5.1: Limits

Let $A \subset \mathbb{R}$ and let $x$ be a limit point of $A .$ In the following, we will let $S(A, x)$ denote the set of all convergent sequences $\left\{x_{n}\right\}_{n \in I}$ such that $x_{n} \in A$ for all $n \in I, x_{n} \neq x$ for all $n \in I,$ and $\lim _{n \rightarrow \infty} x_{n}=x .$ We will let $S^{+}(A, x)$ be the subset of $S(A, x)$ of sequences $\left\{x_{n}\right\}_{n \in I}$ for which $x_{n}>x$ for all $n \in I$ and $S^{-}(A, x)$ be the subset of $S(A, x)$ of sequences $\left\{x_{n}\right\}_{n \in I}$ for which $x_{n}<x$ for all $n \in I .$

Definition

Let $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R}, L \in \mathbb{R},$ and suppose $a$ is a limit point of $D .$ We say the limit of $f$ as $x$ approaches $a$ is $L,$ denoted

$$\lim _{x \rightarrow a} f(x)=L ,$$

if for every sequence $\left\{x_{n}\right\}_{n \in I} \in S(D, a)$,

$$\lim _{n \rightarrow \infty} f\left(x_{n}\right)=L .$$

If $S^{+}(D, a) \neq \emptyset,$ we say the limit from the right of $f$ as $x$ approaches $a$ is $L,$ denoted

$$\lim _{x \rightarrow a+} f(x)=L ,$$

if for every sequence $\left\{x_{n}\right\}_{n \in I} \in S^{+}(D, a)$,

$$\lim _{n \rightarrow \infty} f\left(x_{n}\right)=L ,$$

and, if $S^{-}(D, a) \neq \emptyset,$ we say the limit from the left of $f$ as $x$ approaches $a$ is $L,$ denoted

$$\lim _{x \rightarrow a-} f(x)=L ,$$

if for every sequence $\left\{x_{n}\right\}_{n \in I} \in S^{-}(D, a)$,

$$\lim _{n \rightarrow \infty} f\left(x_{n}\right)=L .$$

We may also denote

$$\lim _{x \rightarrow a} f(x)=L$$

by writing

$$f(x) \rightarrow L \text { as } x \rightarrow a .$$

Similarly, we may denote

$$\lim _{x \rightarrow a^{+}} f(x)=L$$

by writing

$$f(x) \rightarrow L \text { as } x \downarrow a$$

and

$$\lim _{x \rightarrow a^{-}} f(x)=L$$

by writing

$$f(x) \rightarrow L \text { as } x \uparrow a$$

We also let

$$f(a+)=\lim _{x \rightarrow a^{+}} f(x)$$

and

$$f(a-)=\lim _{x \rightarrow a^{-}} f(x).$$

It should be clear that if $\lim _{x \rightarrow a} f(x)=L$ and $S^{+}(D, a) \neq \emptyset,$ then $f(a+)=L$. Similarly, if $\lim _{x \rightarrow a} f(x)=L$ and $S^{-}(D, a) \neq \emptyset,$ then $f(a-)=L$.

Proposition $\PageIndex{1}$

Suppose $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $a$ is a limit point of $D$. If $f(a-)=f(a+)=L,$ then $\lim _{x \rightarrow a} f(x)=L$.

Proof

Suppose $\left\{x_{n}\right\}_{n=m}^{\infty} \in S(D, a) .$ Let

$$J^{-}=\left\{n: n \in \mathbb{Z}, x_{n}<a\right\}$$

and

$$J^{+}=\left\{n: n \in \mathbb{Z}, x_{n}>a\right\}.$$

Suppose $J^{-}$ is empty or finite and let $k=m-1$ if $J^{-}=\emptyset$ and, otherwise, let $k$ be the largest integer in $J^{-} .$ Then $\left\{x_{n}\right\}_{n=k+1}^{\infty} \in S^{+}(D, a),$ and so

$$\lim _{n \rightarrow \infty} f\left(x_{n}\right)=f(a+)=L.$$

A similar argument shows that if $J^{+}$ is empty or finite, then

$$\lim _{n \rightarrow \infty} f\left(x_{n}\right)=f(a-)=L.$$

If neither $J^{-}$ nor $J^{+}$ is finite or empty, then $\left\{x_{n}\right\}_{n \in J}-$ and $\left\{x_{n}\right\}_{n \in J}+$ are subsequences of $\left\{x_{n}\right\}_{n=m}^{\infty}$ with $\left\{x_{n}\right\}_{n \in J^{-}} \in S^{-}(D, a)$ and $\left\{x_{n}\right\}_{n \in J+} \in S^{+}(D, a) .$ Hence, given any $\epsilon>0,$ we may find integers $N$ and $M$ such that

