Limits of Functions and Sequences

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Definition of Limit

To define the derivative and integral of a function, we need the concept of limit, just as we learned in high school. What makes the limit we now treat more advanced than back then is that we now define it.

«ins id=”def1”>Definition 1</ins> Any open interval $(c,d)$ containing a real number $a$ ($c<a«d$) is called a neighborhood of the point $a$.

A neighborhood of $a$ will later be defined more rigorously in analysis and topology, but for now this definition suffices. For convenience, the set obtained from a neighborhood of $a$ by removing $a$ itself is called a deleted neighborhood.

«ins id=”def2”>Definition 2</ins> Consider a function $f$ defined on some deleted neighborhood of $a$. We say that a real number $L$ is the limit of $f$ as $x \to a$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that

\[0 < \lvert x - a \rvert < \delta \implies \lvert f(x) - L \rvert < \epsilon\]

holds. In this case we write $\displaystyle\lim_{x \to a} f(x) = L$.

The intuitive explanation is as follows. In high school we described the limit by saying that $f(x)$ gets infinitely close to $L$, but this cannot serve as a rigorous mathematical definition because “closeness” is not a mathematical concept. It is analogous to saying that the collection of numbers near $L$ does not define a set mathematically.

Intuitively, the above $\epsilon$-$\delta$ definition resolves this by a process of agreement that works for everyone. That is, however close we demand $f(x)$ to be to $L$ (that is, whatever $\epsilon>0$ is given), we can meet the demand as long as we make $x$ sufficiently close (so that $\lvert f(x) - L\rvert < \epsilon$). Let us examine this in the following example.

«ins id=”ex3”>Example 3</ins> In this example we apply the definition of limit to linear and quadratic functions.

First consider the linear function $f(x)=2x-1$ and show that its limit as $x\rightarrow 3$ is $5$. Then

\[\lvert f(x)-L\rvert=\lvert (2x-1)-5\rvert=2\lvert x-3\rvert\]

so if we choose $\delta=\epsilon/2$,

\[0<\lvert x-3\rvert<\delta\implies 0<< 2\lvert x-3\rvert<<2\delta=\epsilon\]

follows.

As another example, consider the quadratic function $g(x)=x^2$ and show that its limit as $x\rightarrow 2$ is $4$. First, as above,

\[\lvert g(x)-L\rvert=\lvert x^2-4\rvert=\lvert x-2\rvert\lvert x+2\rvert\]

compute this. The key point in this calculation is that $\lvert x-2\rvert$ is small near $2$, but $\lvert x+2\rvert$ does not become too large. That is, if we consider a place where $\delta\leq 1$, then $\lvert x+2\rvert«5$ there, and a place with $\delta>1$ is not of interest to us in the first place. Therefore, setting $\delta=\min(1,\epsilon/5)$,

\[0 < \lvert x-2\rvert < \delta \implies \lvert x^2 - 4\rvert < 5\delta \leq \epsilon\]

holds.

As above, the essence of this definition is that we can determine $\epsilon$ by $\delta$, and continuing the intuition from above, the rule of finding a $\delta>0$ that satisfies this condition for any $\epsilon>0$ is exactly what we do when proving the limit of a function.

Properties of Limits

Now let us examine the properties of limits based on this. The first property is that if a limit exists, it is unique.

«ins id=”prop4”>Proposition 4 (Uniqueness of Limit)</ins> If $\displaystyle\lim_{x\to a} f(x) = L$ and $\displaystyle\lim_{x\to a} f(x) = L’$, then $L = L’$.

Proof

Assume for contradiction that $L \neq L’$. Then $\epsilon = \frac{1}{2}\lvert L - L’\rvert > 0$. By Definition 2 there exist $\delta_1, \delta_2 > 0$ corresponding to each such that the following two conditions

\[0 < \lvert x-a\rvert < \delta_1\implies \lvert f(x) - L\rvert < \epsilon,\qquad 0 < \lvert x-a\rvert < \delta_2\implies\lvert f(x) - L'\rvert < \epsilon\]

can be satisfied. Now set $\delta = \min(\delta_1, \delta_2)$; then for $x$ with $0 < \lvert x-a\rvert < \delta$, by the triangle inequality

\[\lvert L - L'\rvert \leq \lvert L - f(x)\rvert + \lvert f(x) - L'\rvert < \epsilon + \epsilon = \lvert L - L'\rvert\]

which is a contradiction. Therefore $L = L’$.

