Real Analysis: Limits
Limits of Real Functions at Infinity
Motivation
The discussion of limits so far has focused on the behavior of a function as the values in its domain approach a particular point. However, of equal interest is what happens to the value of a function, particularly a real function, as the values in its domain grow towards infinity.
Rigorously defining the limit of a real function at infinity requires some extra thought. Since the Euclidean distance function does not work as a metric for the extended real numbers, we are forced to think outside the box if we are to stay true to the topological definition of a limit in terms of open sets. There are two workarounds for this explained in the following two sections, but you may skip them if you are only interested in getting to the practical definition of a limit of a real function at infinity.
No Euclidean Metric Space for $\overline{\mathbb{R}}$
The extended real numbers $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, \infty\}$ cannot be turned into a metric space by using the regular Euclidean distance metric $d(a, b) = |b - a|.$ Even if we tried to use this function as though it were a metric anyway, the only open balls that would contain either $\infty$ or $-\infty$ would be the singleton sets $\{\infty\}$ and $\{-\infty\}.$ This would in turn leave $\infty$ and $-\infty$ as isolated points, which would preclude us from taking limits as them. However, we can clearly imagine that a function can converge to some value as the values in its domain approach $\infty,$ so we will need to be a bit smarter. There are (at least) two possible solutions.
A Different Metric for $\overline{\mathbb{R}}$?
The first solution is to find a metric that does properly handle infinity. For this we need to find a bijection from $\overline{\mathbb{R}}$ to a subset of $\mathbb{R},$ such as $[-1, 1]$. Assuming we had a bijection $h : \overline{\mathbb{R}} \rightarrow [-1, 1],$ we could define a metric for $\mathbb{R}$ as $d(a, b) = |h(b) - h(a)|.$ Such a function would transform $\infty$ and $-\infty$ into limit points, which would in turn allow us to take limits at them. While such functions do exist, they would make all of the actual calculations of limits much more cumbersome. For example, consider the following bijection $h$ between $\overline{\mathbb{R}}$ and $[0, 1]$:
$$h(x) = \left\{ \begin{array}{ll} -1 & x = -\infty\\ \dfrac{-1+\sqrt{4x^2+1}}{2x} & x \in \mathbb{R} - \{0\} \\ 0 & x =0 \\ 1 & x = \infty \end{array} \right.$$
whose inverse is
$$h^{-1}(x) = \left\{ \begin{array}{ll} -\infty & x=-1\\ \dfrac{x}{x^2-1} & x \in (-1,1)\\ \infty & x = 1\end{array} \right.$$
We can see that $B_{1}(-\infty) = [-\infty, 0).$ But throwing that gnarly rational function into the mix when calculating something as simple as $\lim\limits_{x \rightarrow \infty} \frac{1}{x}$ seems excessive when we know the answer should be $0.$
A Topological Space for $\overline{\mathbb{R}}$
The second solution is to use a different definition of open sets. This involves briefly leaving the world of metric topology and reaching into the world of general topology. Recall that in metric topology the open sets in $\mathbb{R}$ are those sets that are made out of open balls, which in $\mathbb{R}$ are open intervals. However, since $\overline{\mathbb{R}}$ is not a metric space with the standard Euclidean metric, the construction does not seamlessly generalize to open intervals containing $\infty,$ but instead breaks down entirely as explained above.
The solution is to switch from using open sets made out of open balls to simply using open sets made out of arbitrary unions of open intervals, where $(a, b),$ $[-\infty, b),$ and $(a, \infty]$ for all $a, b \in \mathbb{R}$ are defined to be be the open intervals in $\overline{\mathbb{R}}.$ This construction of open sets is called the order topology on $\overline{\mathbb{R}},$ and the standard one generated from open balls is called the metric topology. Unlike the would-be metric topology on $\overline{\mathbb{R}}$ using the Euclidean distance metric, the order topology on $\overline{\mathbb{R}}$ treats $\infty$ and $-\infty$ as limit points.
