3.2: Limit Theorems

Here we state and prove various theorems that facilitate the computation of general limits.

Definition $\PageIndex{1}$

Let $f, g: D \rightarrow \mathbb{R}$ and let $c$ be a constant. The functions $f+g$, $fg$, and $cf$ are respectively defined as functions from $D$ to $\mathbb{R}$ by

$(f+g)(x)=f(x)+g(x)$,

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

$(c f)(x)=c f(x)$

for $x \in D$. Let $\widetilde{D}=\{x \in D: g(x) \neq 0\}$. The function $\dfrac{f}{g}$ is defined as a function from $\widetilde{D}$ to $\mathbb{R}$ by

$$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}$$

for $x \in \widetilde{D}$.

Theorem $\PageIndex{1}$

Let $f, g: D \rightarrow \mathbb{R}$ and let $c \in \mathbb{R}$. Suppose $\bar{x}$ is a limit point of $D$ and

$$\lim _{x \rightarrow \bar{x}} f(x)=\ell, \lim _{x \rightarrow \bar{x}} g(x)=m.$$

Then

1. $\lim _{x \rightarrow \bar{x}}(f+g)(x)=\ell+m$,
2. $\lim _{x \rightarrow \bar{x}}(f g)(x)=\ell m$,
3. $\lim _{x \rightarrow \bar{x}}(c f)(x)=c \ell$,
4. $\lim _{x \rightarrow \bar{x}}\left(\dfrac{f}{g}\right)(x)=\dfrac{\ell}{m}$ provided that $m \neq 0$.

Proof

Let us first prove (a). Let $\left\{x_{n}\right\}$ be a sequence in $D$ that converges to $\bar{x}$ and $x_{n} \neq \bar{x}$ for every $n$. By Theorem 3.1.2,

$$\lim _{n \rightarrow \infty} f\left(x_{n}\right)=\ell \text { and } \lim _{n \rightarrow \infty} g\left(x_{n}\right)=m.$$

It follows from Theorem 2.2.1 that

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

Applying Theorem 3.1.2 again, we get $\lim _{x \rightarrow \bar{x}}(f+g)(x)=\ell+m$. The proofs of (b) and (c) are similar.

Let us now show that if $m \neq 0$, then $\bar{x}$ is a limit point of $\widetilde{D}$. Since $\bar{x}$ is a limit point of $D$, there is a sequence $\left\{u_{k}\right\}$ in $D$ converging to $\bar{x}$ such that $u_{k} \neq \bar{x}$ for every $k$. Since $m \neq 0$, it follows from an easy application of Theorem 3.1.6 that there exists $\delta > 0$ with

$g(x) \neq 0$ whenever $0<|x-\bar{x}|<\delta, x \in D.$

This implies

$x \in \widetilde{D}$ whenever $0<|x-\bar{x}|<\delta, x \in D.$

Then $u_{k} \in \widetilde{D}$ for all $k$ sufficiently large, and hence $\bar{x}$ is a limit point of $\widetilde{D}$. The rest of the proof of (d) can be completed easily following the proof of (a). $\square$

Example $\PageIndex{1}$

Consider $f: \mathbb{R} \backslash\{-7\} \rightarrow \mathbb{R}$ given by $f(x)=\dfrac{x^{2}+2 x-3}{x+7}$.

Solution

Then, combining all parts of Theorem 3.2.1, we get

$$\begin{align*} \lim _{x \rightarrow-2} f(x) &=\dfrac{\lim _{x \rightarrow-2}\left(x^{2}+2 x-3\right)}{\lim _{x \rightarrow-2}(x+7)}=\dfrac{\lim _{x \rightarrow-2} x^{2}+\lim _{x \rightarrow-2} 2 x-\lim _{x \rightarrow-2} 3}{\lim _{x \rightarrow-2} x+\lim _{x \rightarrow-2} 7} \\[4pt] &=\dfrac{\left(\lim _{x \rightarrow-2} x\right)^{2}+2 \lim _{x \rightarrow-2} x-\lim _{x \rightarrow-2} 3}{\lim _{x \rightarrow-2} x+\lim _{x \rightarrow-2} 7}=\dfrac{(-2)^{2}+2(-2)-3}{-2+7}=-\dfrac{3}{5} \end{align*}$$

