Prove $\lim\limits_{n \to \infty} n/(n+3)=1$ I have trouble with proving $\lim\limits_{n \to \infty} n/(n+3)=1$.
I know may divide the numerator and the denominator by $n$ as a step, but I think it might be wrong.
 A: Note that when $n \to \infty \implies \lim\limits_{n \to \infty} \dfrac{1}{n} =0$.
Then we have that $$ \lim_{n \to \infty} \frac{n}{n+3} = \lim_{n \to \infty} \frac{1}{1 + \frac{3}{n}} = \frac{1}{1 + 0} =1 $$
I guess this will be fine for you.
A: If you divide $\frac n{n+3}$ numerator and denominator by $n$ you get $\frac 1{ 1+ \frac 3n}$ and if you accept that $\lim \frac 1{1+3\frac 1n} = \frac 1{1 + 3\lim \frac 1n}$ and if you accept that $\lim \frac 1n = 0$ you can claime $\lim \frac n{n+3} =\lim \frac 1{1  + 3\cdot \frac 1n} = \frac 1{1+ 3\lim \frac 1n} =\frac 1{1 + 3\cdot 0} = \frac 11 = 1$.
Alternatively You can do $\lim \frac n{n+3} = \lim\frac {(n+3) -3}{n+3} =\lim_{n\to \infty} 1 -\frac 3{n+3} =\lim_{m-3=n\to \infty} 1 -\frac 3{m} = \lim_{m\to \infty} 1 -\frac 13{m} = 1-0 = 0$.
but in each case you must justify that you can put limits "inside" functions like that (i.e. that you have a theorem that $f$ is continuous and $a_n \to L$ then $\lim f(a_n) = f(\lim a_n)$.  That requires you have proven the theorem and that you can verify that $f$ is continuous and that $a_n \to L$.)
You can also to an $\epsilon-N$ proof.  For $\epsilon > 0$ then $\frac n{n+3} -1 < \epsilon $ will be assure if $n-(n+3) < \epsilon(n+3)$ or in other words if $3-3\epsilon < \epsilon n$ or if $n> \frac 3{\epsilon} - 3$.
(i.e.  For any $\epsilon: 0 < \epsilon < 1$ then for any $n > \frac 3{\epsilon}-3$ then $|\frac n{n+3} - 1| < \epsilon$.  So $\lim_{n\to \infty} \frac n{n+3} = 1$.
A: Hint:
$$\frac{n}{n+3}=\frac{n+3-3}{n+3}=1-\frac{3}{n+3}$$