Everything we have proven about metric topology by using open sets is still valid for any other topology. Thus our generality has paid off, as an expansion of the definition of a limit in terms of open sets reveals the intuitively simple structure we would expect. And unlike the metric topology, this structure keeps concrete calculations as simple as possible. As a result, the definition we provide now is the one implied by the order topology on $\overline{\mathbb{R}}.$
Limits of Real Functions at Infinity
If $f$ is a function from $\overline{\mathbb{R}}$ to $\mathbb{R}$ for every $\varepsilon > 0$ there exists an $r \in \mathbb{R}$ such that $|f(x) - L| < \varepsilon$ whenever $x > r,$ then the limit of $f$ at infinity equals $L.$ This fact is expressed mathematically as
$$\lim\limits_{x \rightarrow \infty} f(x) = L.$$
Likewise, when the limit of $f$ at infinity is $L,$ we can say that $f$ converges to $L.$ If the limit at infinity is not defined, then $f$ diverges.
An identical definition exists for limits at negative infinity. If for every $\varepsilon > 0$ there exists an $r \in \mathbb{R}$ such that $|f(x) - L| < \varepsilon$ whenever $x < r,$ then the limit of $f$ at negative infinity equals $L.$ This fact is expressed similarly mathematically as
$$\lim\limits_{x \rightarrow -\infty} f(x) = L.$$
Note: Any real function $f$ can be extended to be a function of the extended reals and have any values whatsoever filled in for $f(\infty)$ and $f(-\infty),$ as the value of a function at its limit point does not matter. Thus the foregoing definitions may treat $f$ as a real function without any meaningful cost in precision.
Algebraic Limit Theorem for Limits at Infinity
The algebraic limit theorem for real functions for limits at real points can be extended to cover limits at infinity. If $f$ and $g$ are real functions with limits at infinity, then the algebraic limit theorem for limits at infinity states the following:
-
$\lim\limits_{x \rightarrow \infty} af(x) = a\lim\limits_{x \rightarrow \infty}f(x).$
-
$\lim\limits_{x \rightarrow \infty} \left(f(x) + g(x)\right) = \lim\limits_{x \rightarrow \infty} f(x) + \lim\limits_{x \rightarrow \infty} g(x).$
-
$\lim\limits_{x \rightarrow \infty} \left(f(x) - g(x)\right) = \lim\limits_{x \rightarrow \infty} f(x) - \lim\limits_{x \rightarrow \infty} g(x).$
-
$\lim\limits_{x \rightarrow \infty} \left(f(x)g(x)\right) = \left(\lim\limits_{x \rightarrow \infty} f(x)\right)\left(\lim\limits_{x \rightarrow \infty} g(x)\right).$
-
$\lim\limits_{x \rightarrow \infty} \dfrac{1}{g(x)} = \dfrac{1}{\lim\limits_{x \rightarrow \infty} g(x)}$ when $\lim\limits_{x \rightarrow \infty} g(x) \neq 0.$
-
Corollary: $\lim\limits_{x \rightarrow \infty} \dfrac{f(x)}{g(x)} = \dfrac{\lim\limits_{x \rightarrow \infty} f(x)}{\lim\limits_{x \rightarrow \infty} g(x)}$ when $\lim\limits_{x \rightarrow \infty} g(x) \neq 0.$
-
Squeeze Theorem for Limits at Infinity
Just like limits of real functions at real values, limits of real functions at infinity have a squeeze theorem. If $f,$ $g,$ and $h$ are all functions from $\overline{\mathbb{R}}$ to $\mathbb{R}$ such that $f(x) \leq g(x) \leq h(x)$ for all $x > \delta$ for some $\delta \in \mathbb{R}$ and $\lim\limits_{x \rightarrow \infty} f(x) = \lim\limits_{x \rightarrow \infty} h(x),$ then $\lim\limits_{x \rightarrow \infty} g(x) = \infty.$