Example $\PageIndex{2}$

We proceed in the same way to compute the following limit.

$$\lim _{x \rightarrow 0} \dfrac{1+(2 x-1)^{2}}{x^{2}+7}. \nonumber$$

Solution

$$\begin{align*} \lim _{x \rightarrow 0} \dfrac{1+(2 x-1)^{2}}{x^{2}+7} &=\dfrac{\lim _{x \rightarrow 0} 1+\lim _{x \rightarrow 0}(2 x-1)^{2}}{\lim _{x \rightarrow 0} x^{2}+\lim _{x \rightarrow 0} 7} \\[4pt] &=\dfrac{1+1}{0+7}=\dfrac{2}{7}. \end{align*}$$

Example $\PageIndex{3}$

We now consider

$$\lim _{x \rightarrow-1} \dfrac{x^{2}+6 x+5}{x+1}. \nonumber$$

Solution

Since the limit of the denominator $0$ we cannot apply directly part (d) of Theorem 3.2.1. Instead, we first simplify the expression keeping in mind that in the definition of limit we never need to evaluate the expression at the limit point itself. In this case, this means we may assume that $x \neq-1$. For any such $x$ we have

$$\dfrac{x^{2}+6 x+5}{x+1}=\dfrac{(x+1)(x+5)}{x+1}=x+5. \nonumber$$

Therefore,

$$\lim _{x \rightarrow-1} \dfrac{x^{2}+6 x+5}{x+1}=\lim _{x \rightarrow-1} x+5=4. \nonumber$$

Theorem $\PageIndex{2}$: Cauchy's criterion

Let $f: D \rightarrow \mathbb{R}$ and let $\bar{x}$ be a limit point of $D$. Then $f$ has a limit at $\bar{x}$ if and only if for any $\varepsilon>0$, there exists $\delta > 0$ such that

$|f(r)-f(s)|<\varepsilon$ whenever $r, s \in D$ and $0<|r-\bar{x}|<\delta, 0<|s-\bar{x}|<\delta .$

Proof

Suppose $\lim _{x \rightarrow \bar{x}} f(x)=\ell$. Given $\varepsilon>0$, there exists $\delta > 0$ such that

$|f(x)-\ell|<\dfrac{\varepsilon}{2}$ whenever $x \in D$ and $0<|x-\bar{x}|<\delta .$

Thus, for $r, s \in D$ with $0<|r-\bar{x}|<\delta$ and $0<|s-\bar{x}|<\delta$, we have

$$|f(r)-f(s)| \leq|f(r)-\ell|+|\ell-f(s)|<\varepsilon .$$

Let us prove the converse. Fix a sequence $\left\{u_{n}\right\}$ in $D$ such with $\lim _{n \rightarrow \infty} u_{n}=\bar{x}$ and $u_{n} \neq \bar{x}$ for every $n$. Given $\varepsilon>0$, there exists $\delta > 0$ such that

$|f(r)-f(s)|<\varepsilon$ whenever $r, s \in D$ and $0<|r-\bar{x}|<\delta$, $0<|s-\bar{x}|<\delta .$

Then there exists $N \in \mathbb{N}$ satisfying

$0<\left|u_{n}-\bar{x}\right|<\delta$ for all $n \geq N .$

This implies

$\left|f\left(u_{n}\right)-f\left(u_{m}\right)\right|<\varepsilon$ for all $m, n \geq N .$

Thus, $\left\{f\left(u_{n}\right)\right\}$ is a Cauchy sequence, and hence there exists $\ell \in \mathbb{R}$ such that

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

We now prove that $f$ has limit $\ell$ at $\bar{x}$ using Theorem 3.1.2. Let $\left\{x_{n}\right\}$ be a sequence in $D$ such that $\lim _{n \rightarrow \infty} x_{n}=\bar{x}$ and $x_{n} \neq \bar{x}$ for every $n$. By the previous argument, there exists $\ell^{\prime} \in \mathbb{R}$ such that

