Real analysis - epsilon definition of limit of sequence So I'm completely lost on this so far:
The question reads use the epsilon definition to prove the following:
$$\lim_{n\to\infty}(n^2+n+1)/(2n^2+3) = 1/2$$
so we have to show that for any $\epsilon>0$, there is an $M>0$ such that $n>M \Rightarrow |a_n-L|<\epsilon$
normally when I solve these I set an $\epsilon>0$, and then try to find an M(that depends on ε) such that n>M satisfies the definition.. anyway this is what I tried
\begin{align*} \left |\frac{n^2 + n + 1}{2n^2+3} - \frac{1}{2}\right| < \epsilon &\Rightarrow \left|\frac{n^2+n+1}{2n^2+3}-\frac{\frac{1}{2}(2n^2+3)}{2n^2+3}\right| < \epsilon \\ &\Rightarrow \left|\frac{n-\frac{1}{2}}{2n^2+3}\right| < \epsilon \\  \end{align*} 
now at this point everything I've done has led nowhere .. I can't seem to isolate the n and so I'm hoping there is maybe some trick I've missed or something.. 
I don't necessarily want a full rigorous answer, just a thought on what I might try from there (or maybe a different strategy altogether) ... Thanks
 A: Hint: Look at  $\left|\frac{n-\frac{1}{2}}{2n^2+3}\right|$. The top is $\le n$, the bottom is $\ge 2n^2$, and therefore $\dots$
Remark: It may be worthwhile to look at the problem in detail. We want to show that given any $\epsilon \gt 0$, we can find an $M$ such that a certain inequality holds if $n>M$.  
OK, we have been given an $\epsilon$.  Now let's forget about $\epsilon$ for a while.   
We are interested in the size of 
$$\left |\frac{n^2 + n + 1}{2n^2+3} - \frac{1}{2}\right|.$$
Manipulation showed that this is equal to
$$\left|\frac{n-\frac{1}{2}}{2n^2+3}\right|.\tag{$1$}$$
With a bit less work  we could have simply used
$$\left|\frac{2n-1}{2(2n^2+3)}\right|.$$
It is intuitively clear that the above quantity is real small if $n$ is large, since the top is big but the bottom is much bigger. 
Please note that we are not asked to find the smallest $M$ such that beyond it Expression $(1)$ is less than $\epsilon$.  So we can afford to give away a lot. By the Hint, Expression $(1)$, if $n\ge 1$, is $\le \frac{1}{2n}$. This can be made $\lt \epsilon$ as long as $n \gt 1/(2\epsilon)$. 
You were trying to "isolate the $n$." This is not necessary, and is often hopelessly difficult.  But in fact in this case it could be done. Set our slightly complicated expression equal to $\epsilon$. Multiply through by the denominator $2n^2+3$. We get a quadratic equation in $n$, and can solve. However, one should not be even thinking in terms of isolating $n$! 
