Look familiar?

Let $\displaystyle \frac 1m = - \frac {2}{3n}$ .

$\displaystyle \lim_{n\ \to\ \infty} \bigg(1 - \frac{2}{3n} \bigg)^{6n} = \lim_{m\ \to\ \infty} \bigg(1 + \frac{1}{m} \bigg)^{- 4m} = e^{- 4}$

Let $x=\lim_{n\rightarrow \infty} \left( 1 - \frac{2}{3n} \right)^{6n}$ The limit at this point is indeterminate of the form $1^\infty$ . Taking $\ln$ of both sides we obtain $\ln x = \lim_{n\rightarrow \infty} \left(6n \ln \left( 1 - \frac{2}{3n} \right)\right)$ The limit at this point is indeterminate of the form $\infty\cdot 0$ . We bring down $6n$ to the denominator, so that $\ln x = \lim_{n\rightarrow \infty} \frac{\ln \left( 1 - \dfrac{2}{3n} \right)}{\dfrac{1}{6n}}$ Now, the limit at this point is indeterminate of the form $\dfrac{0}{0}$ . We can now apply l'Hopital's rule once, and obtain $\ln x \stackrel{\mathrm{LHR}}{=} \lim_{n\rightarrow \infty} \frac{f'(x)}{g'(x)}=\lim_{n\rightarrow \infty}\frac{ \dfrac{1}{1-\dfrac{2}{3n}}\cdot \dfrac{2}{3n^2} }{-\dfrac{1}{6n^2}} = \lim_{n\rightarrow \infty} \frac{\dfrac{2}{3n^2 - 2n}}{-\dfrac{1}{6n^2}} =\lim_{n\rightarrow \infty} \frac{-12n^2}{3n^2 - 2n} = \lim_{n\rightarrow \infty} \frac{-12}{3 - \dfrac{2}{n}} = \frac{-12}{3-0}= -4$ Therefore $\ln x=-4$ , and $\boxed{x=e^{-4}}$