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

Fix any $\varepsilon > 0$ and let $\delta > 0$ satisfy (3.3). There exists $K \in \mathbb{N}$ such that

$\left|u_{n}-\bar{x}\right|<\delta$ and $\left|x_{n}-\bar{x}\right|<\delta$

for all $n \geq K$. Then $\left|f\left(u_{n}\right)-f\left(x_{n}\right)\right|<\varepsilon$ for such $n$. Letting $n \rightarrow \infty$, we have $\left|\ell-\ell^{\prime}\right| \leq \varepsilon$. Thus, $\ell=\ell^{\prime}$ since $\varepsilon$ is arbitrary. It now follows from Theorem 3.1.2 that $\lim _{x \rightarrow \bar{x}} f(x)=\ell$. $\square$

The rest of this section discussed some special limits and their properties.

Definition $\PageIndex{2}$: Left Limit Point and Right Limit Point

Let $a \in \mathbb{R}$ and $\delta > 0$. Define

$$B_{-}(a ; \delta)=(a-\delta, a) \text { and } B_{+}(a ; \delta)=(a, a+\delta) .$$

Given a subset $A$ of $\mathbb{R}$, we say that $a$ is a left limit point of $A$ if for any $\delta > 0$, $B_{-}(a ; \delta)$ contains an infinite number of elements of $A$. Similarly, $a$ is called a right limit point of $A$ if for any $\delta>0$, $B_{+}(a ; \delta)$ contains an infinite number of elements of $A$.

It follows from the definition that $a$ is a limit point of $A$ if and only if it is a left limit point of $A$ or it is a right limit point of $A$.

Definition $\PageIndex{3}$

(One-sided limits) Let $f: D \rightarrow \mathbb{R}$ and let $\bar{x}$ be a left limit point of $D$. We write

$$\lim _{x \rightarrow \bar{x}^{-}} f(x)=\ell$$

if for every $\varepsilon > 0$, there exists $\delta > 0$ such that

$$|f(x)-\ell|<\varepsilon \text { for all } x \in B_{-}(\bar{x} ; \delta) .$$

We say that $\ell$ is the left-handed limit of $f$ at $\bar{x}$. The right-hand limit of $f$ at $\bar{x}$ can bee defined in a similar way and is denoted $\lim _{x \rightarrow \bar{x}^{+}} f(x)$.

Example $\PageIndex{4}$

Consider the function $f: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}$ given by

$$f(x)=\dfrac{|x|}{x} .$$

Solution

Let $\bar{x} = 0$. Note first that $0$ is a limit point of the set $D=\mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}$. Since, for $x>0$, we have $f(x)=x / x=1$, we have

$$\lim _{x \rightarrow \bar{x}^{+}} f(x)=\lim _{x \rightarrow 0^{+}} 1=1 .$$

Similarly, for $x<0$ we have $f(x)=-x / x=-1$. Therefore,

$$\lim _{x \rightarrow \bar{x}^{-}} f(x)=\lim _{x \rightarrow 0^{-}}-1=-1 .$$

Example $\PageIndex{5}$

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by

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

Solution

We have

$$\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}} x^{2}-1=0 ,$$

and

$$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} x+4=3 ,$$

The following theorem follows directly from the definition of one-sided limits.

Theorem $\PageIndex{3}$

Let $f: D \rightarrow \mathbb{R}$ and let $\bar{x}$ be both a left limit point of $D$ and a right limit point of $D$. Then

$$\lim _{x \rightarrow \bar{x}} f(x)=\ell$$

if and only if

$$\lim _{x \rightarrow \bar{x}^{+}} f(x)=\ell \text { and } \lim _{x \rightarrow \bar{x}^{-}} f(x)=\ell .$$

Proof

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Example $\PageIndex{6}$

It follows from Example 3.2.4 that $\lim _{x \rightarrow 0} \dfrac{|x|}{x}$ does not exists, since the one-sided limits do not agree.