Meanwhile, Definition 2 can in principle only be used when a candidate $L$ for the limit is given and we want to show that the limit is actually $L$. That is, it is not a tool that tells us what the limit of a function is. For this the following proposition is useful.

«ins id=”prop5”>Proposition 5 (Limit Laws)</ins> Let $\displaystyle\lim_{x\to a} f(x) = L$ and $\displaystyle\lim_{x\to a} g(x) = M$. Then

1. $\displaystyle\lim_{x\to a} \bigl(f(x) + g(x)\bigr) = L + M$,
2. for any constant $c$, $\displaystyle\lim_{x\to a} cf(x) = cL$,
3. $\displaystyle\lim_{x\to a} f(x)g(x) = LM$,
4. if $M \neq 0$, $\displaystyle\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{L}{M}$

hold.

Proof

1. Given $\epsilon > 0$, from the definition of limit for $f$ and $g$ we can obtain $\delta_1, \delta_2 > 0$ corresponding to $\epsilon/2$. Then setting $\delta = \min(\delta_1,\delta_2)$, when $0 < \lvert x-a\rvert < \delta$,

   \[\lvert (f(x)+g(x)) - (L+M)\rvert \leq \lvert f(x)-L\rvert + \lvert g(x)-M\rvert < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\]

   holds.

2. If $c=0$, any $\delta$ works so it is trivial. If $c \neq 0$, take the $\delta$ corresponding to $\epsilon/\lvert c\rvert$.

3. We use the following inequality:

   \[\lvert f(x)g(x) - LM\rvert = \lvert f(x)(g(x)-M) + M(f(x)-L)\rvert \leq \lvert f(x)\rvert\,\lvert g(x)-M\rvert + \lvert M\rvert\,\lvert f(x)-L\rvert\]

   Intuitively, as $x$ goes to $a$, both $\lvert g(x)-M\rvert$ and $\lvert f(x)-L\rvert$ go to $0$, so if we can guarantee that the prefactors $\lvert f(x)\rvert, \lvert g(x)\rvert$ are finite, we can make this smaller than $\epsilon$ by a calculation similar to 1 above.

   The trick is to set $\epsilon=1$ and apply Definition 2 to $f$ and $g$ respectively. Then there exist suitable $\delta_1, \delta_2$ such that

   \[0<\lvert x-a\rvert<\delta_1\implies 0<\lvert f(x)-L\rvert<<1\implies \lvert f(x)\rvert<< \lvert L\rvert+1\]

   and similarly we can choose $\delta$ so that $\lvert g(x)\rvert <\lvert M\rvert+1$. Now choose $\delta$ so that all of these two conditions and the following two conditions

   \[\lvert g(x)-M\rvert < \frac{\epsilon}{2(\lvert L\rvert+1)},\qquad \lvert f(x)-L\rvert < \frac{\epsilon}{2(\lvert M\rvert+1)}\]

   hold simultaneously.

4. It suffices to show $1/g(x) \to 1/M$ and then apply 3. Since $M \neq 0$, we have $\lvert g(x)\rvert > \lvert M\rvert/2$ in a neighborhood of $a$, and

\[\left\lvert \frac{1}{g(x)} - \frac{1}{M}\right\rvert = \frac{\lvert g(x)-M\rvert}{\lvert g(x)\rvert\,\lvert M\rvert} < \frac{2}{\lvert M\rvert^2}\lvert g(x)-M\rvert\]

so it suffices to make $\lvert g(x)-M\rvert$ sufficiently small.

Then the following holds.

«ins id=”cor6”>Corollary 6 (Limits of Powers and Roots)</ins> If $\displaystyle\lim_{x\to a} f(x) = L$, then

1. for any positive integer $k$, $\displaystyle\lim_{x\to a} \bigl(f(x)\bigr)^k = L^k$,
2. if $L > 0$, for any positive integer $k$, $\displaystyle\lim_{x\to a} \sqrt[k]{f(x)} = \sqrt[k]{L}$

hold.

