Limit of the sequence $a_n=\frac{n}{n+1}$ Please help me check limit of the sequence $a_n=\frac{n}{n+1}$
 A: HINT: If you look at the numbers $\frac{n}{n+1}$ for large $n$, you should be able to make a good guess. To prove your guess, note that $$\frac{n}{n+1}=1-\frac1{n+1}\;.$$
A: $|a_n|=|\frac{n}{n+1}|$
$=|\frac{n+1-1}{n+1}|$
$=|\frac{n+1}{n+1}-\frac{1}{n+1}|$
$=|1-\frac{1}{n+1}|$
$1<n+1$ for $n\in N$ $\Rightarrow$ $\frac{1}{n+1}<1$ $\Rightarrow$ $1-\frac{1}{n+1}<1$.
Definitely
$|a_n|<1$ $\Rightarrow$ $-1<a_n<1$
A: Consider the function $f(x)=\frac{x}{x+1}$. We have $\lim_{x \to \infty} \frac{x}{x+1}=\lim_{x \to \infty} \frac {1}{1+(1/x)}$. Do you see how to finish?
A: It is a consequence of the unboundedness of the natural numbers that, for any $\epsilon >0$, there exists a natural number $n$ for which $\dfrac 1 n <\epsilon$.  This is used to prove the "basic" limit,
$$\tag 1 \lim\limits_{n\to \infty}\frac 1 n = 0$$
The expression
$\dfrac n {n+1}$ can be written as 
$$1-\frac 1 {n+1}$$
The algebra of limits says that if $\lim\limits_{n\to \infty}a_n=a$ and $\lim\limits_{n\to \infty} b_n=b$ , then $\lim\limits_{n\to \infty} c_n = a+b$ where $c_n=a_n+b_n$. It is straightforward that $\lim\limits_{n\to \infty} 1 =1$, so you need to show 
$$\tag 2 \lim_{n\to \infty}\frac {-1} {n+1} $$
exists and equals something. Can you do this using $(1)$? Hint: if $n$ is a natural number, so is $n+1$, and $$\left|-\frac 1 {n+1}-0\right|=\frac 1 {n+1}$$
