Limit of functions of a real variable

Summary

This lesson turns into a precise definition the idea that the values of a function approach a number when the variable approaches a point. It shows why the intuitive idea is not enough, introduces accumulation points, which indicate where it makes sense to ask about a limit, and states the epsilon-delta definition, which reads as a challenge and its response. It writes the definition and its negation as quantified statements, and proves that the limit, if it exists, is unique, that it depends only on the values near the point, and that it forces the function to remain bounded and to keep its sign near that point. The problems construct the delta for polynomial functions, roots and quotients, prove that certain limits do not exist, and find the largest delta that responds to a given epsilon.

Learning objectives

Explain the idea of the limit of a function at a point and distinguish it from the value of the function at that point.
State the epsilon-delta definition of limit, justify the requirement that the point be an accumulation point of the domain, and write the definition and its negation with quantifiers.
Prove from the definition the value of the limit of simple algebraic functions, constructing a delta for each epsilon.
Prove the uniqueness of the limit and its local properties, and apply the negation of the definition to prove that a limit does not exist.

The idea of limit

Algebra and geometry study the value a function takes at each point; calculus also studies what happens to its values when the variable approaches a point, even if the function is not defined at that point. The need is an old one. The slope of the line through two points of a curve is computed with a difference quotient; the slope of the tangent line ought to be the value of that quotient when the two points merge, but then the quotient is $\frac{0}{0}$ , which is not a number. Likewise, the average velocity of a moving body is the quotient of the distance traveled by the time elapsed, and the velocity at an instant is what that quotient comes to be when the elapsed time shrinks to nothing. In both cases one asks about the values of a function near a point where it is not defined.

A simple example shows the situation. The following function is not defined at $x = 1$ , because its denominator vanishes there; nevertheless, its values at points near $1$ are close to $2$ :

f(x) = \frac{x^2 - 1}{x - 1}, \qquad f(0.9) = 1.9, \quad f(0.99) = 1.99, \quad f(1.01) = 2.01, \quad f(1.1) = 2.1

The reason is that, for $x \neq 1$ , the numerator factors as $(x - 1)(x + 1)$ and the quotient equals $x + 1$ ; the graph of $f$ is the line $y = x + 1$ with a hole at the point $(1, 2)$ . We then say that the limit of $f$ as $x$ tends to $1$ is $2$ , and we write:

\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2

However, the phrase “the values approach $2$ ” does not work as a definition, for two reasons. The first is that approaching is not enough: when $x$ goes from $0.9$ to $0.99$ , the value of $f$ also approaches $3$ , because $1.99$ is closer to $3$ than $1.9$ is, and yet $3$ is not the limit. What sets $2$ apart is that the values of $f$ come as close as we like to it, and not merely closer than before. The second reason is that a table examines finitely many points, and a function may behave differently between them: the function $\sin\left(\frac{\pi}{x}\right)$ takes the value $0$ at $x = 0.1$ , at $x = 0.01$ and at every point of the form $\frac{1}{n}$ , and yet it has no limit at $0$ , because between those points it oscillates between $-1$ and $1$ ; proposed problem 3 proves this for $\sin\left(\frac{1}{x}\right)$ , with the same argument. If this is so, the definition must speak of all nearby points at once and fix with numbers what “as close as we like” means. The tools for this are the absolute value, which measures distances, and the triangle inequality, studied in the lesson on the real line.

Accumulation points

Before defining the limit, we must specify at which points it makes sense to ask about it. The point $a$ need not belong to the domain of the function, as the example above shows; but there must be points of the domain as close to $a$ as we like, since otherwise there would be no values of the function to observe.

DefinitionAccumulation point

Let $A$ be a set of real numbers. A real number $a$ is an accumulation point of $A$ if every open interval centered at $a$ contains some element of $A$ other than $a$ :

(\forall \delta > 0)(\exists x \in A)(0 < \abs{x - a} < \delta)

An element of $A$ that is not an accumulation point of $A$ is called an isolated point of $A$ .

