Proof that if $x_n \rightarrow \infty$ then $\frac{x_n}{x_n +1}$ converges Prove that if $x_n \rightarrow \infty$ then the sequence $\frac{x_n}{x_n +1}$ converges.
Here is my attempt:
We know that $x_n \rightarrow \infty$ so for every $M$ there exists an $N$ such that for all $k \geq N,$ $x_k \geq M$. Write $\lvert \frac{M}{M+1}\rvert < \epsilon $ for $\epsilon > 0.$ Solving, we get $\frac{1}{\epsilon} < M.$
How do I connect this back to $N$? Or is this a complete proof, given that I have found a relation between $\epsilon$ and $M$?
 A: For all $n \ge N$, 
$$\left\lvert 1 - \frac{x_n}{1 + x_n}\right\rvert = \left\lvert \frac{1}{1 + x_n}\right\rvert = \frac{1}{1 + x_n} \le \frac{1}{1 + M} < \frac{1}{M}$$
Since $M$ is an arbitrary positive number, so is $\frac{1}{ M}$. Thus $\frac{x_n}{1 + x_n}$ converges to $1$.
A: Hint:
\begin{align}
\left|\frac{x_n}{1+x_n} - \frac{x_m}{1+x_m} \right| = \left| \frac{x_n-x_m}{(1+x_n)(1+x_m)}\right|\leq \frac{2\max(|x_n|, |x_m|)}{(1+x_n)(1+x_m)}<
\frac{2}{\min(|1+x_n|, |1+x_m|)}
\end{align}
A: Consider the following calculation: $\frac{x_n}{x_n + 1} = \frac{x_n}{x_n} \cdot \frac{1}{1 + \frac{1}{x_n}} = \frac{1}{1 + \frac{1}{x_n}}$. Letting $n \to \infty$ gives therefore: $\lim_{n \to \infty} \frac{x_n}{x_n + 1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{x_n}} = \frac{\lim_{n \to \infty} 1}{\lim_{n \to \infty} (1 + \frac{1}{x_n})} = \frac{1}{1 + 0} = \frac{1}{1} = 1$; The third-to-last equality is due to the fact that $\lim_{n \to \infty} x_n = \infty$, so the reciprocal term $\frac{1}{x_n}$ will tend to zero as $n$ tends to infinity.