$$\left|f\left(x_{n}\right)-L\right|<\epsilon$$

whenever $n \in\left\{j: j \in J^{-}, j>N\right\}$ and

$$\left|f\left(x_{n}\right)-L\right|<\epsilon$$

whenever $n \in\left\{j: j \in J^{+}, j>M\right\} .$ Let $P$ be the larger of $N$ and $M .$ Since $J^{-} \cup J^{+}=\left\{j: j \in \mathbb{Z}^{+}, j \geq m\right\},$ it follows that

$$\left|f\left(x_{n}\right)-L\right|<\epsilon$$

whenever $n>P .$ Hence $\lim _{n \rightarrow \infty} f\left(x_{n}\right)=L,$ and so $\lim _{x \rightarrow a} f(x)=L$. $\quad$ Q.E.D.

Proposition $\PageIndex{2}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D,$ and $f: D \rightarrow \mathbb{R}$. If $\lim _{x \rightarrow a} f(x)=L$ and $\alpha \in \mathbb{R},$ then

$$\lim _{x \rightarrow a} \alpha f(x)=\alpha L.$$

Proof

Suppose $\left\{x_{n}\right\}_{n \in I} \in S(D, a) .$ Then

$$\lim _{n \rightarrow \infty} \alpha f\left(x_{n}\right)=\alpha \lim _{n \rightarrow \infty} f\left(x_{n}\right)=\alpha L.$$

Hence $\lim _{x \rightarrow a} \alpha f(x)=\alpha L$. $\quad$ Q.E.D.

Proposition $\PageIndex{3}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D, f: D \rightarrow \mathbb{R},$ and $g: D \rightarrow \mathbb{R} .$ If $\lim _{x \rightarrow a} f(x)=L$ and $\lim _{x \rightarrow a} g(x)=M,$ then

$$\lim _{x \rightarrow a}(f(x)+g(x))=L+M.$$

Proof

Suppose $\left\{x_{n}\right\}_{n \in I} \in S(D, a) .$ Then

$$\lim _{n \rightarrow \infty}\left(f\left(x_{n}\right)+g\left(x_{n}\right)\right)=\lim _{n \rightarrow \infty} f\left(x_{n}\right)+\lim _{n \rightarrow \infty} g\left(x_{n}\right)=L+M.$$

Hence $\lim _{x \rightarrow a}(f(x)+g(x))=L+M$. $\quad$ Q.E.D.

Proposition $\PageIndex{4}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D, f: D \rightarrow \mathbb{R},$ and $g: D \rightarrow \mathbb{R} .$ If $\lim _{x \rightarrow a} f(x)=L$ and $\lim _{x \rightarrow a} g(x)=M,$ then

$$\lim _{x \rightarrow a} f(x) g(x)=L M.$$

Exercise $\PageIndex{1}$

Prove the previous proposition.

Proposition $\PageIndex{5}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D, f: D \rightarrow \mathbb{R}$, $g: D \rightarrow \mathbb{R},$ and $g(x) \neq 0$ for all $x \in D .$ If $\lim _{x \rightarrow a} f(x)=L, \lim _{x \rightarrow a} g(x)=M,$ and $M \neq 0,$ then

$$\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{L}{M}.$$

Exercise $\PageIndex{2}$

Prove the previous proposition.

Proposition $\PageIndex{6}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D, f: D \rightarrow \mathbb{R},$ and $f(x) \geq 0$ for all $x \in D .$ If $\lim _{x \rightarrow a} f(x)=L,$ then

$$\lim _{x \rightarrow a} \sqrt{f(x)}=\sqrt{L}.$$

Exercise $\PageIndex{3}$

Prove the previous proposition.

Given $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $A \subset D,$ we let

$$f(A)=\{y: y=f(x) \text { for some } x \in A\}.$$

In particular, $f(D)$ denotes the range of $f$.