Proof

1. Apply 3 of Proposition 5 (Limit Laws) and run induction on $k$.

2. First, since $L > 0$, choose $\delta_1 > 0$ corresponding to $\epsilon_1 = L/2$; then

   \[0 < \lvert x-a\rvert < \delta_1\implies\lvert f(x)-L\rvert < L/2\]

   so $f(x) > L/2 > 0$. Meanwhile, for any positive real numbers $u,v$, considering the expansion

   \[u - v = \bigl(u^{1/k}-v^{1/k}\bigr)\bigl(u^{(k-1)/k}+u^{(k-2)/k}v^{1/k}+\cdots+v^{(k-1)/k}\bigr)\]

   each term of the second factor on the right is greater than $\min(u,v)^{(k-1)/k}$, so

   \[\bigl\lvert u^{1/k}-v^{1/k}\bigr\rvert \leq \frac{\lvert u-v\rvert}{k\,\min(u,v)^{(k-1)/k}}\]

   holds. Therefore if $0 < \lvert x-a\rvert < \delta_1$, then $0<L/2 < \min(f(x), L)$, so substituting $f(x)=u$, $L=v$,

   \[\bigl\lvert \sqrt[k]{f(x)}-\sqrt[k]{L}\bigr\rvert \leq \frac{\lvert f(x)-L\rvert}{k\,(L/2)^{(k-1)/k}}\]

   Now for any $\epsilon > 0$, choose $\delta_2 > 0$ corresponding to $k\,(L/2)^{(k-1)/k}\,\epsilon$ and set $\delta = \min(\delta_1,\delta_2)$; then when $0 < \lvert x-a\rvert < \delta$, the right-hand side becomes smaller than $\epsilon$.

By combining these laws, the limit of a polynomial can be computed by separating it into the limits of each term. The key is the following example.

«ins id=”ex7”>Example 7</ins> For any real number $a$,

\[\lim_{x\rightarrow a}x=a\]

This is obtained by taking $\delta=\epsilon$. Also, for any real number $c$,

\[\lim_{x\rightarrow a}c=c\]

This holds for any $\delta$ whatsoever.

Then for any polynomial

\[f(x)=c_nx^n+\cdots +c_1x+c_0\]

by applying the first law of Proposition 5 (Limit Laws) repeatedly we obtain

\[\lim_{x\rightarrow a}f(x)=\lim_{x \rightarrow a}c_nx^n+\cdots +\lim_{x\rightarrow a}c_1x+ \lim_{x\rightarrow a} c_0\]

and then applying the second law of Proposition 5 (Limit Laws) and Corollary 6 (Limits of Powers and Roots) we convert this to

\[\lim_{x\rightarrow a}f(x)=c_n(\lim_{x\rightarrow a}x)^n+\cdots c_1\lim_{x\rightarrow a}x+\lim_{x\rightarrow a}c_0\]

and then apply Example 7. In a similar way, one can show that the limit of any rational function formed as a ratio of polynomials is obtained by taking the limits of the numerator and denominator separately and forming their quotient, provided the limit of the denominator is not $0$.

Squeeze Theorem and Order of Limits

Limits that cannot be computed by limit laws alone are often handled by trapping them with inequalities. The key tool for this method is the squeeze theorem.

«ins id=”prop8”>Proposition 8 (Squeeze Theorem)</ins> If $g(x) \leq f(x) \leq h(x)$ in a deleted neighborhood of a real number $a$ and $\displaystyle\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L$, then $\displaystyle\lim_{x\to a} f(x) = L$.

Proof

For $\epsilon > 0$, collect $\delta_1, \delta_2$ obtained from the definition of limit for $g$ and $h$ and the radius $\delta_3$ of the neighborhood where $g \leq f \leq h$ holds, and set $\delta = \min(\delta_1,\delta_2,\delta_3)$. If $0 < \lvert x-a\rvert < \delta$, then $L - \epsilon < g(x) \leq f(x) \leq h(x) < L + \epsilon$, so $\lvert f(x) - L\rvert < \epsilon$.

Another basic fact about how inequalities and limits fit together is that limits preserve order.

«ins id=”prop9”>Proposition 9 (Preservation of Order under Limits)</ins> If $f(x) \leq g(x)$ in a neighborhood of $a$ (excluding $a$) and the two limits $L = \displaystyle\lim_{x\to a}f(x)$, $M = \displaystyle\lim_{x\to a}g(x)$ exist, then $L \leq M$.

Proof

Assume $L > M$ and set $\epsilon = \frac{1}{2}(L - M) > 0$. In a sufficiently small neighborhood, $f(x) > L - \epsilon = \frac{L+M}{2}$ and $g(x) < M + \epsilon = \frac{L+M}{2}$, so $f(x) > g(x)$, contradicting the assumption. Therefore $L \leq M$.

