limits of the sequence $n/(n+1)$ Given the problem:
Determine the limits of the sequnce $\{x_n\}^
\infty_{
n=1}$
$$x_n = \frac{n}{n+1}$$
The solution to this is:  
step1:  
$\lim\limits_{n \rightarrow \infty} x_n = \lim\limits_{n \rightarrow \infty} \frac{n}{n + 1}$  
step2:  
$=\lim\limits_{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}$  
step3:  
$=\frac{1}{1 + \lim\limits_{n \rightarrow \infty} \frac{1}{n}}$  
step4:  
$=\frac{1}{1 + 0}$  
step5:  
$=1$
I get how you go from step 2 to 5 but I don't understand how you go from step 1 to 2.
Again, I'm stuck on the basic highschool math.
Please help
 A: Divide the numerator and denominator by $n$. Why is this legal, in other words, why does this leave your fraction unchanged? 

Because $$\frac {\frac a n} {\frac b n}=\frac {a \cdot \frac 1 n} {b \cdot \frac 1 n}=\frac a b$$ where the last equality is because $\dfrac 1 n$'s get cancelled. 

Further, remember the fact that: 
$$\frac{a+b}{n}=\frac a n+\frac b n$$
A: This is just algebraic manipulation from step 1 to step 2. Since $n \neq 0$, we can do the following. I will write it out in full detail so that you are clear on the steps involved.
$$\begin{eqnarray*} \frac{n}{n+1} &=& n \left(\frac{1}{n+1}\right)\\
&=& (n^{-1})^{-1} \left(\frac{1}{n+1}\right)\\
&=& \left(\frac{1}{n}\right)^{-1} \left(\frac{1}{n+1}\right) \\
&=&\frac{1}{\left(\frac{1}{n}\right)}\left(\frac{1}{n+1}\right)\\
&=& \frac{1}{ \left(\frac{1}{n}\right)(n+1)}\\
&=& \frac{1}{\left(\frac{n+1}{n}\right)} \\
&=& \frac{1}{\left( \frac{n}{n} + \frac{1}{n}         \right)}\\
&=& \frac{1}{ \left( 1 + \frac{1}{n} \right)}
\end{eqnarray*}$$
as desired.
A: You can also do this way,
$$\begin{eqnarray*} \lim_ {n\to \infty}x_n &=& \lim_{n\to \infty} \frac{n+1-1}{n+1}\\
&=& \lim_{n\to \infty}1-\frac{1}{n+1}\\
&=& 1-\lim_{n\to \infty}\frac{1}{n+1}\\
&=& 1 \end{eqnarray*}$$
because as $n \to \infty$, we have $\frac{1}{n+1} \to 0 .$