Proposition $\PageIndex{7}$

Suppose $D \subset \mathbb{R}, E \subset \mathbb{R}, a$ is a limit point of $D, g: D \rightarrow \mathbb{R}$, $f: E \rightarrow \mathbb{R},$ and $g(D) \subset E .$ Moreover, suppose $\lim _{x \rightarrow a} g(x)=b$ and, for some $\epsilon>0$, $g(x) \neq b$ for all $x \in(a-\epsilon, a+\epsilon) \cap D .$ If $\lim _{x \rightarrow b} f(x)=L,$ then

$$\lim _{x \rightarrow a} f \circ g(x)=L.$$

Proof

Suppose $\left\{x_{n}\right\}_{n \in I} \in S(D, a) .$ Then

$$\lim _{n \rightarrow \infty} g\left(x_{n}\right)=b.$$

Let $N \in \mathbb{Z}^{+}$ such that $\left|x_{n}-a\right|<\epsilon$ whenever $n>N .$ Then

$$\left\{g\left(x_{n}\right)\right\}_{n=N+1}^{\infty} \in S(E, b),$$

so

$$\lim _{n \rightarrow \infty} f\left(g\left(x_{n}\right)\right)=L.$$

Thus $\lim _{x \rightarrow a} f \circ g(x)=L$. $\quad$ Q.E.D.

Example $\PageIndex{1}$

Let

$$g(x)=\left\{\begin{array}{ll}{0,} & {\text { if } x \neq 0,} \\ {1,} & {\text { if } x=0.}\end{array}\right.$$

If $f(x)=g(x),$ then

$$f \circ g(x)=\left\{\begin{array}{ll}{1,} & {\text { if } x \neq 0,} \\ {0,} & {\text { if } x=0.}\end{array}\right.$$

Hence $\lim _{x \rightarrow 0} f \circ g(x)=1,$ although $\lim _{x \rightarrow 0} g(x)=0$ and $\lim _{x \rightarrow 0} f(x)=0$.

5.1.1 Limits of Polynomials and Rational Functions

Example $\PageIndex{2}$

If $c \in \mathbb{R}$ and $f: \mathbb{R} \rightarrow \mathbb{R}$ is given by $f(x)=c$ for all $x \in \mathbb{R}$, then clearly $\lim _{x \rightarrow a} f(x)=c$ for any $a \in \mathbb{R}$.

Example $\PageIndex{3}$

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x)=x$ for all $x \in \mathbb{R} .$ If, for any $a \in \mathbb{R},\left\{x_{n}\right\}_{n \in I} \in S(\mathbb{R}, a),$ then

$$\lim _{n \rightarrow \infty} f\left(x_{n}\right)=\lim _{n \rightarrow \infty} x_{n}=a.$$

Hence $\lim _{x \rightarrow a} x=a$.

Example $\PageIndex{4}$

Suppose $n \in \mathbb{Z}^{+}$ and $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x)=x^{n}$. Then

$$\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} x^{n}=\prod_{i=1}^{n} \lim _{x \rightarrow a} x=a^{n}.$$

Definition

If $n \in \mathbb{Z}, n \geq 0,$ and $b_{0}, b_{1}, \ldots, b_{n}$ are real numbers with $b_{n} \neq 0,$ then we call the function $p: \mathbb{R} \rightarrow \mathbb{R}$ defined by

$$p(x)=b_{n} x^{n}+b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0}$$

a polynomial of degree $n$.

Exercise $\PageIndex{4}$

Show that if $f$ is a polynomial and $a \in \mathbb{R},$ then $\lim _{x \rightarrow a} f(x)=f(a)$.

Definition

Suppose $p$ and $q$ are polynomials and

$$D=\{x: x \in \mathbb{R}, q(x) \neq 0\}.$$

We call the function $r: D \rightarrow \mathbb{R}$ defined by

$$r(x)=\frac{p(x)}{q(x)}$$

a rational function.

Exercise $\PageIndex{5}$

Show that if $f$ is a rational function and $a$ is in the domain of $f,$ then $\lim _{x \rightarrow a} f(x)=f(a)$.