Solution

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Definition $\PageIndex{4}$

(monotonicity) Let $f:(a, b) \rightarrow \mathbb{R}$.

1. We say that $f$ is increasing on $(a, b)$ if, for all $x_{1}, x_{2} \in(a, b)$, $x_{1}<x_{2}$ implies $f\left(x_{1}\right) \leq f\left(x_{2}\right)$.
2. We say that $f$ is decreasing on $(a, b)$ if, for all $x_{1}, x_{2} \in(a, b)$, $x_{1}<x_{2}$ implies $f\left(x_{1}\right) \geq f\left(x_{2}\right)$.

If $f$ is increasing or decreasing on $(a, b)$, we say that $f$ is monotone on this interval. Strict monotonicity can be defined similarly using strict inequalities: $f\left(x_{1}\right)<f\left(x_{2}\right)$ in (1) and $f\left(x_{1}\right)>f\left(x_{2}\right)$ in (2).

Theorem $\PageIndex{4}$

Suppose $f:(a, b) \rightarrow \mathbb{R}$ is increasing on $(a, b)$ and $\bar{x} \in(a, b)$. Then $\lim _{x \rightarrow \bar{x}^{-}} f(x)$ and $\lim _{x \rightarrow \bar{x}^{+}} f(x)$ exist. Moreover,

$$\sup _{a<x<\bar{x}} f(x)=\lim _{x \rightarrow \bar{x}^{-}} f(x) \leq f(\bar{x}) \leq \lim _{x \rightarrow \bar{x}^{+}} f(x)=\inf _{\bar{x}<x<b} f(x) .$$

Proof

Since $f(x) \leq f(\bar{x})$ for all $x \in(a, \bar{x})$, the set

$$\{f(x): x \in(a, \bar{x})\}$$

is nonempty and bounded above. Thus,

$$\ell=\sup _{a<x<\bar{x}} f(x)$$

is a real number. We will show that $\lim _{x \rightarrow \bar{x}^{-}} f(x)=\ell$. For any $\varepsilon > 0$, by the definition of the least upper bound, there exists $a<x_{1}<\bar{x}$ such that

$$\ell-\varepsilon<f\left(x_{1}\right) .$$

Let $\delta=\bar{x}-x_{1}>0$. Using the increasing monotonicity, we get

$$\ell-\varepsilon<f\left(x_{1}\right) \leq f(x) \leq \ell<\ell+\varepsilon \text { for all } x \in\left(x_{1}, \bar{x}\right)=B_{-}(\bar{x} ; \delta) .$$

Therefore, $\lim _{x \rightarrow \bar{x}^{-}} f(x)=\ell$. The rest of the proof of the theorem is similar. $\square$

Let

$$B_{0}(\bar{x} ; \delta)=B_{-}(\bar{x} ; \delta) \cup B_{+}(\bar{x} ; \delta)=(\bar{x}-\delta, \bar{x}+\delta) \backslash\{\bar{x}\} .$$

Definition $\PageIndex{5}$

(infinite limits) Let $f: D \rightarrow \mathbb{R}$ and let $\bar{x}$ be a limit point of $D$. We write

$$\lim _{x \rightarrow \bar{x}} f(x)=\infty]$$

if for every $M \in \mathbb{R}$, there exists $\delta > 0$ such that

$$f(x)>M \text { for all } x \in B_{0}(\bar{x} ; \delta) \cap D .$$

Similarly, we write

$$\lim _{x \rightarrow \bar{x}} f(x)=-\infty$$

if for every $M \in \mathbb{R}$, there exists $\delta > 0$ such that

$$f(x)<M \text { for all } x \in B_{0}(\bar{x} ; \delta) \cap D .$$

Infinite limits of functions have similar properties to those of sequences from Chapter 2 (see Definition 2.3.2 and Theorem 2.3.6).

