4.5 Limit laws

We now explore limit laws for functions, which tell us how limits behave under sums, products and quotients. These laws are completely analogous to the limit laws for sequences (see Theorem 2.38).

Theorem 4.49 (Limit laws).

Let $I\subseteq\mathbb{R}$ be an interval and $a\in I$ . Let $f,g\colon E\to\mathbb{R}$ where either $E=I$ or $E=I\setminus\{a\}$ .

Suppose that $\displaystyle\lim_{x\to a}f(x)$ and $\displaystyle\lim_{x\to a}g(x)$ exist. Then

1

The limit $\displaystyle\lim_{x\to a}(f+g)(x)$ exists and

$\displaystyle\lim_{x\to a}(f+g)(x)=\lim_{x\to a}f(x)+\lim_{x\to a}g(x);$
2

The limit $\displaystyle\lim_{x\to a}(f\cdot g)(x)$ exists and

$\displaystyle\lim_{x\to a}(f\cdot g)(x)=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x);$
3

If $g(x)\neq 0$ for all $x\in I$ and $\displaystyle\lim_{x\to a}g(x)\neq 0$ , then $\displaystyle\lim_{x\to a}(f/g)(x)$ exists and

$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\displaystyle\lim_{x\to a}f(x)}{% \displaystyle\lim_{x\to a}g(x)}.$

Exercise 4.50.

Prove Theorem 4.49. Your arguments should be very similar to those used to prove the corresponding limit laws for sequences in Theorem 2.38.

The limit laws can be converted into statements about continuous functions.

Corollary 4.51.

Let $I\subseteq\mathbb{R}$ be an interval and $f$ , $g\colon I\to\mathbb{R}$ be continuous.

1

The function $f+g\colon I\to\mathbb{R}$ is continuous;
2

The function $f\cdot g\colon I\to\mathbb{R}$ is continuous;
3

If $g(x)\neq 0$ for all $x\in I$ , then the function $(f/g)\colon I\to\mathbb{R}$ is continuous.

Exercise 4.52.

Using Lemma 4.42, show Theorem 4.49 implies Corollary 4.51.

Hint: for the product law in part 2, the result of Exercise 4.48 is useful.

Corollary 4.51 provides us with a new gizmo which often makes the task of determining whether a given function is continuous a lot simpler.

Example 4.53 (Polynomials).

Let $p\colon\mathbb{R}\to\mathbb{R}$ be a polynomial function, so that there exists some $d\in\mathbb{N}$ and coefficients $c_{0}$ , $c_{1}$ , $\dots$ , $c_{d}\in\mathbb{R}$ such that

p(x):=c_{d}x^{d}+c_{d-1}x^{d-1}+\cdots+c_{1}x+c_{0}\qquad\text{for all $x\in% \mathbb{R}$.}

It is straightforward to check that constant functions and the linear monomial $\iota\colon\mathbb{R}\to\mathbb{R}$ , $\iota(x):=x$ for all $x\in\mathbb{R}$ are continuous. Since $p$ can be written in terms of sums and products of these functions, it follows from Corollary 4.51 that $p$ is continuous.

Exercise 4.54.

Show that $\displaystyle f(x):=\frac{\sin^{2}x+10x^{5}+\cos x}{\sin^{4}x+5x^{2}+2}$ for all $x\in\mathbb{R}$ is continuous.

Another way to combine functions is to form compositions.

Theorem 4.55 (Composition law for limits).

Let $I$ , $J\subseteq\mathbb{R}$ be intervals, $a\in I$ , $b\in J$ and $f\colon I\to J\setminus\{b\}$ and $g\colon J\to\mathbb{R}$ be such that

\lim_{x\to a}f(x)=b\qquad\text{and}\qquad\lim_{y\to b}g(y)=\ell\qquad\text{% exist.}

Then $\displaystyle\lim_{x\to a}g\circ f(x)$ exists and

\lim_{x\to a}g\circ f(x)=\lim_{y\to b}g(y)=\ell.

Proof.

Let $\varepsilon>0$ be given. Since $\displaystyle\lim_{y\to b}g(y)=\ell$ , there exists some $\eta>0$ such that

(4.9) (4.9) if

y\in J

satisfies

0<|y-b|<\eta

, then

|g(y)-\ell|<\varepsilon

Since $\displaystyle\lim_{x\to a}f(x)=b$ , we can the apply the definition of the limit, using the number $\eta>0$ in place of $\varepsilon$ , to see that there exists some $\delta>0$ such that

(4.10) (4.10) if

x\in I

satisfies

0<|x-a|<\delta

, then

0<|f(x)-b|<\eta

where $|f(x)-b|>0$ comes from the fact that $b$ is not in the image of $f$ .

Let $x\in I$ satisfy $0<|x-a|<\delta$ . Then (4.10) implies that $0<|f(x)-b|<\eta$ . Consequently, we may take $y=f(x)$ in (4.9) to give $|g(f(x))-\ell|=|g\circ f(x)-\ell|<\varepsilon$ . Thus, by the $\varepsilon$ - $\delta$ definition of a limit, $\lim_{x\to a}g\circ f(x)=\ell$ , as required. ∎

The composition law can also be converted into statements about continuity.

Corollary 4.56.

Let $I$ , $J\subseteq\mathbb{R}$ be intervals and suppose $f\colon I\to J$ is continuous at a point $a\in I$ and $g\colon J\to\mathbb{R}$ is continuous at $f(a)\in J$ . Then the composition $g\circ f\colon I\to\mathbb{R}$ is continuous at $a$ .

Proof.

This follows from a slight modification of the proof of Theorem 4.55. ∎

Example 4.57.

Recall the function $f\colon\mathbb{R}\to\mathbb{R}$ from Example 4.36, given by

f(x):=\begin{cases}\sin(1/x)&\text{if $x\neq 0$,}\\ 0&\text{if $x=0$.}\end{cases}

We showed in Example 4.36 that $f$ is not continuous at $0$ . We can use Corollary 4.56 to show $f$ is continuous at $a$ for all $a\in\mathbb{R}\setminus\{0\}$ .

Proof.

We know from Exercise 4.28 that the reciprocal function $r(x):=1/x$ a continuous function on $(0,\infty)$ and from Lemma 4.33 that $\sin\colon\mathbb{R}\to\mathbb{R}$ is continuous. Hence, by Corollary 4.56, $f$ is continuous at $a$ for all $a\in(0,\infty)$ . By a similar argument, we can see that $f$ is also continuous for all $a\in(-\infty,0)$ . Thus the desired result follows. ∎

Exercise 4.58.

Show that $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=\cos\Big{(}\frac{1}{\sin x+2}\Big{)}$ for all $x\in\mathbb{R}$ is continuous.