Limit $ \lim_{n\to\infty} \left(\frac{3n^2-n+1}{2n^2+n+1}\right)^{\large \frac{n^2}{1-n}}$ 
I'm required to compute the limit of the following sequence:
  $$ \lim_{n\to\infty} \left(\frac{3n^2-n+1}{2n^2+n+1}\right)^{\large \frac{n^2}{1-n}}$$

I'm not sure as to how to approach it. I've tried tackling it in a few ways, but none of them led me to a form I know how to tackle. 
Let $a_n$ denote the base.
$$ \lim_{n\to\infty}(a_n)^{\frac{n^2}{1-n}}= \left(\lim_{n\to\infty}(a_n)^{\large \frac{1}{1-n}}\right)^{n^2}=\lim_{n\to\infty} \left(\sqrt[1-n]{a_n}\right)^{n^2}=1^{\infty} $$
From there though I don't know how to compute: 
$$e^{\lim_{n\to\infty} \left(\sqrt[n-1]{a_n}n^2\right)}$$
I tried using the ln function though I couldn't get any further. I'd be glad for help :)
 A: You can estimate:
$$\frac{2n^2+n+1}{3n^2-n+1}<\frac{3}{4}, n>7;$$
$$0<\left(\frac{3n^2-n+1}{2n^2+n+1}\right)^{\large \frac{n^2}{1-n}}=\left(\frac{2n^2+n+1}{3n^2-n+1}\right)^{\large \frac{n^2}{n-1}}<\left(\frac34\right)^{\large \frac{n^2}{n-1}}=\left(\frac34\right)^{n+1+\large \frac{1}{n-1}}\to 0.$$
A: HINT
$$\left(\frac{3n^2-n+1}{2n^2+n+1}\right)^{\frac{n^2}{1-n}}\sim c\cdot \left(\frac32\right)^{-n}$$
indeed
$$\left(\frac{3n^2-n+1}{2n^2+n+1}\right)^{\frac{n^2}{1-n}}=\left(\frac32\right)^{-n}\left(\frac{2n^2-\frac23n+\frac23}{2n^2+n+1}\right)^{\frac{n^2}{1-n}}$$
and
$$\left(\frac{2n^2-\frac23n+\frac23}{2n^2+n+1}\right)^{\frac{n^2}{1-n}}=\left[\left(1-\frac{\frac53n+\frac13}{2n^2+n+1}\right)^{\frac{2n^2+n+1}{\frac53n+\frac13}}\right]^{\frac{\frac53n^3+\frac13n^2}{(2n^2+n+1)(1-n)}}\to\left(\frac1e\right)^{-\frac56}$$
A: You can't reduce this to the limit form of the exponential function, because the limit of the content of the bracket is 3/2. But the exponent becomes more as $-n$ as $n\to\infty$; so you have the the reciprocal of a positive-exponentially increasing number.
