10.2 Limits of Functions

Our definition of tangent to a curve is going to be based on the idea of limit. The word limit was used in mathematics long before the definition we will give was thought of. One finds statements like `` The limit of a regular polygon when the number of sides becomes infinite, is a circle." Early definitions of limit often involved the ideas of time or motion. Our definition will be purely mathematical.

10.5   Definition (Interior points and approachable points.) Let $S$ be a subset of $\mbox{{\bf R}}$. A point $x \in S$ is an interior point of $S$ if there is some positive number $\epsilon$ such that the interval $(x-\epsilon ,x+\epsilon )$ is a subset of $S$. A point $x\in R$ is an approachable point from $S$ if there is some positive number $\epsilon$ such that either $(x-\epsilon ,x)\subset S$ or $(x,x+\epsilon )\subset S$. (Without loss of generality we could replace `` $\epsilon$" in this definition by $${ {1\over N}}$$ for some $N\in\mbox{${\mbox{{\bf Z}}}^{+}$}$.)

Note that interior points of $S$ must belong to $S$. Approachable points of $S$ need not belong to $S$. Any interior point of $S$ is approachable from $S$.

10.6   Example. If $S$ is the open interval $(0,1)$ then every point of $S$ is an interior point of $S$. The points that are approachable from $S$ are the points in the closed interval $[0,1]$.

If $T$ is the closed interval $[0,1]$ then the points that are approachable from $T$ are exactly the points in $T$, and the interior points of $T$ are the points in the open interval $(0,1)$.

10.7   Definition (Limit of a function.) Let $f$ be a real valued function such that $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$. Let $a\in\mbox{{\bf R}}$ and let $L\in\mbox{{\bf R}}$. We say

$$\lim_{x\to a} f(x)=L \index{limit}$$ (10.8)

if

1)

$a$ is approachable from $\mbox{{\rm dom}}(f)$, and

2)

For every sequence $\{x_n\}$ in $\mbox{{\rm dom}}(f)\setminus\{a\}$

$$\{x_n\}\to a\hspace{1em}\mbox{$\Longrightarrow$}\hspace{1em}\{ f(x_n)\}\to L.$$

Note that the value of $f(a)$ (if it exists) has no influence on the meaning of $${\lim_{x\to a}}f(x)=L$$. Also the `` $x$" in (10.8) is a dummy variable, and can be replaced by any other symbol that has no assigned meaning.

10.9   Example. For all $a\in\mbox{{\bf R}}$ we have

$$\lim_{x\to a} x = a.$$

Also

$$\lim_{x\to a}\cos (x)=\cos (a)$$

and

$$\lim_{x\to a}\sin (x)=\sin (a),$$

by lemma 9.34. Also

$$\lim_{x\to 0} {{\sin x}\over x}=1$$

by theorem 9.37.

10.10   Example. $${\lim_{x\to 0}{x\over {\vert x\vert}}}$$ is not defined. Let $${x_n = {{(-1)^n}\over n}}$$. Then $\{x_n\}$ is a sequence in $\mbox{{\bf R}}\setminus\{0\}$, and $\{x_n\} \to 0$ and $${ {{x_n}\over {\vert x_n\vert}} = { {{(-1)^n}\over n}\over {( {1\over n} )} }=(-1)^n}$$. We know there is no number $L$ such that $\{ (-1)^n\}\to L$.

10.11   Example. Let $f$ be the spike function

$$f(x) = \cases{ 0 & if $x \in \mbox{{\bf R}}\setminus \{{1 \over 2}\}$ \cr 1 & if $x = {1\over 2}$. }$$

Then $${\lim_{x\to {1\over 2}} f(x) = 0}$$, since if $\{x_n\}$ is a generic sequence in $\mbox{{\rm dom}}(f) \setminus \{{1\over 2}\}$, then $\{f(x_n)\}$ is the constant sequence $\{0\}$.

10.12   Example. The limit

$$\lim_{x \to 0} (\sqrt{x} + \sqrt{-x})$$

does not exist. If $f(x) = \sqrt{x} + \sqrt{-x}$, then the domain of $f$ consists of the single point $0$, and $0$ is not approachable from $\mbox{{\rm dom}}(f)$. If we did not have condition 1) in our definition, we would have

$$\lim_{x \to 0} \sqrt{x} + \sqrt{-x} = 0 \mbox{ and } \lim_{x \to 0} \sqrt{x} + \sqrt{-x} = \pi,$$

which would not be a good thing. (If there are no sequences in $\mbox{{\rm dom}}(f)\setminus\{a\}$, then

$$\mbox{for every sequence $\{x_n\}$ in } \mbox{{\rm dom}}(f) \setminus \{a\} \mbox{[statement about $\{x_n\}$]}$$

is true, no matter what $[$statement about $\{x_n\}{]}$ is.)

