Show that the $\lim_{n\to \infty} \frac n{n+1} = 1$

Question

Show that the $\lim_{n\to \infty} \frac n{n+1} = 1$ using the limit definition.

So far I have:

Let $\varepsilon > 0$, there exists $N \ge 0$ such that,

$|a_n-L|< \varepsilon$, for all $n \ge N$.

$\left|\frac n{n+1} -1\right| < \varepsilon$. I have ended up with $\left|\frac 1{n+1} \right| \le \frac n n = 1$

(which is $< \varepsilon$, for all $n$)

I know there needs to be an n at the end, so for all $n >$ 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

Answer 1

Using that for $A,B>0$

$$A<B \iff \frac 1 A > \frac 1 B$$

we have that $\forall \varepsilon>0$

$$\left| \frac1{n+1}\right|<\varepsilon \iff n+1 >\frac1 \varepsilon \iff n>\frac1 \varepsilon -1$$

then we can take any $N\ge \frac1 \varepsilon -1$ to comply the definition of limit.

Answer 2

\begin{align*} \epsilon > \left|\frac{n}{n+1} - 1\right| = \left|-\frac{1}{n+1}\right| = \frac{1}{n+1} \hspace{10pt} \Longrightarrow \hspace{10pt} n > \frac{1}{\epsilon} -1 \end{align*}