# Find the limit (proof). (assignment)

This is my assignment question.

Find $\lim_{n \to \infty} (2n^2+10n+5)/n^2$.

My attempt: suppose $\lim_{n \to \infty} (2n^2+10n+5)/n^2=2$

Then, $|\frac {2n^2+10n+5}{n^2}-2|= \frac {10n+5}{n^2}\le \frac {15n}{n^2}= \frac {15}{n}$.

Then, let $M \in N$ and $\frac {15}M< \varepsilon$. Then, for $\varepsilon >0$, there exists $M$ such that $|\frac {2n^2+10n+5}{n^2}-2|= \frac {10n+5}{n^2}\le \frac {15n}{n^2}= \frac {15}{n}\le\frac {15}{M}<\varepsilon$ for $n\ge M$.

My question is (1) Can I say "suppose $\lim_{n \to \infty} (2n^2+10n+5)/n^2=2$" before finding the limit ? (2)$\frac {10n+5}{n^2}\le \frac {15n}{n^2}$. I am also not sure whether it is true.

• yes of course you need to claim that lim=2, indeed this kind of proof are not used for the calculation of the limit but to get confident with the definition and to prove the simplest base case
– user
Mar 23, 2018 at 8:27

(1) You can say: " I claim that the limit $=2$".

Then you prove this claim. Your proof above is fine.

(2) for $n \in \mathbb N$ we have $5 \le 5n$ , hence $10n+5 \le 15n$ and $\frac {10n+5}{n^2}\le \frac {15n}{n^2}$ follows.

You could just do $$\lim_{n \to \infty}\frac{2n^2+10n+5}{n^2}=\lim_{n\to\infty}\frac{2+\frac{10}n+\frac5{n^2}}{1}=\frac{2+0+0}{1}=2$$ since both $\dfrac1n$ and $\dfrac1{n^2}$ tend to $0$ as $n\to\infty$.

• I think the OP is asking for clarifications about a specific proof by definition.
– user
Mar 23, 2018 at 8:26
• I know the OP used the definition of a limit to find it, but the question didn't ask for a specific approach so I thought it would be fine to give OP a different (and faster) approach to the problem. Mar 23, 2018 at 15:24

Of course you can, because you never use this assumption in the proof !

You are indeed working on $|f(n)-2|$ and come to establish it $<\epsilon$, but nowhere do you use the fact that $\lim f(n)=2$.

Here $2$ is just a tentative value (which you found by educated guess), and you prove it to be the right choice.

If there was a line in your proof saying "knowing that $\lim f(n)=2$, then ...", this would of course be a circular argument.

Yes of course we need to claim that the limit is equal to 2, indeed this kind of proof are not used for the calculation of the limit but to get confident with the definition and to prove the simplest base case

For the proof the logic is firstly fix $\epsilon>0$ and then find $N$ such that for $n\ge N$

$$|\frac {2n^2+10n+5}{n^2}-2|<\epsilon\implies10n+5<\epsilon n^2\implies\epsilon n^2-10n-5>0 \\\implies n>\frac{10+\sqrt{100+20 \epsilon}}{2\epsilon}$$

which complete the proof.