TITLE: Using the definition of limit to evaluate $\lim_{n \to \infty}\frac{3n+1}{n}$
QUESTION [3 upvotes]: I am trying to to understand the limit definition.
For the sequence $\dfrac{3n+1}{n}$ I can divide by the highest polynomial in denominator and get that 3 may be the limit.
Using the limit definition $$\left|\dfrac{3n+1}{n}-3\right|<\epsilon \implies \left|\dfrac{3n+1-3n}{n}\right|<\epsilon \implies \dfrac{1}{n}< \epsilon \rightarrow \dfrac{1}{\epsilon}<n $$ 
which means that whatever $\epsilon$ smallest as I want the will be $N<n$ that from then on all the element of the sequence will be close to $3$.
Now if I take a wrong limit like $2$ I get $$\left|\frac{3n+1}{n}-2\right|<\epsilon \implies \left|\frac{3n+1-2n}{n}\right|<\epsilon \implies\frac{n+1}{n}<\epsilon$$
Now because I can not isolate $n$ that mean that there is no $N<n$ that from than on all the element of the sequence will be close to $2$?

REPLY [2 votes]: Note that $\frac{n+1}{n}=1+\frac 1n$. You'll have some trouble getting this to be less than $\epsilon$, since it's always greater than $1$ for positive integers $n$!
Let's say you want to find $\lim_{n \to \infty} \frac{p(n)}{q(n)}$, where $p$ and $q$ are polynomials of the same degree. Let's write $p(x) = a_0x^k + \dots a_k$ and $q(x) = b_0 x^k + \dots + b_k$. Then I claim the limit of $\frac{p(n)}{q(n)}$ is $a_0/b_0$. (I can write a proof if you want, but you can probably find it in your textbook.) In particular, when you're trying to use the $\varepsilon$-definition to see that $\lim \frac{p(n)}{q(n)} \to L$, you have an intermediate step where you write $\|\frac{p(n)}{q(n)} - L\| = \|\frac{p_2(n)}{q_2(n)}\|$ for some other polynomials $p_2$ and $q_2$. (In your example above, when you picked $L = 2$, you got $p_2(n) = n+1$ and $q_2(n) = n$.) So if you want $\frac{p(n)}{q(n)}$ to have any hope of converging to $L$, this new thing had better converge to zero - so $p_2$ had better have degree less than $q_2$! And that's the problem when you picked $L=2$ - the two polynomials had the same degree.

REPLY [1 votes]: Logic:
$\frac{3n+1}{n}$ well, $3n+1$ is only a little bit bigger than $3n$, take n being 100, then $3n+1$ is 301, which is basically 300, take 1,000,000, then 3,000,001 is basically 3,000,000
So $\frac{3n+1}{n}\approx\frac{3n}{n}=3$
That's the logic.
Formal:
Don't bother doing $\forall\epsilon>0\exists N\in\mathbb{N}:n>N\implies|a_n-a|<\epsilon$ where $a_n$ is the $n^\text{th}$ term and $a$ is the limit.
You should have done convergence of the sum of two convergent sequences, just be like:
BUT WAIT: $\frac{3n+1}{n}=\frac{3n}{n}+\frac{1}{n}=3+\frac{1}{n}$ and note that $\frac{1}{n}\rightarrow 0$ which is an easy one to prove (reply in a comment if you want the proof of this)
Job done!

REPLY [1 votes]: In general, $x_n \to \ell$ as $n \to \infty$ if and only if $x_n - \ell \to 0$ as $n \to \infty$.
So the way to find out if you have the correct limit is:

Do we have $x_n - \ell \to 0$?


In your example, $x_n = \frac{3n+1}n$ and $\ell = 3$. Notice that $x_n - \ell = \frac 1n \to 0$ as $n \to \infty$. This means that for any $\epsilon > 0$, we can find $N$ such that for every $n \ge N$, $\frac 1n \lt \epsilon$.
But $x_n - 2 = \frac{n+1}n \to 1 \ne 0$
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