Exercise $\PageIndex{6}$

Suppose $D \subset \mathbb{R}, a \in D$ is a limit point of $D,$ and $\lim _{x \rightarrow a} f(x)=L$. If $E=D \backslash\{a\}$ and $g: E \rightarrow \mathbb{R}$ is defined by $g(x)=f(x)$ for all $x \in E,$ show that $\lim _{x \rightarrow a} g(x)=L .$

Exercise $\PageIndex{7}$

Evaluate

$$\lim _{x \rightarrow 1} \frac{x^{5}-1}{x^{3}-1}.$$

Exercise $\PageIndex{8}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D, f: D \rightarrow \mathbb{R}, g: D \rightarrow \mathbb{R}$, $h: D \rightarrow \mathbb{R},$ and $f(x) \leq h(x) \leq g(x)$ for all $x \in D .$ If $\lim _{x \rightarrow a} f(x)=L$ and $\lim _{x \rightarrow a} g(x)=L,$ show that $\lim _{x \rightarrow a} h(x)=L .$ (This is the squeeze theorem for limits of functions.)

Note that the above results which have been stated for limits will hold as well for the appropriate one-sided limits, that is, limits from the right or from the left.

Exercise $\PageIndex{9}$

Suppose

$$f(x)=\left\{\begin{array}{ll}{x+1,} & {\text { if } x<0,} \\ {4,} & {\text { if } x=0,} \\ {x^{2},} & {\text { if } x>0.}\end{array}\right.$$

Evaluate $f(0), f(0-),$ and $f(0+) .$ Does $\lim _{x \rightarrow 0} f(x)$ exist?

5.1.2 Equivalent Definitions

Proposition $\PageIndex{8}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D,$ and $f: D \rightarrow \mathbb{R}$. Then $\lim _{x \rightarrow a} f(x)=L$ if and only if for every $\epsilon>0$ there exists a $\delta>0$ such that

$$|f(x)-L|<\epsilon \text { whenever } x \neq a \text { and } x \in(a-\delta, a+\delta) \cap D.$$

Proof

Suppose $\lim _{x \rightarrow a} f(x)=L .$ Suppose there exists an $\epsilon>0$ such that for every $\delta>0$ there exists $x \in(a-\delta, a+\delta) \cap D, x \neq a,$ for which $|f(x)-L| \geq \epsilon$. For $n=1,2,3, \ldots,$ choose

$$x_{n} \in\left(a-\frac{1}{n}, a+\frac{1}{n}\right) \cap D,$$

$x_{n} \neq a,$ such that $\left|f\left(x_{n}\right)-L\right| \geq \epsilon .$ Then $\left\{x_{n}\right\}_{n=1}^{\infty} \in S(D, a),$ but $\left\{f\left(x_{n}\right)\right\}_{n=1}^{\infty}$ does not converge to $L,$ contradicting the assumption that $\lim _{x \rightarrow a} f(x)=L$.

Now suppose that for every $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-L|<\epsilon$ whenever $x \neq a$ and $x \in(a-\delta, a+\delta) \cap D .$ Let $\left\{x_{n}\right\}_{n \in I} \in S(D, a) .$ Given $\epsilon>0$, let $\delta>0$ be such that $|f(x)-L|<\epsilon$ whenever $x \neq a$ and $x \in(a-\delta, a+\delta) \cap D .$ Choose $N \in \mathbb{Z}$ such that $\left|x_{n}-a\right|<\delta$ whenever $n>N .$ Then $\left|f\left(x_{n}\right)-L\right|<\epsilon$ for all $n>N .$ Hence $\lim _{n \rightarrow \infty} f\left(x_{n}\right)=L,$ and so $\lim _{x \rightarrow a} f(x)=L .$ $\quad$ Q.E.D.

The proofs of the next two propositions are analogous.

Proposition $\PageIndex{9}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D, f: D \rightarrow \mathbb{R},$ and $S^{-}(D, a) \neq \emptyset .$ Then $\lim _{x \rightarrow a^{-}} f(x)=L$ if and only if for every $\epsilon>0$ there exists a $\delta>0$ such that

$$|f(x)-L|<\epsilon \text { whenever } x \in(a-\delta, a) \cap D.$$

Proposition $\PageIndex{10}$

Suppose $D \subset \mathbb{R}, a$ is a limit point of $D, f: D \rightarrow \mathbb{R},$ and $S^{+}(D, a) \neq \emptyset .$ Then $\lim _{x \rightarrow a^{+}} f(x)=L$ if and only if for every $\epsilon>0$ there exists a $\delta>0$ such that

$$|f(x)-L|<\epsilon \text { whenever } x \in(a, a+\delta) \cap D.$$

5.1.3 Examples

Example $\PageIndex{5}$

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$$f(x)=\left\{\begin{array}{ll}{1,} & {\text { if } x \text { is rational, }} \\ {0,} & {\text { if } x \text { is irrational. }}\end{array}\right.$$