For example, $1$ is an accumulation point of $\mathbb{R} \setminus \{1\}$ , the domain of the function in the example, although it does not belong to it. Every point of the closed interval $\icc{0}{1}$ is an accumulation point of the open interval $\ioo{0}{1}$ , the endpoints included. The number $0$ is an accumulation point of the set of the numbers $\frac{1}{n}$ with $n$ a natural number, because, by the Archimedean property, for every $\delta > 0$ there is a natural number $n$ with $\frac{1}{n} < \delta$ . On the other hand, the set $\mathbb{N}$ has no accumulation points: the interval with center $a$ and radius $\frac{1}{2}$ contains at most one natural number $m$ ; if $m = a$ , it contains no natural number other than $a$ , and if $m \neq a$ , the interval of radius $\abs{m - a}$ no longer contains any. The condition $0 < \abs{x - a}$ is essential; without it, every element of $A$ would be an accumulation point of $A$ , because each interval centered at $a$ contains $a$ itself.

The definition of limit

With these notions the definition can be given. The phrase “ $f(x)$ is as close to $L$ as we like” is translated as follows: for each distance $\varepsilon > 0$ , however small, the values of $f$ lie at a distance less than $\varepsilon$ from $L$ whenever $x$ is sufficiently close to $a$ , that is, at less than a certain distance $\delta > 0$ , without being equal to $a$ .

DefinitionLimit of a function at a point

Let $f$ be a real function defined on a set $A$ of real numbers, $a$ an accumulation point of $A$ and $L$ a real number. We say that $L$ is the limit of $f$ as $x$ tends to $a$ , and we write $\lim_{x \to a} f(x) = L$ , if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that every $x$ in $A$ that satisfies $0 < \abs{x - a} < \delta$ also satisfies $\abs{f(x) - L} < \varepsilon$ :

\lim_{x \to a} f(x) = L \leftrightarrow (\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)

Geometrically, the condition $\abs{f(x) - L} < \varepsilon$ says that the point $(x, f(x))$ of the graph lies in the horizontal band between the lines $y = L - \varepsilon$ and $y = L + \varepsilon$ ; the condition $0 < \abs{x - a} < \delta$ says that $x$ lies in the interval with center $a$ and radius $\delta$ , without being its center. The definition can be read as a game between two players: the first proposes an $\varepsilon$ , the radius of a band around $L$ , as narrow as they wish; the second must respond with a $\delta$ such that the portion of the graph lying over the punctured interval of radius $\delta$ stays inside the band. The limit is $L$ if the second player can respond to every challenge.

ExampleThe challenge and the response

The function $g$ equals $x^2$ at every $x \neq 1$ , and at $x = 1$ it is assigned, on purpose, a value that does not fit with the others: $g(1) = \frac{5}{2}$ . We see how to respond to two challenges for the limit $L = 1$ at the point $a = 1$ .

Proof

$(\forall x \neq 1)\left(g(x) = x^2\right), \qquad g(1) = \frac{5}{2}$
The function is defined on the whole line. Near $1$ , its values are close to $1$ , except the value at $x = 1$ , which stands apart.
$\abs{g(x) - 1} < \frac{4}{5} \leftrightarrow \frac{1}{5} < g(x) < \frac{9}{5}$
First challenge: $\varepsilon = \frac{4}{5}$ . The values of $g$ must stay in the band that goes from $\frac{1}{5}$ to $\frac{9}{5}$ .
$\delta = \min\left(1 - \sqrt{\frac{1}{5}}, \sqrt{\frac{9}{5}} - 1\right) = \resaltar{\sqrt{\frac{9}{5}} - 1}$
For the positive $x$ other than $1$ , the condition is equivalent to $x$ lying between $\sqrt{\frac{1}{5}}$ and $\sqrt{\frac{9}{5}}$ . The interval is not symmetric about $1$ : the right endpoint is closer. We respond with the smaller of the two distances.
$0 < \abs{x - 1} < \delta \rightarrow \abs{g(x) - 1} < \frac{4}{5}$
Every $x$ whose distance from $1$ is less than $\delta$ , without being $1$ , satisfies the condition: the portion of the graph over the punctured interval stays in the band. The point $(1, g(1))$ lies outside the band, but that does not matter, because the definition excludes $x = 1$ .
$\varepsilon = \frac{2}{5} \rightarrow \delta = \resaltar{\sqrt{\frac{7}{5}} - 1}$
A more demanding challenge, $\varepsilon = \frac{2}{5}$ , requires a smaller response. In general, for $0 < \varepsilon < 1$ the response $\delta = \sqrt{1 + \varepsilon} - 1$ works, and it becomes as small as we like when $\varepsilon$ does: $\delta$ depends on $\varepsilon$ .

