Proving $\lim_{n\rightarrow-\infty}\frac{3x^2+x}{2x^2+1}=\frac{3}{2}$ Using Definition Definition of Limit of Function As n $\rightarrow\infty$: Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a function and let $B\in\mathbb{R}$. If for all $\epsilon>0$, there exists $N>0$ such that $x<-N\Rightarrow |g(x)-B|<\epsilon$, we write $\lim_{n\rightarrow\infty}g(x)=B$.
I want to prove $\lim_{n\rightarrow-\infty}\frac{3x^2+x}{2x^2+1}=\frac{3}{2}$.
Here is my attempt. 
We must find the value $-N<0$ such that: $x<-N\Rightarrow |g(x)-\frac{3}{2}|=|\frac{3x^2+x}{2x^2+1}-\frac{3}{2}|=|\frac{3x^2+x}{2x^2+1}-\frac{3(x^2+\frac{1}{2})}{2(x^2+\frac{1}{2})}|=|\frac{x-\frac{3}{2}}{2x^2+1}|<\epsilon$.
When $N>\frac{3}{2}$, we have: $-x>N\Rightarrow |\frac{-x-\frac{3}{2}}{2x^2+1}|=\frac{-x-\frac{3}{2}}{2x^2+1}<\frac{-x-\frac{3}{2}}{2x^2}<\frac{-x}{2x^2}=\frac{-1}{2x}<\epsilon\Rightarrow -x>\frac{1}{2\epsilon}$. So, choose $N>\frac{1}{2\epsilon}$.
Now, this is the bit where I'm having some issues.
Then, for all $x<-N<\min\{-\frac{1}{2\epsilon},-\frac{3}{2}\}$, we have $|\frac{x-\frac{3}{2}}{2x^2+1}|=\frac{-x+\frac{3}{2}}{2x^2+1}<\frac{-x+\frac{3}{2}}{2x^2}=\frac{\frac{1}{2\epsilon}+\frac{3\epsilon}{2\epsilon}}{\frac{9}{2}}$... I can't quite get this inequality to work the way I need it to.
Any help appreciated.

You want the $x^2$ in the bottom to become dominant.  And the negatives are confusing things.  Bout $x < -\frac 32$ so $-x \ge  \frac 32 > 0$ so 
$\frac {-x + \frac 32}{2x^2} = \frac {1 + \frac 3{-2x}}{-2x}$
Replace $y = -x$ and $y > \max(\frac 32, \frac 1{2\epsilon})$  and $\frac 1y <  \min (\frac 23,  2\epsilon)$.
$\frac {1+ \frac 3{2y}}{2y} < \frac 1y\frac {1 + \frac 32*\frac 1y}2< (2\epsilon)\frac {1 + \frac 32\frac 32}2= 2\epsilon$
So ... some minor arithmetic error slipped in.  But it's minor and fixable. If we chose $N> \max (\frac 32, \frac 1\epsilon)$  it'd probably work.
