Limit as n approaches $\infty$ of $\frac{x}{x+n}$. How does one go about proving 
$$\lim_{n\rightarrow \infty}\frac{x}{x+n}=0$$
for any positive x.  Intuitively this is pretty obvious.  I'm assuming this is a squeeze theorem question where
$$\frac{1}{x+n}\leq \frac{x}{x+n}<\frac{x}{x}$$
but this doesn't really get us anywhere.
 A: Note that $x$ doesn't necessarily have to be positive. When $x=0$, the problem is reduced to $\lim_{n\to\infty}\frac{1}{n}=0$ which is straightforward and easy. Hence, we're left to prove it for non-zero real numbers.
Now take $x \in \mathbb{R}-\{0\}$. Take $m \in \mathbb{N}$ to be sufficiently large such that $x + m > 0$.
We now have:
$$\frac{1}{n-m+1} \leq \frac{1}{x+n}\leq \frac{1}{n-m} \hspace{10px}(\text{when }n >m)$$
Using the Archemedean property of the real line, for any given $\epsilon >0$, you can find $t \in \mathbb{N}$ such that $\frac{1}{t} < \frac{\epsilon}{|x|}$. Now, set $N=t+m$. This proves that
$$\forall \epsilon >0, \exists N \in \mathbb{N}: n\geq N \implies |\frac{1}{x+n}|<\frac{\epsilon}{|x|}$$
$$\forall \epsilon >0, \exists N \in \mathbb{N}: n\geq N \implies |\frac{x}{x+n}|<\epsilon$$
Hence, $\lim_{n\to\infty}\frac{x}{x+n}=0$.
A: For 
$x > 0 \tag 1$
and 
$n > 0, \tag 2$
we have
$\dfrac{x}{x + n} = \dfrac{x / n}{x / n + 1}; \tag 3$
note that
$1 + x/n > 1, \tag 4$
or 
$(1 + x/n)^{-1} < 1; \tag 5$
pick $\epsilon > 0$; then for $n$ sufficiently large,
$\dfrac{x}{n} < \epsilon; \tag 6$
also, from (5) and (6) together in collusion,
$\dfrac{x/n}{1 + x/n} = (1 + x/n)^{-1} (x/n) < x/n < \epsilon; \tag 7$
thus
$\dfrac{x}{x + n} = \dfrac{x/n}{1 + x/n} < \epsilon; \tag 8$
this shows that by taking $n$ large enough, we have $x/(x + n)$ arbitrarily small; hence
$\displaystyle \lim_{n \to \infty} \dfrac{x}{x + n} = 0. \tag 9$
A: I think you are somehow thinking that we are using the letter $x$ to represent a fixed positive number that $x$ is variable.  It isn't.  $x$ is a fixed positive number.
For any $\epsilon > 0$ then $\frac x{x+n} < \epsilon \iff$
$x < \epsilon (x+n) \iff$
$x - \epsilon x < \epsilon n \iff$
$n > \frac {x(1-\epsilon)}\epsilon$
And that's that. 
For any $\epsilon > 0$ if $M > \frac {x(1-\epsilon)}\epsilon$ then $n> M$ means $\frac {x}{x+n} < \frac {x}{x + \frac {x(1-\epsilon)}\epsilon}=\epsilon$.
.....
Another way of putting this is:
$\lim_{n\to \infty} \frac x{x+n} = x\lim_{n\to \infty} \frac 1{x+n} = x\lim_{x+n\to \infty} \frac 1{x+n} = x\lim_{m=x+n\to \infty}\frac 1m = x*0 = 0$.
A: $x>0$, real, $n \in \mathbb{Z^+}$.
$0 < \dfrac{x}{x+n} \lt \dfrac {x}{n}$.
Let $\epsilon >0$ be given.
Archimedean principle:
There is a $n_0 \in \mathbb{Z^+}$ such that
$n_0 > \dfrac{x}{\epsilon}$.
For $n \ge n_0$ :
$|\dfrac{x}{x+n}| \lt \dfrac{x}{n} \lt \dfrac{x}{n_0} \lt  x (\dfrac{\epsilon}{x}) = \epsilon.$.