Problems
Algebraic Limit Theorem for Limits at Infinity I: Show that $\lim\limits_{x \rightarrow \infty} af(x) = a\lim\limits_{x \rightarrow \infty}f(x).$
Assume $f$ is a real function where $\lim\limits_{x \rightarrow \infty} f(x) = L.$ First note that if $a = 0,$ then $\lim\limits_{x \rightarrow \infty} 0f(x) = 0 = 0\lim\limits_{x \rightarrow \infty} f(x).$ If $a \neq 0,$ then for every $\frac{\varepsilon}{|a|} > 0$ there exists an $r \in \mathbb{R}$ such that $|f(x) - L| < \frac{\varepsilon}{|a|}$ whenever $x > r.$ Therefore $|a||f(x) - L| < \varepsilon,$ and so $|af(x) - aL| < \varepsilon.$ Therefore $\lim\limits_{x \rightarrow \infty} af(x) = a\lim\limits_{x \rightarrow \infty} f(x).$
Algebraic Limit Theorem for Limits at Infinity II: Show that $\lim\limits_{x \rightarrow \infty} \left(f(x) + g(x)\right) = \lim\limits_{x \rightarrow \infty} f(x) + \lim\limits_{x \rightarrow \infty} g(x).$
Let $f$ and $g$ be real functions with limits at infinity of $p$ and $q,$ respectively. Then for any $\frac{\varepsilon}{2} > 0$ there exists an $r_f \in \mathbb{R}$ such that $|f(x) - p| < \frac{\varepsilon}{2}$ whenever $x > r_f,$ and there also exists an $r_g \in \mathbb{R}$ such that $|g(x) - q| < \frac{\varepsilon}{2}$ whenever $x > r_g.$ Pick $r = \text{max}\{r_f, r_g\}.$ Then $|f(x) - p| < \frac{\varepsilon}{2}$ and $|g(x) - q| < \frac{\varepsilon}{2}$ whenever $x > r.$ By the properties of absolute value, we see that
$ |(f(x) + g(x)) - (p + q)| = |(f(x) - p) + (g(x) - q)| \\ |(f(x) + g(x)) - (p + q)| < |f(x) - p| + |g(x) - q| \\ |(f(x) + g(x)) - (p + q)| < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} \\ |(f(x) + g(x)) - (p + q)| < \varepsilon $
Therefore $\lim\limits_{x \rightarrow \infty}\left(f(x) + g(x)\right) = \lim\limits_{x \rightarrow \infty} f(x) + \lim\limits_{x \rightarrow \infty} g(x).$
Algebraic Limit Theorem for Limits at Infinity III: Show that $\lim\limits_{x \rightarrow \infty} (f(x) - g(x)) = \lim\limits_{x \rightarrow \infty} f(x) - \lim\limits_{x \rightarrow \infty} g(x).$
Assume $f$ and $g$ are real functions with limits at infinity. First, note that $f(x) - g(x) = f(x) + (-g(x)).$ The result follows by the algebraic limit theorem:
$\lim\limits_{x \rightarrow \infty} (f(x) - g(x)) = \lim\limits_{x \rightarrow \infty} (f(x) + (-g(x))) \\ \lim\limits_{x \rightarrow \infty} (f(x) - g(x)) = \lim\limits_{x \rightarrow \infty} f(x) + \lim\limits_{x \rightarrow \infty} (-g(x)) \\ \lim\limits_{x \rightarrow \infty} (f(x) - g(x)) = \lim\limits_{x \rightarrow \infty} f(x) - \lim\limits_{x \rightarrow \infty} g(x)$
Algebraic Limit Theorem for Limits at Infinity IV: Show that $\lim\limits_{x \rightarrow \infty} f(x)g(x) = \lim\limits_{x \rightarrow \infty} f(x) \lim\limits_{x \rightarrow \infty} g(x).$
Assume $\lim\limits_{x \rightarrow \infty} f(x) = a$ and $\lim\limits_{x \rightarrow \infty} g(x) = b.$ We would like to to show for any $\varepsilon$ there exists an $r \in \mathbb{R}$ such that $|f(x)g(x) - ab| < \varepsilon$ whenever $x > r.$
We want to connect the limit of the products to the product of the limits. Since the limit of the products is what we are trying to solve for, we must start by examining the product of the limits.
