<s>[INST] 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.
 [/INST] Let $o<\epsilon <1/2$. Then $x <-\frac 1 {\epsilon} $ implies $|g(x)-2|=\frac {|x-3/2|} {2x^{2}+1} \leq \frac {|x|+3/2} {2x^{2}}=\frac {|x|} {2x^{2}}+\frac 3 {4x^{2}}=\frac {1} {2|x|}+\frac 3 {4x^{2}}< {\epsilon /2}+\frac {3\epsilon ^{2}} 4<\epsilon$.
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