Note that a strict inequality $f < g$ does not give a strict inequality $L < M$. For instance $f(x) = 0 < x^2 = g(x)$ ($x \neq 0$), but both limits as $x \to 0$ are equal to $0$. That is, inequalities may weaken under limits.

The most famous application of the squeeze theorem is the following trigonometric limit, which is used decisively when treating trigonometric functions in differentiation.

«ins id=”ex10”>Example 10</ins> $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = 1$. For $0 < x < \pi/2$, comparing the areas of the unit circle gives the inequality

\[\frac{1}{2}\sin x \leq \frac{1}{2}x \leq \frac{1}{2}\tan x\]

That is, $\sin x \leq x \leq \tan x$.

Now dividing both sides of this inequality by $\sin x > 0$ and taking reciprocals, we know

\[\cos x \leq \frac{\sin x}{x} \leq 1\]

Our claim is that $\cos x \to 1$. For this, using the half-angle formula for trigonometric functions and $\sin\frac{x}{2} \leq \frac{x}{2}$ obtained above,

\[0 \leq \lvert 1 - \cos x\rvert = \left\lvert2\sin^2\frac{x}{2}\right\rvert \leq 2\left(\frac{x}{2}\right)^2 = \frac{x^2}{2}\]

so

\[-\frac{x^2}{2}\leq 1 - \cos x \leq \frac{x^2}{2}\]

and applying the squeeze theorem we know $\cos x \to 1$. Now applying the squeeze theorem again to the preceding inequality, we know that the limit of $\frac{\sin x}{x}$ is $1$.

The following example is also classical.

«ins id=”ex11”>Example 11</ins> $\displaystyle\lim_{x\to 0} x\sin\frac{1}{x} = 0$. Since $\bigl\lvert x\sin\frac{1}{x}\bigr\rvert \leq \lvert x\rvert$,

\[-\lvert x\rvert \leq x\sin\frac1x \leq \lvert x\rvert\]

and both ends go to $0$. On the other hand, $\sin\frac1x$ itself has no limit as $x \to 0$, because $x$ oscillates infinitely between $-1$ and $1$ as it approaches $0$. The factor $x$ pressing this oscillation to $0$ is the contribution of the squeeze theorem.

One-Sided Limits and Limits at Infinity

The limits so far were the case where $x$ approaches $a$ from both sides. By restricting the direction of approach to one side, or extending the definition to the case where $x$ or $f(x)$ becomes infinitely large, we can describe the shape of a function more finely.

«ins id=”def12”>Definition 12</ins> For a real number $a$ and a function $f$, suppose that for some $c > 0$, $f$ is defined on $(a, a+c)$. We say that a real number $L$ is the right limit of $f$ as $x \to a^+$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that

\[a < x < a+\delta \implies \lvert f(x) - L\rvert < \epsilon\]

holds. In this case we write $\displaystyle\lim_{x\to a^+} f(x) = L$. Similarly, when $f$ is defined on $(a-c, a)$, the left limit as $x \to a^-$ is defined by

\[a-\delta < x < a \implies \lvert f(x) - L\rvert < \epsilon\]

and we write $\displaystyle\lim_{x\to a^-} f(x) = L$.

The existence of the limit $\displaystyle\lim_{x\to a} f(x)$ is equivalent to both one-sided limits existing and being equal to each other. For example, $f(x) = \dfrac{\lvert x\rvert}{x}$ has right limit $1$ as $x \to 0^+$ and left limit $-1$ as $x \to 0^-$, so the limit as $x \to 0$ does not exist. A point where the two one-sided limits are finite but different is called a jump discontinuity of the function (§Continuous Functions).

«ins id=”def13”>Definition 13</ins> For a function $f$ defined on a deleted neighborhood of a real number $a$, $\displaystyle\lim_{x\to a} f(x) = \infty$ means that for every $M > 0$ there exists a $\delta > 0$ such that if $0 < \lvert x-a\rvert < \delta$ then $f(x) > M$. Similarly, $\displaystyle\lim_{x\to a} f(x) = -\infty$ means that for every $M > 0$ there exists a $\delta > 0$ such that if $0 < \lvert x-a\rvert < \delta$ then $f(x) < -M$.

For example, $\displaystyle\lim_{x\to 0}\frac{1}{x^2} = \infty$, and in this case the line $x = 0$ is called a vertical asymptote of the graph.

