Ok so I have been trying for days already to find a solution to this all around the web and in math books but to no success. The problem is to evaluate a limit of a function composed by polynomial functions and an exponential function:
$$\lim_{x \to +\infty} \left(\frac{3x+2}{3x-2}\right)^{2x}$$
I know from a software that the solution is $\exp\left(\dfrac{8}{3}\right)$, but I can't reach this.
One thing I did to try to find the limit was:
$$\lim_{x \to +\infty}\left(\frac{3x+2}{3x-2}\right)^{2x}=\lim_{x \to +\infty}\exp\left(\ln\left(\left(\frac{3x+2}{3x-2}\right)^{2x}\right)\right)\\
=\lim_{x \to +\infty}\exp\left(2x\ln\left(\frac{3x+2}{3x-2}\right)\right)
=\exp{\left(\lim_{x \to +\infty}2x\times\lim_{x \to +\infty}\ln\left(\frac{3x+2}{3x-2}\right)\right)}$$
But this doesn't work because the limit to the right goes to zero while the one on the left goes to infinity. I tried other things too, but the problem only gets more complicated and a solution seems to get farther and farther away.
Best Answer
Write $$ \frac{3x+2}{3x-2}=\frac{3x-2+4}{3x-2}=1+\frac{4}{3x-2} $$ and then make the substitution $t=3x-2$, so $x=(t+2)/3$ and your limit becomes $$ \lim_{t\to\infty}\left(1+\frac{4}{t}\right)^{2\frac{t+2}{3}}= \left(\lim_{t\to\infty} \left(1+\frac{4}{t}\right)^t\cdot \lim_{t\to\infty}\left(1+\frac{4}{t}\right)^2 \right)^{2/3} $$