In this course we will not care much about functions like $\sqrt{x} + \sqrt{-x}$.

10.13   Example. I will show that

$$\lim_{x\to a}\sqrt x=\sqrt a$$ (10.14)

for all $a\in\mbox{{\bf R}}_{\geq 0}$.

Case 1: Suppose $a \in \mbox{${\mbox{{\bf R}}}^{+}$}$. Let $\{x_n\}$ be a generic sequence in $\mbox{${\mbox{{\bf R}}}^{+}$}\setminus\{a\}$ such that $\{x_n\}\to a$. Then

$$0 \leq \vert\sqrt x_n-\sqrt a\vert=\Big\vert {{\sqrt x_n-\sqr... ...\over {\sqrt x_n+\sqrt a}}<{{\vert x_n-a\vert}\over {\sqrt a}}.$$

Now, since $\{x_n\}\to a$, we have

$$\lim \left\{ {\vert x_n-a\vert\over \sqrt a} \right\} = {1 \over \sqrt{a}} \lim\{\vert x_n - a\vert\} = 0,$$

so by the squeezing rule $\lim \{ \sqrt {x_n} - \sqrt{a} \} = 0$ which is equivalent to

$$\lim \{\sqrt{x_n} \} = \sqrt{a}.$$

This proves (10.14) when $a>0$.

Case 2: Suppose $a=0$. The domain of the square root function is $[0,\infty)$, and $0$ is approachable from this set.

Let $\{x_n\}$ be a sequence in $\mbox{${\mbox{{\bf R}}}^{+}$}$ such that $\{x_n\} \to 0$. To show that $\{\sqrt {x_n}\}\to 0$, I'll use the definition of limit. Let $\epsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$. Then $\epsilon^2\in\mbox{${\mbox{{\bf R}}}^{+}$}$, so by the definition of convergence, there is an $N(\epsilon^2)\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that for all $n\in\mbox{{\bf Z}}_{\geq N(\epsilon^2)}$ we have $(x_n=\vert x_n - 0\vert<\epsilon^2)$. Then for all $n\in\mbox{{\bf Z}}_{\geq N(\epsilon^2)}$ we have $(\sqrt{x_n}=\vert\sqrt{ x_n} - 0\vert<\epsilon)$ and hence $\{\sqrt x_n\}\to 0$. $\diamondsuit$

Many of our rules for limits of sequences have immediate corollaries as rules for limits of functions. For example, suppose $f,g$ are real valued functions with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$ and $\mbox{{\rm dom}}(g)\subset\mbox{{\bf R}}$. Suppose $${\lim_{x\to a}f(x)=L}$$ and $${\lim_{x\to a} g(x)=M}$$. Let $\{x_n\}$ be a generic sequence in $\Big(\mbox{{\rm dom}} (f)\cap\mbox{{\rm dom}}(g)\Big)\setminus\{a\}$ such that $\{x_n\}\to a$. Then $\{x_n\}$ is a sequence in $\mbox{{\rm dom}}(f)\setminus\{a\}$ and $\{x_n\}\to a$, so

$$\{f(x_n)\}\to L.$$

Also $\{x_n\}$ is a sequence in $\mbox{{\rm dom}}(g)\setminus\{a\}$ and $\{x_n\}\to a$ so

$$\{g(x_n)\}\to M.$$

By the sum and product rules for sequences, for any $c \in \mbox{{\bf R}}$

$$\begin{eqnarray*} \{(f\pm g)(x_n)\} &=& \{f(x_n)\pm g(x_n)\}\to L\pm M,\\ \{(fg)(x_n)\} &=& \{f(x_n)g(x_n)\}\to LM, \end{eqnarray*}$$

and

$$\{(cf)(x_n)\} = \{ c\cdot f(x_n)\}\to cL,$$

and thus we've proved that

$$\begin{eqnarray*} \lim_{x\to a}(f\pm g)(x) &=& L\pm M=\lim_{x\to a}f(x)\pm \lim_... ...{x\to a}(fg)(x) &=& LM = \lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x) \end{eqnarray*}$$

and

$$\lim_{x\to a}(cf)(x) = cL =c\lim_{x\to a}f(x).$$

Moreover if $a\in\mbox{{\rm dom}}\displaystyle { ({f\over g})}$ (so that $g(a)\neq 0$), and if $x_n\in\mbox{{\rm dom}}\displaystyle { ({f\over g})}$ for all $x_n$ (so that $g(x_n)\neq 0$ for all $n$), it follows from the quotient rule for sequences that

$$\Big\{ ({f\over g})(x_n)\Big\} =\Big\{ {{f(x_n)}\over {g(x_n)}}\Big\}\to {L\over M},$$

so that

$$\lim_{x\to a} {{f(x)}\over {g(x)}}={\displaystyle {\lim_{x\to... ...yle {\lim_{x\to a}g(x)} } \mbox{ if } \lim_{x\to a}g(x)\not= 0.$$

