Vtyjkyuk

How does one go about proving
$$lim_nrightarrow inftyfracxx+n=0$$
for any positive x. Intuitively this is pretty obvious. I'm assuming this is a squeeze theorem question where
$$frac1x+nleq fracxx+n<fracxx$$
but this doesn't really get us anywhere.

For

$x > 0 tag 1$

and

$n > 0, tag 2$

we have

$dfracxx + n = dfracx / nx / n + 1; tag 3$

note that

$1 + x/n > 1, tag 4$

or

$(1 + x/n)^-1 < 1; tag 5$

pick $epsilon > 0$; then for $n$ sufficiently large,

$dfracxn < epsilon; tag 6$

also, from (5) and (6) together in collusion,

$dfracx/n1 + x/n = (1 + x/n)^-1 (x/n) < x/n < epsilon; tag 7$

thus

$dfracxx + n = dfracx/n1 + x/n < epsilon; tag 8$

this shows that by taking $n$ large enough, we have $x/(x + n)$ arbitrarily small; hence

$displaystyle lim_n to infty dfracxx + n = 0. tag 9$

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 $frac xx+n < epsilon iff$

$x < epsilon (x+n) iff$

$x - epsilon x < epsilon n iff$

$n > frac x(1-epsilon)epsilon$

And that's that.

For any $epsilon > 0$ if $M > frac x(1-epsilon)epsilon$ then $n> M$ means $frac xx+n < frac xx + frac x(1-epsilon)epsilon=epsilon$.

.....

Another way of putting this is:

$lim_nto infty frac xx+n = xlim_nto infty frac 1x+n = xlim_x+nto infty frac 1x+n = xlim_m=x+nto inftyfrac 1m = x*0 = 0$.

Note that $x$ doesn't necessarily have to be positive. When $x=0$, the problem is reduced to $lim_ntoinftyfrac1n=0$ which is straightforward and easy. Hence, we're left to prove it for non-zero real numbers.

Now take $x in mathbbR-0$. Take $m in mathbbN$ to be sufficiently large such that $x + m > 0$.

We now have:

$$frac1n-m+1 leq frac1x+nleq frac1n-m hspace10px(textwhen n >m)$$

Using the Archemedean property of the real line, for any given $epsilon >0$, you can find $t in mathbbN$ such that $frac1t < fracepsilon$. Now, set $N=t+m$. This proves that

$$forall epsilon >0, exists N in mathbbN: ngeq N implies |frac1x+n|<fracepsilon$$

$$forall epsilon >0, exists N in mathbbN: ngeq N implies |fracxx+n|<epsilon$$

Hence, $lim_ntoinftyfracxx+n=0$.

$x>0$, real, $n in mathbbZ^+$.

$0 < dfracxx+n lt dfrac xn$.

Let $epsilon >0$ be given.

Archimedean principle:

There is a $n_0 in mathbbZ^+$ such that

$n_0 > dfracxepsilon$.

For $n ge n_0$ :

$|dfracxx+n| lt dfracxn lt dfracxn_0 lt x (dfracepsilonx) = epsilon.$.