Is it correct to say that $\lim_{x \to \infty} \left( \frac{3x+2}{-3x+1}\right)^{2x} = e^2$? Is it correct to say that $$\lim_{x \to \infty} \left( \frac{3x+2}{-3x+1}\right)^{2x} = e^2$$
I am confused since Wolfram gives me the answer on the image below.

 A: The problem from the fact that your limit is. $=e^{2xlog({{(3x+2)}\over{-3x+1}})}$. But $log({{(3x+2)}\over{-3x+1}}$ does not make sense as $x$ goes towards infinity since the limit of $(3x+2)/(-3x+1)$ is negative.
A: It looks like WA had a little problem. I have no idea what it means by $e^{2i0\text{ to }\pi}$, but even its own approximation of the terms as:
$$e^{2i\pi x + 2 +O(1/x)}$$
Would indicate the limit is points on the circle of radius $e^2$, rather than points of the circle of radius $1$, as its answer would seem to imply.
That said, as noted by myself and others, the limit is problematic if the limit is taken over $x$ real, as opposed to $x$ an integer. While it isn't formal, custom is for $\lim_{x\to\infty}$ to mean $x$ is allowed to take non-integer values, and we choose a variable like $m$ or $n$ if we want the limit to be interpreted as an integer limit. If $x$ is restricted to integers, then the limit is definitely $e^2$. 
Wolfram alpha seems to assume that all limits are real limits.
