Tricky Indeterminate Power With Indeterminate Solution

Question

I have learn in L'hospital rule about indeterminate power, we will take $\ln$ to find the limit.
But in this question, I suspect there is no solution to the limit.
$\lim_{x\to \infty}({\frac {2-3x}{1+6x})^x=\lim_{x\to \infty}\exp(x\ln(\frac{2-3x}{1+6x}))}$
Where it also equal $\lim_{x\to\infty}\exp(x\ln(\frac{-1}{2}+\frac{5}{2(1+6x)}))$
Then the exponent can be rewritten into, $\exp(\frac{\ln(\frac{-1}{2}+\frac{5}{2(1+6x)})}{\frac{1}{x}})$ but this is not indeterminate form.
Because as $x\to\infty$, $\ln(\frac{-1}{2}+\frac{5}{2(1+6x)})\to\ln(\frac{-1}{2})$, but $\ln\frac{-1}{2}$ is not defined. So, can I say that the limit is indeterminate?
I am trying to show that the limit is indeterminate.