Show that the $\lim_{n\to \infty} \frac n{n+1} = 1$ Show that the $\lim_{n\to \infty} \frac n{n+1} = 1$ using the limit definition.
So far I have:
Let $\varepsilon &gt; 0$, there exists $N \ge 0$ such that,
$|a_n-L|&lt; \varepsilon$, for all $n \ge N$.
$\left|\frac n{n+1} -1\right| &lt; \varepsilon$. I have ended up with $\left|\frac 1{n+1} \right|  \le \frac n n = 1$
(which is $&lt; \varepsilon$, for all $n$)
I know there needs to be an n at the end, so for all $n &gt;$ something $\varepsilon$. I am not sure what to do after $n/n$, or is that enough to show the limit of the sequence $= 1$.
Thanks
Using that for $A,B&gt;0$
$$A&lt;B \iff \frac 1 A &gt; \frac 1 B$$
we have that $\forall \varepsilon&gt;0$
$$\left| \frac1{n+1}\right|&lt;\varepsilon \iff n+1 &gt;\frac1 \varepsilon \iff n&gt;\frac1 \varepsilon  -1$$
then we can take any $N\ge \frac1 \varepsilon  -1$ to comply the definition of limit.
&quot; I have ended up with |1/(n+1)| ≤ n/n = 1&quot;: no, you end up with
$$\left|\frac1{n+1}\right|&lt;\epsilon$$ and it is no problem to find an $N$ above which this holds.