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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

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Wow thanks a lot guys ... The hint gave me a good idea of what I could try and the remark increased my understanding of exactly what we are trying to show.. I'm not sure how to comment on an answer, but if you read this thanks again :P –  Ronald Jun 2 '12 at 15:16
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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 $n$!

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Alright so I thought I had it.. but now I see I had it the wrong way round.. from $\left|\frac{n-\frac{1}{2}}{2n^2+3}\right|$ I thought since $n-\frac{1}{2} < n$ and $2n^2+3 > 2n^2$ we could then go $\left |\frac{n}{2n^2}\right|< \left|\frac{n-\frac{1}{2}}{2n^2+3}\right|$ giving an easy problem, but I now see that in fact, $\left |\frac{n}{2n^2}\right|> \left|\frac{n-\frac{1}{2}}{2n^2+3}\right|$ and we can't really use that since we have $\left|\frac{n-\frac{1}{2}}{2n^2+3}\right| < \epsilon$ sigh is something going wrong in my brain ? does what i say even make sense?? –  Ronald Jun 2 '12 at 16:47
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No, where you say easy problem you are turned around. Fact, $\left|\frac{n}{2n^2}\right| \gt $ you know what. Therefore, **if we make sure** that $\left|\frac{n}{2n^2}\right|<\epsilon$, then for even more sure you know what will be $\lt \epsilon$. And we can easily make sure that $\left|\frac{n}{2n^2}\right|\lt \epsilon$. Please assure me that the logic has become clear! –  André Nicolas Jun 2 '12 at 16:58
    
Ok now I see.. so we are basically showing $\frac{n-1/2}{2n^2+3} < \frac{n}{2n^2} < \epsilon $ and so by showing $\frac{n}{2n^2}$ converges to zero, $\frac{n-1/2}{2n^2+3}$ definitely converges to zero and in fact the value we pick for $M$ where $n>M$ is higher than it needs to be thats ok because like you say we dont need to find the smallest $M$ ?? is this idea correct ..? –  Ronald Jun 2 '12 at 17:14
    
So then what if we had it the other way around .. for example $\frac{n+1/2}{2n^2-3}$ such that the top is $> n$ and bottom is $< 2n^2$ ?? could we then say $\frac{n}{2n^2} < \frac{n+1/2}{2n^2-3} < \epsilon$ that is what I originally thought but now I'm not too sure –  Ronald Jun 2 '12 at 17:15
    
Your comment "Ok now," yes, idea correct. Your $(n+1/2)/(2n^2-3)$ question is important, you **will**need to know how to handle it. I would say top is $\lt 2n$. For $n>2$, bottom is $\gt n^2$. So ratio is $\lt \frac{2n}{n^2}$. Now if we make this less than $\epsilon$, we are fine. –  André Nicolas Jun 2 '12 at 17:27
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