The example shows three features of the definition worth keeping in mind. First, $\delta$ depends on $\varepsilon$ , and in general also on the point $a$ ; if a $\delta$ responds to a challenge, any smaller positive number also responds, so it suffices to find one, and the largest is not needed. Second, the value of the function at $a$ , if it exists, plays no role: the condition $0 < \abs{x - a}$ excludes it, and that is why the limit of $g$ at $1$ is $1$ even though $g(1)$ equals $\frac{5}{2}$ . Third, only the points $x$ of the domain count; the others are not examined.

ExampleA linear function

We prove that $\lim_{x \to 2} (3x - 1) = 5$ . Given $\varepsilon > 0$ , we must ensure that $\abs{(3x - 1) - 5} < \varepsilon$ . The expression simplifies and shows its relation to $\abs{x - 2}$ :

\abs{(3x - 1) - 5} = \abs{3x - 6} = 3\abs{x - 2}

It therefore suffices that $3\abs{x - 2} < \varepsilon$ , that is, that $\abs{x - 2} < \frac{\varepsilon}{3}$ . We respond with $\delta = \frac{\varepsilon}{3}$ : if $0 < \abs{x - 2} < \frac{\varepsilon}{3}$ , then $\abs{(3x - 1) - 5} < 3 \cdot \frac{\varepsilon}{3} = \varepsilon$ . Likewise, for a constant function $c$ any $\delta$ responds, because $\abs{c - c} = 0 < \varepsilon$ ; and for the identity function $\delta = \varepsilon$ responds, because the required condition, $\abs{x - a} < \varepsilon$ , is the same as the one assumed. Therefore:

\lim_{x \to a} c = c, \qquad \lim_{x \to a} x = a

The method of the example is the general one: we write $\abs{f(x) - L}$ so that the factor $\abs{x - a}$ appears, and we bound the rest. When the rest is not a constant, we bound it by restricting $\delta$ in advance to a fixed value, for example $\delta \leq 1$ ; this is the technique of the following example.

ExampleA square

We prove that $\lim_{x \to 3} x^2 = 9$ . Let $\varepsilon > 0$ .

Proof

$\abs{x^2 - 9} = \abs{x - 3} \cdot \dato{f}{\abs{x + 3}}$
We factor the difference of squares; the factor $\abs{x - 3}$ appears, which will be made small, multiplied by a factor that depends on $x$ .
$\abs{x - 3} < 1 \rightarrow 5 < x + 3 < 7 \rightarrow \dato{c}{\abs{x + 3} < 7}$
If we require $\abs{x - 3} < 1$ , then $x$ lies between $2$ and $4$ , and the second factor is bounded by $7$ .
$\delta = \min\left(1, \frac{\varepsilon}{7}\right)$
With that restriction, it suffices in addition that $7\abs{x - 3} < \varepsilon$ . We choose $\delta$ as the smaller of the two numbers, so that both conditions hold.
$\abs{x^2 - 9} = \abs{x - 3}\abs{x + 3} < \frac{\varepsilon}{7} \cdot 7 = \resaltar{\varepsilon}$
If $0 < \abs{x - 3} < \delta$ , then $\abs{x - 3} < 1$ and $\abs{x - 3} < \frac{\varepsilon}{7}$ ; the required inequality follows.

The limit from the beginning of the lesson is proved in the same way, with an observation that will be used often: since the definition does not examine the point $a$ , we may simplify the expression of $f$ at the points other than $a$ . For $x \neq 1$ , the quotient $\frac{x^2 - 1}{x - 1}$ equals $x + 1$ , and then $\abs{f(x) - 2} = \abs{x - 1}$ ; $\delta = \varepsilon$ responds. Canceling the factor $x - 1$ is legitimate precisely because $x$ never equals $1$ in the definition.

The definition as a quantified statement

The definition of limit is a statement with three alternating quantifiers: for every $\varepsilon$ , there exists $\delta$ , for every $x$ . Its logical structure is that of the statements studied by predicate logic, and knowing how to handle it is half the work with limits. Two consequences follow from it.

The first is that the order of the quantifiers matters. In the definition, $\delta$ is chosen after $\varepsilon$ is known, and that is why it may depend on it. If we wrote first $(\exists \delta > 0)$ and then $(\forall \varepsilon > 0)$ , one and the same $\delta$ would have to work for all $\varepsilon$ at once, which is a much stronger claim, one that proposed problem 4 asks you to characterize.