Example $\PageIndex{7}$

We show that $\lim _{x \rightarrow 1} \dfrac{1}{(x-1)^{2}}=\infty$ directly from Definition 3.2.5.

Solution

Let $M \in \mathbb{R}$. We want to find $\delta > 0$ that will guarantee $\dfrac{1}{(x-1)^{2}}>M$ whenever $0<|x-1|<\delta$. As in the case of finite limits, we work backwards from $\dfrac{1}{(x-1)^{2}}>M$ to an inequality for $|x-1|$. To simplify calculations, note that $|M|+1>M$. Next note that $\dfrac{1}{(x-1)^{2}}>|M|+1$, is equivalent to $\sqrt{\dfrac{1}{|M|+1}>|x-1|}$.

Now, choose $\delta$ such that $0<\delta<\sqrt{\dfrac{1}{|M|+1}}$. Then, if $0<|x-1|<\delta$ we have

$$\dfrac{1}{(x-1)^{2}}>\dfrac{1}{\delta^{2}}>\dfrac{1}{\dfrac{1}{|M|+1}}=|M|+1>M ,$$

as desired.

Definition $\PageIndex{6}$

(limits at infinity) Let $f: D \rightarrow \mathbb{R}$, where $D$ is not bounded above. We write

$$\lim _{x \rightarrow \infty} f(x)=\ell$$

if for every $\varepsilon > 0$, there exists $c \in \mathbb{R}$ such that

$$|f(x)-\ell|<\varepsilon \text { for all } x>c, x \in D .$$

Let $f: D \rightarrow \mathbb{R}$, where $D$ is not bounded below. We write

$$\lim _{x \rightarrow-\infty} f(x)=\ell$$

if for every $\varepsilon > 0$, there exists $c \in \mathbb{R}$ such that

$$|f(x)-\ell|<\varepsilon \text { for all } x<c, x \in D .$$

We can also define

$$\lim _{x \rightarrow \infty} f(x)=\pm \infty \text { and } \lim _{x \rightarrow-\infty} f(x)=\pm \infty$$

in a similar way.

Example $\PageIndex{8}$

We prove form the definition that

$$\lim _{x \rightarrow-\infty} \dfrac{3 x^{2}+x}{2 x^{2}+1}=\dfrac{3}{2} .$$

Solution

The approach is similar to that for sequences, with the difference that $x$ need not be an integer.

Let $\varepsilon > 0$. We want to indentify $c$ so that

$$\left|\dfrac{3 x^{2}+x}{2 x^{2}+1}-\dfrac{3}{2}\right|<\varepsilon ,$$

for all $x<c$.

Now, $\text { v, }\left|\dfrac{3 x^{2}+x}{2 x^{2}+1}-\dfrac{3}{2}\right|=\dfrac{|2 x-3|}{2\left(2 x^{2}+1\right)}$.

Therefore, simplifying, 3.5 is equivalent to

$$\dfrac{1}{\varepsilon}<\dfrac{2\left(2 x^{2}+1\right)}{|2 x-3|} .$$

We first restrict $x$ to be less than $0$, so $|2 x-3|>3$. Then, since $\dfrac{4 x^{2}}{3}<\dfrac{2\left(2 x^{2}+1\right)}{|2 x-3|}$, 3.6 will be guaranteed if $1 / \varepsilon<4 x^{2} / 3$ or equivalently $\sqrt{3 /(4 \varepsilon)}<|x|$. We set $c<\min \{0,-\sqrt{3 /(4 \varepsilon)}\}$. Then, if $x<c$, we have $\sqrt{3 /(4 \varepsilon)}<-x=|x|$. Thus, $1 / \varepsilon<\dfrac{2\left(2 x^{2}+1\right)}{|2 x-3|}$ and, hence,

$$\left|\dfrac{3 x^{2}+x}{2 x^{2}+1}-\dfrac{3}{2}\right|=\dfrac{|2 x-3|}{2\left(2 x^{2}+1\right)}<\varepsilon .$$

Exercise $\PageIndex{1}$

Find the following limits:

