Calculating a limit to infinity on e 
$$\lim_{x\to0^+}(1+3x)^\frac{-2}{3x^3}$$

Okay so I got to the point where I have
$$\lim _{t\to \infty }\left(\left(1+\frac{1}{t}\right)^{t}\right)^{-18t^{2}}$$
Now, my instructors don't allow us to use L'Hopital and we haven't proven anything about situations such that
$$\lim _{x\to \infty }\left(f\left(x\right)^{g\left(x\right)}\right)$$
so is my only option is to use squeeze rule? since I know the expression is basically $e^{-18t^{2}}$
and we know that $2 < e < 3$..
is there a simpler way to solve this? if so then how? if not then how do I prove $2^{-18t^{2}} < e^{-18t^{2}} < 3^{-18t^{2}}$?
 A: Alternative resolution
$$L=\underset{x\to 0^+}{\text{lim}}(3 x+1)^{-\frac{2}{3 x^3}}$$
consider
$$\log L = \log \left(\underset{x\to 0^+}{\text{lim}}(3 x+1)^{-\frac{2}{3 x^3}}\right)=$$
$$=\underset{x\to 0^+}{\text{lim}}\left(-\frac{2}{3 x^3}\log(1+3 x)\right)$$
As $\log (1+3x)\sim 3x $ as $x\to 0^+$ we have
$$ \log L=\underset{x\to 0^+}{\text{lim}}\left(-\frac{2\cdot 3x}{3 x^3}\right)=-\infty$$
if $\log L\to -\infty$ then $L=0$.
Hope this is useful
A: [Hint] : $$(1+3x)^{-2/3x^{3}}=e^{\displaystyle -\frac{2}{3x^{3}}\ln(1+3x)}$$ this is assuming $x>-\frac{1}{3}$
Now let $u=3x$, we get :
$$
\lim_{t\to 0^{+}}\displaystyle e^\frac{\displaystyle-2\ln(1+t)}{\displaystyle t^3}=\lim_{t\to 0^{+}}\displaystyle e^{\frac{\displaystyle-2\ln(1+t)}{\displaystyle t}\cdot\displaystyle\frac{1}{t^{2}}}
$$
A: IDC :)
We did learn AOL of pow in lecture 14, so this is valied:
$\lim _{x\to \infty }\left(f\left(x\right)^{g\left(x\right)}\right)$
so why not just define $g(x) = {-18t^{2}}\space $ and use AOL to get $\frac{1}{e^\infty} = 0$ ?