The second is that the negation of the definition is obtained with the rules for negating quantifiers: the negation of “for every” is “there exists some... that does not”, and that of “there exists” is “for every... not”. This yields what it means for $L$ not to be the limit.

PropositionNegation of the definition of limit

Let $f$ be a function defined on $A$ , $a$ an accumulation point of $A$ and $L$ a real number. The number $L$ is not the limit of $f$ as $x$ tends to $a$ if and only if there exists an $\varepsilon > 0$ such that, for every $\delta > 0$ , there is some $x$ in $A$ with $0 < \abs{x - a} < \delta$ and $\abs{f(x) - L} \geq \varepsilon$ . The negation is moved inward, quantifier by quantifier, until it reaches the implication.

Proof

$\neg (\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)$
We negate the whole definition.
$(\exists \varepsilon > 0) \resaltar{\neg} (\exists \delta > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)$
The negation of a universal statement is the existence of a case in which it fails.
$(\exists \varepsilon > 0)(\forall \delta > 0) \resaltar{\neg} (\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)$
The negation of an existential statement is that all cases fail.
$(\exists \varepsilon > 0)(\forall \delta > 0)(\exists x \in A) \resaltar{\neg} \left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)$
We negate a universal statement again.
$(\exists \varepsilon > 0)(\forall \delta > 0)(\exists x \in A)\left(0 < \abs{x - a} < \delta \wedge \resaltar{\abs{f(x) - L} \geq \varepsilon}\right)$
An implication is false if and only if its antecedent is true and its consequent false; and the negation of $\abs{f(x) - L} < \varepsilon$ is $\abs{f(x) - L} \geq \varepsilon$ , by trichotomy.

In terms of the game, $L$ is not the limit if the first player has a challenge to which no response works: however small the proposed $\delta$ , there is a point of the punctured interval whose value leaves the band. To prove that a function has no limit at $a$ , one must show that no real number $L$ is its limit, as is done in solved problem 4.

Fundamental properties

The notation $\lim_{x \to a} f(x) = L$ speaks of the limit, as if it were a single number. That requires a proof: that two different numbers cannot both satisfy the definition. This is where the requirement that $a$ be an accumulation point of the domain comes in. If it were not, there would exist a $\delta_0 > 0$ such that no $x$ of the domain would satisfy $0 < \abs{x - a} < \delta_0$ ; by always responding with that $\delta_0$ , the implication in the definition would be true for every $L$ , because its antecedent would never hold, and every number would be a limit.

TheoremUniqueness of the limit

Let $f$ be a function defined on $A$ and $a$ an accumulation point of $A$ . If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} f(x) = L'$ , then $L = L'$ . We argue by contradiction: we suppose that the limits are different, for example $L < L'$ , and we choose an $\varepsilon$ so small that the bands of radius $\varepsilon$ around $L$ and around $L'$ do not touch.

Proof

$\dato{e}{\varepsilon = \frac{L' - L}{2} > 0}$
Suppose $L < L'$ , and let $\varepsilon$ be half the distance between them; it is positive.
$0 < \abs{x - a} < \delta_1 \rightarrow \abs{f(x) - L} < \varepsilon, \qquad 0 < \abs{x - a} < \delta_2 \rightarrow \abs{f(x) - L'} < \varepsilon$
The bands of radius $\varepsilon$ around $L$ and around $L'$ have no common points, and we apply the definition to each limit with that $\varepsilon$ : we obtain two responses, $\delta_1$ and $\delta_2$ , which in general are different.
$\delta = \min(\delta_1, \delta_2), \quad (\exists x \in A)\left(0 < \abs{x - a} < \delta\right)$
For both conditions to hold at once, we take the smaller of the two radii. Since $a$ is an accumulation point of $A$ , the punctured interval of radius $\delta$ contains some $x$ of the domain.
$\dato{d}{\abs{f(x) - L} < \varepsilon \wedge \abs{f(x) - L'} < \varepsilon}$
For that $x$ both inequalities hold: the value $f(x)$ would lie in both bands at once.
$L' - L \leq \abs{L' - f(x)} + \abs{f(x) - L} < 2\varepsilon = \resaltar{L' - L}$
By the triangle inequality, the distance between $L$ and $L'$ would be less than $2\varepsilon$ , which is precisely that distance: a contradiction. Therefore $L < L'$ cannot hold; likewise, $L' < L$ cannot hold, and by trichotomy, $L = L'$ .

