If you know that
[tex]e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x[/tex]
then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving [tex]e[/tex].
For starters, we have
[tex]\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}[/tex]
Let [tex]y=\dfrac34(x+1)[/tex]. Then as [tex]x\to\infty[/tex], we also have [tex]y\to\infty[/tex], and
[tex]2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2[/tex]
So in terms of [tex]y[/tex], the limit is equivalent to
[tex]\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}[/tex]
Now use some of the properties of limits: the above is the same as
[tex]\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}[/tex]
The first limit is trivial; [tex]\dfrac1y\to0[/tex], so its value is 1. The second limit comes out to
[tex]\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}[/tex]
To see why this is the case, replace [tex]y=-z[/tex], so that [tex]z\to-\infty[/tex] as [tex]y\to\infty[/tex], and
[tex]\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e[/tex]
Then the limit we're talking about has a value of
[tex]\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}[/tex]
# # #
Another way to do this without knowing the definition of [tex]e[/tex] as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write
[tex]\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)[/tex]
(where the notation means [tex]\exp(x)=e^x[/tex], just to get everything on one line).
Recall that
[tex]\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)[/tex]
if [tex]f[/tex] is continuous at [tex]x=c[/tex]. [tex]\exp(x)[/tex] is continuous everywhere, so we have
[tex]\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)[/tex]
For the remaining limit, write
[tex]\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}[/tex]
Now as [tex]x\to\infty[/tex], both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to
[tex]\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83[/tex]
and our original limit comes out to the same value as before, [tex]\exp\left(-\frac83\right)=\boxed{e^{-8/3}}[/tex].
