Calculate the limit: $\lim_{x\to+\infty}(\frac{x^2 -x +1}{x^2})^{\frac{-3x^3}{2x^2-1}}$ without de l'Hôpital rule I was wondering how can I calculate the limit:
$$\lim_{x\to+\infty}\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}$$
without de l'Hôpital rule.
I tried to reconduct the limit at the well known one:
$$\lim_{x\to+\infty}\left(1+\frac1x\right)^x = e$$
Now I'm concentrating only on the part of the limit with still the Indeterminate Form, I reached this form elevating $e$ to the neperian logarithm of the function, trying to get rid of the $1^\infty$ I.F.
but, at the end of the day, I could only obtain:
$$\begin{align}\lim_{x\to+\infty}\ln\left(\frac{1}{x^2}(1+x^2-x)\right) &= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} \ln\left(1 + \frac{1}{\frac{1}{x^2-x}}\right) \\&= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} (x^2-x) \ln\left(\left(1 + \frac{1}{\frac{1}{x^2-x}}\right)^{\frac{1}{x^2-x}}\right)\end{align}$$
But then defining  
$$t= \frac{1}{x^2-x}$$
the limit
$$\lim_{t\to 0}  \ln\left(\left(1 + \frac{1}{t}\right)^{t}\right)$$
goes no more to
$$ \ln(e)$$
because now $$t \to 0$$
Can you please give me some help?

$$\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}=\left(1+\frac{1-x}{x^2}\right)^{\frac{x^2}{1-x}\frac{-3x(1-x)}{2x^2-1}}=\left[\left(1+\frac{1}{\frac{x^2}{1-x}}\right)^{\frac{x^2}{1-x}}\right]^{\frac{3x^2-3x}{2x^2-1}}\to e^\frac32$$