Consider $(f(x) - a)$ and $(g(x) - b)$, the expressions we evaluate for each individual limit. Taking their product gives
$(f(x) - a)(g(x) - b) = f(x)g(x) - bf(x) - ag(x) + ab $
We can now solve for $f(x)g(x) - ab$, the expression we would like to evaluate for the product of the limits:
$f(x)g(x) - ab = (f(x) - a)(g(x) - b) + bf(x) + ag(x) - 2ab$
We rearrange the righthand side to be fully in terms of differences between sequences and their limits:
$f(x)g(x) - ab = (f(x) - a)(g(x) - b) + bf(x) - ab + ag(x) - ab \\ f(x)g(x) - ab = (f(x) - a)(g(x) - b) + b(f(x) - a) + a(g(x) - b) $
We now take the absolute value of both sides and apply inequalities to isolate the limit terms:
$ |f(x)g(x) - ab| = |(f(x) - a)(g(x) - b) + b(f(x) - a) + a(g(x) - b)| \\ |f(x)g(x) - ab| \leq |(f(x) - a)(g(x) - b)| + |b(f(x) - a)| + |a(g(x) - b)| \\ |f(x)g(x) - ab| \leq |f(x) - a||g(x) - b| + |b||f(x) - a| + |a||g(x) - b| \\ $
Since $\lim\limits_{x \rightarrow \infty} f(x) = a$ and $\lim\limits_{x \rightarrow \infty}g(x) = b$, we can make the following claims:
-
For every $\sqrt{\frac{\varepsilon}{2}} > 0$, there exists an $r_1 \in \mathbb{R}$ such that $|f(x) - a| < \sqrt{\frac{\varepsilon}{2}}.$ for all $x > r_1.$
-
For every $\sqrt{\frac{\varepsilon}{2}} > 0$, there exists an $r_2 \in \mathbb{R}$ such that $|g(x) - b| < \sqrt{\frac{\varepsilon}{2}}$ for all $x > r_2.$
-
For every $\dfrac{\varepsilon}{4} > 0$, there exists a $r_3 \in \mathbb{R}$ such that $|b||f(x) - a| < \dfrac{\varepsilon}{4}$ for all $x > r_3.$
-
For every $\dfrac{\varepsilon}{4} > 0$, there exists a $r_4 \in \mathbb{R}$ such that $|a||g(x) - b| < \dfrac{\varepsilon}{4}$ for all $x > r_4.$
Take $r = \text{max}\{r_1, r_2, r_3, r_4\}$. Then we can substitute each value in for all $x > r$:
$ |f(x)g(x) - ab| < \left(\sqrt{\frac{\varepsilon}{2}}\right)\left(\sqrt{\frac{\varepsilon}{2}}\right) + \dfrac{\varepsilon}{4} + \dfrac{\varepsilon}{4} \\ |f(x)g(x) - ab| < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{4} + \dfrac{\varepsilon}{4} \\ |f(x)g(x) - ab| < \varepsilon $
Therefore $\lim\limits_{n \rightarrow \infty} f(x)g(x) = \lim\limits_{x \rightarrow \infty} f(x) \lim\limits_{x \rightarrow \infty} g(x).$
-
Algebraic Limit Theorem for Limits at Infinity V: Show that $\lim\limits_{x \rightarrow \infty} \dfrac{1}{g(x)} = \dfrac{1}{\lim\limits_{x \rightarrow \infty} g(x)}$ when $\lim\limits_{x \rightarrow \infty} g(x) \neq 0.$
Let $\lim\limits_{x \rightarrow \infty} g(x) = L$ where $L \neq 0.$ Consider the difference of the reciprocals:
$\dfrac{1}{g(x)} - \dfrac{1}{L}$
Multiply the first term by $\frac{L}{L}$ and the second term by $\frac{g(x)}{g(x)}$ to bring out the $L - g(x)$ term in the numerator:
$ \dfrac{1}{g(x)} - \dfrac{1}{L} = \dfrac{L}{Lg(x)} - \dfrac{g(x)}{Lg(x)} \\ \dfrac{1}{g(x)} - \dfrac{1}{L} = \dfrac{L - g(x)}{Lg(x)} \\ $
Taking the absolute value of both sides lets us use the $\varepsilon$ bound for the limit of $g$:
$ \left|\dfrac{1}{g(x)} - \dfrac{1}{L}\right| = \left|\dfrac{L - g(x)}{Lg(x)}\right| \\ \left|\dfrac{1}{g(x)} - \dfrac{1}{L}\right| = \dfrac{|L - g(x)|}{|Lg(x)|} \\ \left|\dfrac{1}{g(x)} - \dfrac{1}{L}\right| = \dfrac{|g(x) - L|}{|Lg(x)|} \\ \left|\dfrac{1}{g(x)} - \dfrac{1}{L}\right| = \dfrac{|g(x) - L|}{|L||g(x)|} \\ $