The following lemma formalizes the observation that allowed us to simplify the quotient $\frac{x^2 - 1}{x - 1}$ : the limit at $a$ depends only on the values of the function near $a$ , and not on the value it takes at $a$ nor on those far from it.

LemmaThe limit depends only on nearby values

Let $f$ and $g$ be functions defined on $A$ , $a$ an accumulation point of $A$ , and $r > 0$ such that $f(x) = g(x)$ for every $x$ in $A$ with $0 < \abs{x - a} < r$ . If $\lim_{x \to a} f(x) = L$ , then $\lim_{x \to a} g(x) = L$ . It suffices to take the response that works for $f$ and, if necessary, to reduce it to $r$ .

Proof

$\dato{f}{0 < \abs{x - a} < \delta_1 \rightarrow \abs{f(x) - L} < \varepsilon}$
Let $\varepsilon > 0$ . By hypothesis, there exists a response $\delta_1$ for $f$ .
$\delta = \min(\delta_1, r)$
We take the smaller of that response and the radius within which the functions coincide.
$\abs{g(x) - L} = \abs{\resaltar{f(x)} - L} < \varepsilon$
If $x$ is in $A$ and $0 < \abs{x - a} < \delta$ , then $g(x) = f(x)$ , because $\delta \leq r$ , and $f(x)$ lies in the band, because $\delta \leq \delta_1$ .

The following two properties say that the limit imposes conditions on the function near the point. The first forces it to remain bounded; the second, to have the sign of the limit when the latter is not zero. Both are obtained by applying the definition with an $\varepsilon$ chosen in advance, and they will be the tools for proving the product and quotient rules in the lesson on the algebra of limits.

PropositionLocal boundedness and preservation of sign

Let $f$ be a function defined on $A$ , $a$ an accumulation point of $A$ , and $\lim_{x \to a} f(x) = L$ . Then there exists a $\delta > 0$ such that, for every $x$ in $A$ with $0 < \abs{x - a} < \delta$ , we have $\abs{f(x)} < \abs{L} + 1$ . If moreover $L \neq 0$ , there exists a $\delta > 0$ such that, for those same $x$ , $f(x)$ has the sign of $L$ and $\abs{f(x)} > \frac{\abs{L}}{2}$ . Each claim is the definition applied to a specific challenge: $\varepsilon = 1$ for the first and $\varepsilon = \frac{\abs{L}}{2}$ for the second.

Proof

$0 < \abs{x - a} < \delta \rightarrow \dato{u}{\abs{f(x) - L} < 1}$
With $\varepsilon = 1$ , the definition gives a $\delta$ such that the values of $f$ differ from $L$ by less than $1$ .
$\abs{f(x)} \leq \abs{f(x) - L} + \abs{L} < \resaltar{\abs{L} + 1}$
By the triangle inequality, the absolute value of $f(x)$ does not exceed that of $L$ plus that distance.
$\dato{m}{L - \frac{\abs{L}}{2} < f(x) < L + \frac{\abs{L}}{2}}$
If $L \neq 0$ , with $\varepsilon = \frac{\abs{L}}{2}$ , which is positive, the definition gives another $\delta$ , and $f(x)$ lies in the interval with center $L$ and radius $\frac{\abs{L}}{2}$ .
$L > 0 \rightarrow f(x) > \resaltar{\frac{L}{2}} > 0, \qquad L < 0 \rightarrow f(x) < \resaltar{\frac{L}{2}} < 0$
If $L > 0$ , the left endpoint is $\frac{L}{2}$ , which is positive; if $L < 0$ , the right endpoint is $\frac{L}{2}$ , which is negative. In both cases, $f(x)$ has the sign of $L$ and is farther from $0$ than $\frac{\abs{L}}{2}$ .

With the definition one can prove the value of any limit, but the work grows with the complexity of the function. The lesson on the algebra of limits proves, starting from this one, the rules for the sum, the product, the quotient, powers and roots, which make it possible to compute most limits without going back each time to look for a $\delta$ .

Solved problems

The problems increase in difficulty: the first three construct $\delta$ for a polynomial, a root and a quotient, each with its own way of bounding the factor that accompanies $\abs{x - a}$ ; the fourth uses the negation of the definition to prove that a limit does not exist; and the last finds the largest $\delta$ that responds to a given challenge.

Solved problem 1A quadratic polynomial

Prove from the definition that $\lim_{x \to 2} (x^2 + x) = 6$ .

