calculus - Using the definition of limit to evaluate $lim_{n to infty}frac{3n+1}{n}$

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}which means that whatever $\epsilon$ smallest as I want the will be $N

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

Answer

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 $x_n - \ell \to 0$?

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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 $x_n - 2 = \frac{n+1}n \to 1 \ne 0$