^x\right]\)

\(\displaystyle \L=\: \lim_{x\to\infty} \left[x\cdot\ln\left(\frac{3x-2}{3x+2}\right)\right]\)

\(\displaystyle \L=\;\lim_{x\to\infty} \left[\frac{\ln\left(\frac{3x-2}{3x+2}\right)}{\frac{1}{x}}\right]\)

\(\displaystyle \L=\:\lim_{x\to\infty}\left[ \frac{\ln(3x-2)\,-\,\ln(3x+2)}{\frac{1}{x}}\right] \;\;\) . . . Good!

Then applying L'Hopital, we get:

. . \(\displaystyle \L\frac{\frac{3}{3x-2}\,-\,\frac{3}{3x+2}}{-\frac{1}{x^2}} \;=\;\frac{\frac{12}{(3x-2)(3x+2)}}{-\frac{1}{x^2}}\;=\;\frac{-12x^2}{9x^2\,-\,4}\)

Divide top and bottom by \(\displaystyle x^2:\;\;\L\frac{\frac{-12x^2}{x^2}}{\frac{9x^2}{x^2}\,-\,\frac{4}{x^2}} \;=\;\frac{-12}{9\,-\,\frac{4}{x^2}}\)

Therefore: \(\displaystyle \L\:\lim_{x\to\infty}\left(\frac{-12}{9\,-\,\frac{4}{x^2}}\right)\;=\;\frac{-12}{9\,-\,0}\;=\;-\frac{4}{3}\)