1. $\lim _{x \rightarrow 2} \dfrac{3 x^{2}-2 x+5}{x-3}$,
2. $\lim _{x \rightarrow-3} \dfrac{x^{2}+4 x+3}{x^{2}-9}$

Answer

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Exercise $\PageIndex{2}$

Let $f: D \rightarrow \mathbb{R}$ and let $\bar{x}$ is a limit point of $D$. Prove that if $\lim _{x \rightarrow \bar{x}} f(x)$ exists, then

$$\lim _{x \rightarrow \bar{x}}[f(x)]^{n}=\left[\lim _{x \rightarrow \bar{x}} f(x)\right]^{n}, \text { for any } n \in \mathbb{N} .$$

Answer

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Exercise $\PageIndex{3}$

Find the following limits:

1. $\lim _{x \rightarrow 1} \dfrac{\sqrt{x}-1}{x^{2}-1}$,
2. $\lim _{x \rightarrow 1} \dfrac{x^{m}-1}{x^{n}-1}, \text { where } m, n \in \mathbb{N}$,
3. $\lim _{x \rightarrow 1} \dfrac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}, \text { where } m, n \in \mathbb{N}, m, n \geq 2$,
4. $\lim _{x \rightarrow 1} \dfrac{\sqrt{x}-\sqrt[3]{x}}{x-1}$.

Answer

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Exercise $\PageIndex{4}$

Find the following limits:

1. $\lim _{x \rightarrow \infty}\left(\sqrt[3]{x^{3}+3 x^{2}}-\sqrt{x^{2}+1}\right)$.
2. $\lim _{x \rightarrow-\infty}\left(\sqrt[3]{x^{3}+3 x^{2}}-\sqrt{x^{2}+1}\right)$.

Answer

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Exercise $\PageIndex{5}$

Let $f: D \rightarrow \mathbb{R}$ and let $\bar{x}$ be a limit point of $D$. Suppose that

$$|f(x)-f(y)| \leq k|x-y| \text { for all } x, y \in D \backslash\{\bar{x}\} ,$$

where $k \geq 0$ is a constant. Prove that $\lim _{x \rightarrow \bar{x}} f(x)$ exists.

Answer

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Exercise $\PageIndex{6}$

Determine the one-sided limits $\lim _{x \rightarrow 3^{+}}[x]$ and $\lim _{x \rightarrow 3^{-}}[x]$, where $[x]$ denotes the greatest integer that is less than or equal to $x$.

Answer

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Exercise $\PageIndex{7}$

Find each of the following limits if they exist:

1. $\lim _{x \rightarrow 1^{+}} \dfrac{x+1}{x-1}$.
2. $\lim _{x \rightarrow 0^{+}}\left|x^{3} \sin (1 / x)\right|$.
3. $\lim _{x \rightarrow 1}(x-[x])$.

Answer

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Exercise $\PageIndex{8}$

For $a \in \mathbb{R}$, let $f$ be the function given by

$$f(x)=\left\{\begin{array}{ll} x^{2}, & \text { if } x>1 \text{;} \\ a x-1, & \text { if } x \leq 1 \text{.} \end{array}\right .$$

Find the value of $a$ such that $\lim _{x \rightarrow 1} f(x)$ exists.

Answer

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Exercise $\PageIndex{9}$

Determine all values of $\bar{x}$ such that the limit $\lim _{x \rightarrow \bar{x}}(1+x-[x])$ exists.

Answer

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Exercise $\PageIndex{10}$

Let $a, b \in \mathbb{R}$ and suppose $f:(a, b) \rightarrow \mathbb{R}$ is increasing. Prove the following.

1. If $f$ is bounded above, then $\lim _{x \rightarrow b^{-}} f(x)$ exists and is a real number.
2. If $f$ is not bounded above, then $\lim _{x \rightarrow b^{-}} f(x)=\infty$.

State and prove analogous reasults in case $f$ is bounded below and in case that the domain of $f$ is one of $(-\infty, b)$, $(a, \infty)$, or $(-\infty, \infty)$.

Answer

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