Limit as n approaches $\infty$ of $\frac{x}{x+n}$.

Question

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. Edit: The squeeze theorem should be of this form: $$\frac{x}{x+2n}\leq \frac{x}{x+n}<\frac{x}{n}$$ Otherwise we aren't proving it goes to zero.

Answer 1

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

$\displaystyle \lim_{n \to \infty} \dfrac{x}{x + n} = 0. \tag 9$

Answer 2

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, $\lim_{n\to\infty}\frac{x}{x+n}=0$.

Answer 3

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:

$\lim_{n\to \infty} \frac x{x+n} = x\lim_{n\to \infty} \frac 1{x+n} = x\lim_{x+n\to \infty} \frac 1{x+n} = x\lim_{m=x+n\to \infty}\frac 1m = x*0 = 0$.

Answer 4

$x>0$, 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 $n \ge n_0$ :

$|\dfrac{x}{x+n}| \lt \dfrac{x}{n} \lt \dfrac{x}{n_0} \lt x (\dfrac{\epsilon}{x}) = \epsilon.$.