Solution

$\abs{(x^2 + x) - 6} = \abs{x - 2} \cdot \dato{f}{\abs{x + 3}}$
Let $\varepsilon > 0$ . Since $2$ is a root of the polynomial $x^2 + x - 6$ , the latter has the factor $x - 2$ .
$\abs{x - 2} < 1 \rightarrow 4 < x + 3 < 6 \rightarrow \dato{c}{\abs{x + 3} < 6}$
We restrict the radius to $1$ in advance: then $x$ lies between $1$ and $3$ , and the second factor is bounded.
$\delta = \min\left(1, \frac{\varepsilon}{6}\right)$
With that restriction it suffices that $6\abs{x - 2} < \varepsilon$ ; we choose the smaller of the two radii.
$\abs{(x^2 + x) - 6} = \abs{x - 2}\abs{x + 3} < \frac{\varepsilon}{6} \cdot 6 = \resaltar{\varepsilon}$
If $0 < \abs{x - 2} < \delta$ , both conditions hold, and the required inequality follows.

Solved problem 2A square root

Prove from the definition that $\lim_{x \to 4} \sqrt{x} = 2$ , where the function is defined on $\ico{0}{\infty}$ .

Solution

$\sqrt{x} - 2 = \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{\sqrt{x} + 2} = \dato{q}{\frac{x - 4}{\sqrt{x} + 2}}$
Let $\varepsilon > 0$ . To make $x - 4$ appear, we multiply and divide by the conjugate expression, $\sqrt{x} + 2$ , which is positive.
$\abs{\sqrt{x} - 2} = \frac{\abs{x - 4}}{\sqrt{x} + 2} \leq \dato{c}{\frac{\abs{x - 4}}{2}}$
The denominator is at least $2$ , because $\sqrt{x} \geq 0$ ; this time there is no need to restrict the radius, because the bound holds on the whole domain.
$\delta = 2\varepsilon$
It therefore suffices that $\frac{\abs{x - 4}}{2} < \varepsilon$ .
$\abs{\sqrt{x} - 2} \leq \frac{\abs{x - 4}}{2} < \frac{2\varepsilon}{2} = \resaltar{\varepsilon}$
If $x \geq 0$ and $0 < \abs{x - 4} < 2\varepsilon$ , we obtain the required inequality. Only the points $x$ of the domain are examined: the negative ones, where the root is not defined, do not count, even though they are close to $4$ if $\varepsilon$ is large.

Solved problem 3A quotient

Prove from the definition that $\lim_{x \to 3} \frac{1}{x} = \frac{1}{3}$ , where the function is defined on $\mathbb{R} \setminus \{0\}$ .

Solution

$\abs{\frac{1}{x} - \frac{1}{3}} = \abs{\frac{3 - x}{3x}} = \frac{\abs{x - 3}}{\dato{d}{3\abs{x}}}$
Let $\varepsilon > 0$ . We bring the difference to a common denominator.
$\abs{x - 3} < 1 \rightarrow x > 2 \rightarrow \dato{c}{3\abs{x} > 6}$
Now the factor that accompanies $\abs{x - 3}$ is in the denominator, and we must prevent the latter from approaching $0$ . If $\abs{x - 3} < 1$ , then $x > 2$ , and the denominator is greater than $6$ .
$\delta = \min(1, 6\varepsilon)$
With that restriction, the quotient is less than $\frac{\abs{x - 3}}{6}$ , and it suffices that this quantity be less than $\varepsilon$ .
$\abs{\frac{1}{x} - \frac{1}{3}} < \frac{\abs{x - 3}}{6} < \frac{6\varepsilon}{6} = \resaltar{\varepsilon}$
If $0 < \abs{x - 3} < \delta$ , both conditions hold. Without the prior restriction the method would fail: near $x = 0$ the quotient grows without bound.