The righthand side contains the term $|g(x) - L|$ in the numerator, which will be less than any $\varepsilon > 0$ whenever $x > r_1$ for some $r_1 \in \mathbb{R}.$ Pick $\varepsilon = \frac{1}{2}|L|.$ It follows that there exists some $r_1 \in \mathbb{R}$ such that $|g(x) - L| < \frac{1}{2}|L|$ whenever $x > r_1.$ However, rather than substitute this inequality in for the numerator, note that it implies that $|g(x)| > \frac{1}{2}|L|.$ Critically, observe that this also ensures that $g(x) \neq 0$ whenever $x > r_1.$ We substitute this value into the denominator instead:
$ \left|\dfrac{1}{g(x)} - \dfrac{1}{L}\right| < \dfrac{2|g(x) - L|}{|L||L|} \\ \left|\dfrac{1}{g(x)} - \dfrac{1}{L}\right| < \dfrac{2|g(x) - L|}{|L|^2} \\ $
Now pick $\frac{1}{2}|L|^2\varepsilon > 0.$ It follows that there is some $r_2 \in \mathbb{R}$ such that $|g(x) - L| < \frac{1}{2}|L|^2\varepsilon$ whenever $x > r_2.$ Pick $r = \text{max}\{r_1, r_2\}.$ It follows that whenever $x > r,$ the inequality simplifies:
$ \left|\dfrac{1}{g(x)} - \dfrac{1}{L}\right| < \dfrac{2|L|^2\varepsilon}{2|L|^2\varepsilon} \\ \left|\dfrac{1}{g(x)} - \dfrac{1}{L}\right| < \varepsilon $
Therefore $\lim\limits_{x \rightarrow \infty} \dfrac{1}{g(x)} = \dfrac{1}{L} = \dfrac{1}{\lim\limits_{x \rightarrow \infty}g(x)}.$
Squeeze Theorem: Show that if $f,$ $g,$ and $h,$ are real functions such that $f(x) \leq g(x) \leq h(x)$ whenever $x > \delta$ for some $\delta \in \mathbb{R}$ and $\lim\limits_{x \rightarrow \infty} f(x) = L = \lim\limits_{x \rightarrow \infty} h(x),$ then $\lim\limits_{x \rightarrow \infty} g(x) = L.$
Assume $f,$ $g,$ and $h,$ are real functions such that $f(x) < g(x) < h(x)$ whenever $x > \delta$ for some $\delta \in \mathbb{R}$ and $\lim\limits_{x \rightarrow \infty} f(x) = L = \lim\limits_{x \rightarrow \infty} h(x).$ Then for every $\varepsilon > 0$ it follows that there exists an $r_f > \delta$ such that $|f(x) - L| < \varepsilon$ whenever $x > r_f.$ It follows that $-\varepsilon < f(x) - L < \varepsilon$ and therefore $L -\varepsilon < f(x) < L + \varepsilon$ whenever $x > r_f.$ There likewise exists an $r_h > \delta$ such that $|h(x) - L| < \varepsilon$ whenever $x > r_h.$ It follows that $-\varepsilon < h(x) - L < \varepsilon$ and therefore $L - \varepsilon < h(x) < L + \varepsilon$ whenever $x > r_h.$ Pick $r = \text{max}\{r_f, r_h\}.$ Then $|f(x) - L| < \varepsilon$ and $|h(x) - L| < \varepsilon$ whenever $x > r.$ Therefore $L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon,$ and so $|g(x) - L| < \varepsilon.$ Therefore $\lim\limits_{x \rightarrow \infty} g(x) = L.$
Show that $\infty$ and $-\infty$ are limit points of $\overline{\mathbb{R}}$ under the order topology.
Let $S$ be an open set in $\overline{\mathbb{R}}$ containing $\infty.$ It follows that $(a, \infty] \subseteq S$ for some $a \in \mathbb{R},$ as this is the only kind of open interval that contains $\infty.$ Note that $a + 1 \in (a, \infty],$ and $a + 1 \neq \infty.$ Therefore $\left(S - \{\infty\}\right) \cap \overline{\mathbb{R}} \neq \varnothing.$ Therefore $\infty$ is a limit point of $\overline{\mathbb{R}}.$
The proof for $-\infty$ is identical. Let $T$ be an open set in $\overline{\mathbb{R}}$ containing $-\infty.$ It follows that $[-\infty, b) \subseteq T$ for some $b \in \mathbb{R},$ as this is the only kind of open interval that contains $-\infty.$ Note that $b - 1 \in [-\infty, b),$ and $b - 1 \neq -\infty.$ Therefore $\left(T - \{-\infty\}\right) \cap \overline{\mathbb{R}} \neq \varnothing.$ Therefore $-\infty$ is a limit point of $\overline{\mathbb{R}}.$
Equivalent definition: Show that the order topology on $\overline{\mathbb{R}}$ and the definition of a limit in terms of open sets imply and are implied by the $\varepsilon-r$ definition of a limit at infinity.
