# 5.1: Limits

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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 $$\lim _{x \rightarrow 0} x f(x)=0$$.

This page titled 5.1: Limits is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.