Solved problem 4A function that jumps

The sign function, defined on $\mathbb{R} \setminus \{0\}$ , equals $1$ on the positive numbers and $-1$ on the negative ones. Prove that it has no limit as $x$ tends to $0$ .

s(x) = \frac{x}{\abs{x}}

Solution

$(\forall L)(\exists \varepsilon > 0)(\forall \delta > 0)(\exists x)\left(0 < \abs{x} < \delta \wedge \abs{s(x) - L} \geq \varepsilon\right)$
The point $0$ is an accumulation point of the domain. We must prove that no real number $L$ is the limit; by the negation of the definition, it suffices to find, for each $L$ , a challenge to which no $\delta$ responds. We propose $\varepsilon = 1$ for all $L$ .
$s\left(\frac{\delta}{2}\right) = 1, \qquad s\left(-\frac{\delta}{2}\right) = -1$
Let $\delta > 0$ be arbitrary. The punctured interval of radius $\delta$ contains a positive number and a negative number, at which the function equals $1$ and $-1$ .
$2 = \abs{1 - (-1)} \leq \abs{1 - L} + \abs{L - (-1)}$
The distance between $1$ and $-1$ is $2$ ; by the triangle inequality, the distances from $L$ to those two values add up to at least $2$ .
$\abs{s\left(\frac{\delta}{2}\right) - L} \geq 1 \vee \abs{s\left(-\frac{\delta}{2}\right) - L} \geq 1$
It follows that one of the two distances is at least $1$ : if both were less than $1$ , their sum would be less than $2$ .
$\neg (\exists L)\left(\lim_{x \to 0} s(x) = L\right)$
Thus, for each $L$ , the challenge $\varepsilon = 1$ has no response: no band of width $2$ contains both values $1$ and $-1$ when its edges are excluded. The function has no limit at $0$ .

Solved problem 5The largest delta

For the function $f(x) = x^2$ at the point $a = 1$ , with $L = 1$ and the challenge $\varepsilon = \frac{1}{2}$ , determine the largest $\delta$ that responds to the challenge.

Solution

$\abs{x^2 - 1} < \frac{1}{2} \leftrightarrow \frac{1}{2} < x^2 < \frac{3}{2}$
The condition $\abs{x^2 - 1} < \frac{1}{2}$ means that $x^2$ lies between $\frac{1}{2}$ and $\frac{3}{2}$ .
$\sqrt{\frac{1}{2}} < x < \sqrt{\frac{3}{2}}$
Near $1$ , the points $x$ are positive, and the condition is equivalent to $x$ lying between the square roots of the endpoints.
$1 - \sqrt{\frac{1}{2}} \approx 0.293, \qquad \sqrt{\frac{3}{2}} - 1 \approx 0.225$
The distances from $1$ to the endpoints are different: the parabola rises faster to the right of $1$ , and the right endpoint is closer.
$\delta = \sqrt{\frac{3}{2}} - 1 = \resaltar{\frac{\sqrt{6} - 2}{2}}$
The interval with center $1$ and radius $\delta$ must fit inside the interval found; its radius cannot exceed the smaller of the two distances, which responds to the challenge.
$0 < \sqrt{\frac{3}{2}} - 1 < \delta', \qquad \abs{\left(\sqrt{\frac{3}{2}}\right)^2 - 1} = \frac{1}{2}$
No larger radius works: if $\delta' > \delta$ , the distance from the point $x = \sqrt{\frac{3}{2}}$ to $1$ is less than $\delta'$ , and there the distance from $x^2$ to $1$ is exactly $\frac{1}{2}$ , which is not less than $\varepsilon$ .

Proposed problems

The problems increase in difficulty: the first two repeat, with other polynomials, the technique of restricting the radius; the third applies the negation of the definition to a function that oscillates; the fourth examines what happens if the order of the quantifiers is reversed; and the last studies a function that has a limit at only one point. Each one has a hint, which is best opened only after you have attempted the problem.

Proposed problem 1Another quadratic polynomial

Prove from the definition that $\lim_{x \to -1} (2x^2 - 3x) = 5$ .

Hint

Factor $2x^2 - 3x - 5$ : it has the root $-1$ . Restrict the radius to $1$ and bound the other factor on the interval that goes from $-2$ to $0$ .

Answer

Since $2x^2 - 3x - 5 = (x + 1)(2x - 5)$ and, if $\abs{x + 1} < 1$ , then $\abs{2x - 5} < 9$ , a valid response is:

\delta = \min\left(1, \frac{\varepsilon}{9}\right)

Proposed problem 2A cube

Prove from the definition that $\lim_{x \to 1} x^3 = 1$ .

Hint

Use the factorization of the difference of cubes, $x^3 - 1 = (x - 1)(x^2 + x + 1)$ , and bound the second factor when $x$ lies between $0$ and $2$ .