Let $f$ be a function from $(\overline{\mathbb{R}}, \mathcal{T}_{<})$ to $(\mathbb{R}, \mathcal{T}_d)$ where $\mathcal{T}_{<}$ is the order topology on $\overline{\mathbb{R}}$ and $\mathcal{T}_{d}$ is the standard metric topology on $\mathbb{R}.$
Firs, assume $\lim\limits_{x \rightarrow \infty} f(x) = L$ for some $L \in \mathbb{R}.$ Then for every open set $S \subseteq \mathbb{R}$ containing $L,$ there exists an open set $T_0 \subseteq \overline{\mathbb{R}}$ containing $\infty$ such that $f(T_0 - \{\infty\}) \subseteq S.$ Because $S$ is open in $\mathcal{T}_{d},$ there exists a $\delta > 0$ such that $B_{\delta}(L) \subseteq S.$ The definition of a limit implies that there is another set $T_1 \subseteq \overline{\mathbb{R}}$ such that $f(T_1 - \{\infty\}) \subseteq B_{\delta}(L).$ Set $r = \sup (T_1^c)$ and set $T_2 = (r, \infty].$ It follows that $T_2$ is open and that $T_2 \subseteq T_1.$ Whenever $x > r$, it follows that $x \in T_2,$ and therefore that $f(T_2 - \{\infty\}) \subseteq f(T_1 - \{\infty\}) \subseteq B_{\delta}(L) \subseteq S$, and therefore that $d(f(x), L) < \delta.$
Conversely, note that $\infty$ is a limit point of $\overline{\mathbb{R}}$ under the order topology. Next, assume that for every $\varepsilon > 0,$ there exists an $r \in \mathbb{R}$ such that $d(f(x), L) < \varepsilon$ whenever $x > r.$ Set $S = B_{\varepsilon}(L)$ and $T = (r, \infty].$ It follows that $f(T - \{\infty\}) \subseteq B_{\varepsilon}(L).$ Since this is true for every $\varepsilon,$ and every open set $S$ containing $L$ must contain such an open ball containing $L,$ the result follows.
Consider the function $f : \overline{\mathbb{R}} \rightarrow \mathbb{R}$ where $\lim\limits_{x \rightarrow \infty} f(x) = L.$ Show that $\lim\limits_{x \rightarrow \infty} f|\overline{\mathbb{N}}(x) = L,$ where $f|\overline{\mathbb{N}}$ is the restriction of $f$ to the extended natural numbers $\overline{\mathbb{N}} = \mathbb{N} \cup \{\infty\}.$
First, we must show that $\infty$ is a limit point of $\overline{\mathbb{N}}$ under the order topology. Let $S$ be an open set containing $\infty.$ Then it contains some open interval $(a, \infty]$ for some $a \in \mathbb{R},$ as this is the only kind of interval that contains $\infty.$ By the Archimedean principle, there exists some $n \in \mathbb{N}$ such that $n > a.$ Since $n > a,$ it follows that $n \in (a, \infty].$ Therefore $\left(S - \{\infty\}\right) \cap \overline{\mathbb{N}} \neq \varnothing,$ and so $\infty$ is a limit point of $\overline{\mathbb{N}}.$
Next, note that because $\lim\limits_{x \rightarrow \infty} f(x) = L,$ for every $\varepsilon > 0$ there exists an $r \in \mathbb{R}$ such that $|f(x) - L| < \varepsilon$ whenever $x > r.$ By the Archimedean principle, there is an $N \in \mathbb{N}$ such that $N > r.$ Therefore $|f(x) - L| < \varepsilon$ whenever $x > N.$ Therefore $\left|f|\overline{\mathbb{N}}(x) - L\right| < \varepsilon$ whenever $x > n,$ and so $\lim\limits_{x \rightarrow \infty} f|\overline{N}(x) = L.$