Let $a \in \mathbb{R} .$ Since every open interval contains both rational and irrational numbers, for any $\delta>0$ and any choice of $L \in \mathbb{R},$ there will exist $x \in(a-\delta, a+\delta),$ $x \neq a,$ such that

$$|f(x)-L| \geq \frac{1}{2}.$$

Hence $\lim _{x \rightarrow a} f(x)$ does not exist for any real number $a$.

Example $\PageIndex{6}$

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$$f(x)=\left\{\begin{array}{ll}{x,} & {\text { if } x \text { is rational, }} \\ {0,} & {\text { if } x \text { is irrational. }}\end{array}\right.$$

Then $\lim _{x \rightarrow 0} f(x)=0$ since, given $\epsilon>0,|f(x)|<\epsilon$ provided $|x|<\epsilon$.

Exercise $\PageIndex{9}$

Show that if $f$ is as given in the previous example and $a \neq 0$, then $\lim _{x \rightarrow a} f(x)$ does not exist.

Exercise $\PageIndex{11}$

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$$f(x)=\left\{\begin{array}{ll}{\frac{1}{q},} & {\text { if } x \text { is rational and } x=\frac{p}{q},} \\ {0,} & {\text { if } x \text { is irrational, }}\end{array}\right.$$

where $p$ and $q$ are taken to be relatively prime integers with $q>0,$ and we take $q=1$ when $x=0 .$ Show that, for any real number $a, \lim _{x \rightarrow a} f(x)=0$.

Example $\PageIndex{7}$

Define $\varphi:[0,1] \rightarrow[-1,1]$ by

$$\varphi(x)=\left\{\begin{array}{ll}{4 x,} & {\text { if } 0 \leq x \leq \frac{1}{4},} \\ {2-4 x,} & {\text { if } \frac{1}{4}<x<\frac{3}{4},} \\ {4 x-4,} & {\text { if } \frac{3}{4} \leq x \leq 1.}\end{array}\right.$$

Next define $s: \mathbb{R} \rightarrow \mathbb{R}$ by $s(x)=\varphi(x-\lfloor x\rfloor),$ where $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$ (that is, $\lfloor x\rfloor$ is the floor of $x$ ). The function $s$ is an example of a sawtooth function. See the graphs of $\varphi$ and $s$ in Figure $5.1 .1 .$ Note that for any $n \in \mathbb{Z}$,

$$s([n, n+1])=[-1,1].$$

Now let $D=\mathbb{R} \backslash\{0\}$ and define $\sigma: D \rightarrow \mathbb{R}$ by

$$\sigma(x)=s\left(\frac{1}{x}\right).$$

See the graph of $\sigma$ in Figure $5.1 .2 .$ Note that for any $n \in \mathbb{Z}^{+}$,

$$\sigma\left(\left[\frac{1}{n+1}, \frac{1}{n}\right]\right)=s([n, n+1])=[-1,1].$$

Hence for any $\epsilon>0, \sigma((0, \epsilon))=[-1,1],$ and so $\lim _{x \rightarrow 0^{+}} \sigma(x)$ does not exist. Similarly, neither $\lim _{x \rightarrow 0^{-}} \sigma(x)$ nor $\lim _{x \rightarrow 0} \sigma(x)$ exist.

Example $\PageIndex{8}$

Let $s$ be the sawtooth function of the previous example and let $D=\mathbb{R} \backslash\{0\} .$ Define $\psi: D \rightarrow \mathbb{R}$ by

$$\psi(x)=x s\left(\frac{1}{x}\right).$$

See Figure 5.1 .2 for the graph of $\psi .$ Then for all $x \in D$,

$$-|x| \leq \psi(x) \leq|x|,$$

and so $\lim _{x \rightarrow 0} \psi(x)=0$ by the squeeze theorem.

Definition

Let $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R} .$ We say $f$ is bounded if there exists a real number $B$ such that $|f(x)| \leq B$ for all $x \in D$.

Exercise $\PageIndex{12}$

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is bounded. Show that $\lim _{x \rightarrow 0} x f(x)=0$.