Verify my $\epsilon$-$\delta$ limit is correct

Question

Have I got this correct? I understand there is no unique solution.

Find $$\lim_{x\to\infty}\frac{3x^2+1}{x^2-3}$$

Given $\epsilon >0$, $x\geq R>0$ $$\frac{3x^2+1}{x^2-3}-3=\frac{10}{x^2-3}\text{$<$ }\frac{3}{x}<\epsilon$$ $$x>\frac{3}{\epsilon }$$

$$\text{$\therefore$ let $R$>}\frac{3}{\epsilon}$$

$$R=\max\Big\{1,\frac{3} {\epsilon}\Big\}$$

$$\left|\frac{3x^2+1}{x^2-3}-3\right|=\frac{10}{x^2}<\frac{3}{x}\text{$\leq $ }\frac{3}{R}<\epsilon$$

Answer 1

Your exposition lacks argumentation. Here's something more (possibly too) detailed:

Problem. Find $\lim_{x\to\infty}\frac{3x^2+1}{x^2-3}$.

Solution: I claim that the desired limit equals $3$. To prove that indeed $$\tag1\lim_{x\to\infty}\frac{3x^2+1}{x^2-3}=3,$$ I need to show that: For any given $\epsilon>0$, there exists a number $R\in\mathbb R$ such that for all $x\in\mathbb R$ with $x>R$ the inequality $$\tag2\left|\frac{3x^2+1}{x^2-3}-3\right|<\epsilon$$ holds. So let $\epsilon>0$ be given. We observe that $$ \frac{3x^2+1}{x^2-3}-3=\frac{10}{x^2-3}.$$ By a suitable choice of $R$, we can ensure that this expression is between $0$ and $\frac5x$ (The following calculation is guaranteed to succeed somehow or other with any positive numerator in place of $5$ as well; there's nothing magic in it). Indeed, for $x>\sqrt3$, we have $x^2-3>0$ (so that $\frac{10}{x^2-3}>0$) and $x>0$, so $$ \frac{10}{x^2-3}<\frac5x\iff 10x<5x^2-15\iff 10<5(x-1)^2.$$ To ensure the latter, it is sufficeint to have $(x-1)^2>2$. Hence certainly a choice of $R\ge3$ will do what we need: Then $x>R\ge3$ guarantes both $x^2-3>0$ and $(x-1)^2>2$, so $0<\frac{10}{x^2-3}<\frac5x $. Now if we additionally have that $R\ge\frac5\epsilon$, then $x>R$ implies $\frac5x<\frac5R\le\epsilon$, so that $0<\frac{10}{x^2-3}<\epsilon $, and even more so $(2)$ holds. Rereadingwhat we just did, we see that we must ensure that $R\ge 3$ and $R\ge \frac5\epsilon$.

Hence with the choice $$R=\max\left\{3,\frac5\epsilon\right\} $$ we have that $(2)$ holds for all $x>0$. This proofs the claim $81)$.$_\square$

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Important points:

• Formulate the result you want to prove (even if you know the result only afterwards)
• Formulate what needs to actually be shown in order to prove the claim.
• Explicitly state any additional assumptions you make (such as $R\ge3$, $R\ge\frac5\epsilon$ here) in simplifying inequalities
• Beware of cases where the additional assumptions might contradict each other (if I had needed $R<3$ and $R\ge \frac3\epsilon$ instead ...)
• Actually show the claim. Finding a result vs. showing its correctness is usually a 180 degree turn, but if either of these two parts be left out at all, then the actual showing part better be left intact (of course, teachers request to also show the full finding process).
• In general, don't abhor textual fillings. A written "Therefore" or "From this and the original condition we conclude" or "For thsi it would be sufficient to have" tells the reader more than a simple logical connective $\Rightarrow$ or $\Leftrightarrow$, especially when the latter is more often then not left out or not justified or written in the plainly wrong direction.