«ins id=”def14”>Definition 14</ins> For a function $f$ defined on $x$ greater than some real number $N_0$, $\displaystyle\lim_{x\to\infty} f(x) = L$ means that for every $\epsilon > 0$ there exists an $N > N_0$ such that if $x > N$ then $\lvert f(x) - L\rvert < \epsilon$. Similarly, for a function defined on $x$ less than some $N_0$, $\displaystyle\lim_{x\to-\infty} f(x) = L$ means that there exists an $N$ such that if $x < N$ then $\lvert f(x) - L\rvert < \epsilon$.

For example, $\displaystyle\lim_{x\to\infty}\frac{1}{x} = 0$, and for rational functions the highest-degree term dominates the behavior, so $\displaystyle\lim_{x\to\infty}\frac{2x^2 + 1}{3x^2 - x} = \frac{2}{3}$. When such a finite limit $L$ exists, the line $y = L$ becomes a horizontal asymptote of the graph.

«ins id=”ex15”>Example 15 (Resolving Indeterminate Forms)</ins> When direct substitution yields $\frac00$, resolve the indeterminate form by algebraic manipulation. By factoring,

\[\lim_{x\to 2}\frac{x^2 - 4}{x - 2} = \lim_{x\to 2}(x + 2) = 4,\]

by rationalizing the numerator,

\[\lim_{x\to 0}\frac{\sqrt{1+x} - 1}{x} = \lim_{x\to 0}\frac{1}{\sqrt{1+x}+1} = \frac12,\]

and combining with Example 10, $\displaystyle\lim_{x\to 0}\frac{\sin 3x}{x} = \lim_{x\to 0}3\cdot\frac{\sin 3x}{3x} = 3$.

Limits of Sequences

In the limit theory so far, the domain was a subset of the real numbers. Restricting the domain to the set of natural numbers $\mathbb{N}$ gives a sequence, and the limit as $n \to \infty$ becomes a discrete version of the limit of a function at infinity discussed above. Since the elements of $\mathbb{N}$ are discretely distributed unlike $\mathbb{R}$, the only meaningful situation when dealing with limits of sequences is $N\rightarrow\infty$. From this perspective we organize the limit of a sequence in $\epsilon$-$N$ language.

«ins id=”def16”>Definition 16</ins> For a real sequence $(a_n)_{n=1}^\infty$ and a real number $L$, if for every $\epsilon > 0$ there exists a natural number $N$ such that

\[n > N \implies \lvert a_n - L \rvert < \epsilon\]

holds, we call $L$ the limit of $a_n$ as $n \to \infty$ and write $\displaystyle\lim_{n\to\infty} a_n = L$.

The meaning of the definition is clear: no matter what error tolerance $\epsilon$ is given, we can find a point $N$ after which all terms fall within $\epsilon$ of $L$. The $\epsilon$-$N$ definition is exactly what makes precise the phrase from high school, “as $n$ grows, $a_n$ gets close to $L$”, and it is nothing more than applying the definition of the limit of a function at infinity to the domain $\mathbb{N}$.

«ins id=”ex17”>Example 17</ins> Let us verify that $\displaystyle\lim_{n\to\infty}\frac{1}{n}=0$. For any $\epsilon > 0$, choose a natural number $N > \frac{1}{\epsilon}$; then when $n > N$, $\frac{1}{n} < \frac{1}{N} < \epsilon$, so $\bigl\lvert \frac{1}{n}-0\bigr\rvert < \epsilon$.

Then the following is the analogue of Proposition 5 (Limit Laws), and its proof is almost identical.

«ins id=”prop18”>Proposition 18 (Limit Laws for Sequences)</ins> Let $\displaystyle\lim_{n\to\infty} a_n = L$ and $\displaystyle\lim_{n\to\infty} b_n = M$. Then

1. $\displaystyle\lim_{n\to\infty} \bigl(a_n + b_n\bigr) = L + M$,
2. for any constant $c$, $\displaystyle\lim_{n\to\infty} c\,a_n = cL$,
3. $\displaystyle\lim_{n\to\infty} a_n b_n = LM$,
4. if $M \neq 0$ and $b_n \neq 0$ for all $n$, $\displaystyle\lim_{n\to\infty} \frac{a_n}{b_n} = \frac{L}{M}$

hold.

With this we have equipped ourselves with the definition of limit and basic tools. In the next post §Continuous Functions, we use limits to define continuity of functions and discuss properties of continuous functions. Meanwhile, why the intuitive “gaplessness” of the real numbers that we treated informally in this post underpins limit theory—for instance, why a monotone bounded sequence converges—is rigorously founded in analysis in §Completeness of the Reals.