Actually all of the results just claimed are not quite true as stated. For we have

$$\lim_{x\to 0}\sqrt x=0$$

and

$$\lim_{x\to 0}\sqrt{-x}=0$$

but

$$\lim_{x\to 0}\sqrt x+\sqrt{-x} \mbox{ does not exist! }$$

The correct theorem is:

10.15   Theorem (Sum, product, quotient rules for limits.) Let $f,g$
be real valued functions with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$ and $\mbox{{\rm dom}}(g)\subset\mbox{{\bf R}}$, and let $c \in \mbox{{\bf R}}$. Suppose $${\lim_{x\to a}f(x)}$$, and $${\lim_{x\to a}g(x)}$$ both exist. Then if $a$ is approachable from $\mbox{{\rm dom}} (f)\cap \mbox{{\rm dom}}(g)$ we have

$$\begin{eqnarray*} \lim_{x\to a}(f\pm g)(x) &=& \lim_{x\to a}f(x)\pm\lim_{x\to a}... ...o a}g(x)\\ \lim_{x\to a}(cf)(x) &=& c\cdot\lim_{x\to a}f(x).\\ \end{eqnarray*}$$

If in addition $${\lim_{x\to a}g(x)\neq 0}$$ then

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

Proof: Most of the theorem follows from the remarks made above. We will assume the remaining parts.

10.16   Theorem (Inequality rule for limits of functions.) Let $f$ and $g$ be real functions with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$ and $\mbox{{\rm dom}}(g)\subset\mbox{{\bf R}}$. Suppose that

i

$${\lim_{x\to a}f(x)}$$ and $${\lim_{x\to a}g(x)}$$ both exist.

ii

$a$ is approachable from $\mbox{{\rm dom}} (f)\cap \mbox{{\rm dom}}(g)$.

iii

There is a positive number $\epsilon$ such that

$f(x) \leq g(x)$ for all $x$ in $\mbox{{\rm dom}}(f) \cap \mbox{{\rm dom}}(g) \cap(a-\epsilon,a+\epsilon)$.

Then $${\lim_{x\to a} f(x) \leq \lim_{x\to a} g(x)}$$.

Proof: Let $\{x_n\}$ be a sequence in $\Big(\mbox{{\rm dom}}(f)\cap \mbox{{\rm dom}}(g) \cap(a-\epsilon, a+\epsilon)\Big)\setminus \{a\}$ such that $\{x_n\}\to a$. Then $\{x_n\}$ is a sequence in $\mbox{{\rm dom}}(f)\setminus\{a\}$ that converges to $a$, so by the definition of limit of a function,

$$\lim \{f(x_n)\} = \lim_{x\to a} f(x).$$

Similiarly

$$\lim\{ g(x_n)\} = \lim_{x\to a} g(x).$$

Also $f(x_n) \leq g(x_n)$ for all $n$, so it follows from the inequality rule for limits of sequences that $\lim\{f(x_n)\} \leq \lim \{g(x_n)\}$, i.e. $${\lim_{x\to a} f(x) \leq \lim_{x\to a} g(x).\mbox{ $\diamondsuit$}}$$.

10.17   Theorem (Squeezing rule for limits of functions.) Let $f$, $g$ and $h$ be real functions with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$, $\mbox{{\rm dom}}(g)\subset\mbox{{\bf R}}$, and $\mbox{{\rm dom}}(h) \subset \mbox{{\bf R}}$. Suppose that

i

$${\lim_{x\to a}f(x)}$$ and $${\lim_{x\to a} h(x)}$$ both exist and are equal.

ii

$a$ is approachable from $\mbox{{\rm dom}}(f) \cap \mbox{{\rm dom}}(g) \cap \mbox{{\rm dom}}(h)$.

iii

There is a positive number $\epsilon$ such that $f(x) \leq g(x) \leq h(x)$ for all $x$ in
$\mbox{{\rm dom}}(f) \cap \mbox{{\rm dom}}(g)\cap \mbox{{\rm dom}}(h) \cap(a-\epsilon,a+\epsilon)$.

Then $${\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = \lim_{x\to a} h(x)}$$.

Proof: The proof is almost identical to the proof of theorem 10.16.