Answer

If $\abs{x - 1} < 1$ , then $0 < x^2 + x + 1 < 7$ , so a valid response is:

\delta = \min\left(1, \frac{\varepsilon}{7}\right)

Proposed problem 3A function that oscillates

Prove that the function $\sin\left(\frac{1}{x}\right)$ , defined on $\mathbb{R} \setminus \{0\}$ , has no limit as $x$ tends to $0$ .

Hint

Follow the method of solved problem 4. In every punctured interval centered at $0$ there are points where the function equals $1$ and points where it equals $-1$ : look for $x$ of the form $\frac{1}{\frac{\pi}{2} + 2n\pi}$ and $\frac{1}{\frac{3\pi}{2} + 2n\pi}$ , with $n$ a sufficiently large natural number.

Answer

For each $L$ , no $\delta$ responds to the challenge $\varepsilon = 1$ , because for large $n$ the distance from each of the following two points to $0$ is less than $\delta$ , and at them the function equals $1$ and $-1$ :

x_n = \frac{1}{\frac{\pi}{2} + 2n\pi}, \qquad y_n = \frac{1}{\frac{3\pi}{2} + 2n\pi}

Worked solution

Solution

$0 < \frac{1}{\frac{3\pi}{2} + 2n\pi} < \frac{1}{\frac{\pi}{2} + 2n\pi} < \frac{1}{2n\pi} < \resaltar{\delta}$
Let $L$ be a real number and $\delta > 0$ . By the Archimedean property there exists a natural number $n$ with $2n\pi > \frac{1}{\delta}$ ; then the following two points are positive and less than $\delta$ .
$\sin\left(\frac{\pi}{2} + 2n\pi\right) = 1, \qquad \sin\left(\frac{3\pi}{2} + 2n\pi\right) = -1$
At the first, $\frac{1}{x}$ is an angle whose sine is $1$ ; at the second, one whose sine is $-1$ .
$\abs{1 - L} + \abs{L + 1} \geq 2 \rightarrow \abs{1 - L} \geq 1 \vee \abs{-1 - L} \geq 1$
As in solved problem 4, the distances from $L$ to $1$ and to $-1$ add up to at least $2$ , and one of them is at least $1$ . The challenge $\varepsilon = 1$ has no response, for any $L$ .

Proposed problem 4The order of the quantifiers

Let $f$ be a function defined on $A$ , $a$ an accumulation point of $A$ and $L$ a real number. Prove that the following statement, which results from the definition of limit by interchanging the first two quantifiers, holds if and only if $f$ takes the value $L$ at all the points of $A$ in some punctured interval centered at $a$ . Does the function $x^2$ satisfy it at $a = 3$ with $L = 9$ ?

(\exists \delta > 0)(\forall \varepsilon > 0)(\forall x \in A)\left(0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} < \varepsilon\right)

Hint

If a nonnegative number is less than every $\varepsilon > 0$ , it is $0$ : otherwise, it would itself be an $\varepsilon$ that it is not less than.

Answer

Once the $\delta$ of the statement is fixed, each $x$ in the punctured interval satisfies $\abs{f(x) - L} < \varepsilon$ for every $\varepsilon > 0$ , hence $f(x) = L$ ; the converse is immediate. The function $x^2$ does not satisfy it at $3$ , although its limit there is $9$ , because it is not constant on any interval:

\left(3 + \frac{\delta}{2}\right)^2 \neq 9

Worked solution

Solution

$\dato{t}{(\forall \varepsilon > 0)\left(\abs{f(x) - L} < \varepsilon\right)}$
Suppose the statement holds, and let $\delta$ be the radius it asserts. Take an $x$ in $A$ with $0 < \abs{x - a} < \delta$ ; the inequality holds for every positive $\varepsilon$ .
$\abs{f(x) - L} < \abs{f(x) - L} \rightarrow \bot, \qquad \resaltar{f(x) = L}$
If $\abs{f(x) - L}$ were positive, we could take $\varepsilon = \abs{f(x) - L}$ , and the inequality would say that a number is less than itself. Therefore it is $0$ , and $f(x) = L$ .
$0 < \abs{x - a} < \delta \rightarrow \abs{f(x) - L} = 0 < \varepsilon$
Conversely, if $f$ takes the value $L$ at the points of $A$ in a punctured interval of radius $\delta$ , that same $\delta$ works for every $\varepsilon$ , because the distance is $0$ .
$0 < \abs{x - 3} < 6 \rightarrow x^2 \neq 9$
The function $x^2$ does not take the value $9$ at any point other than $3$ near it, so it does not satisfy the statement, although it satisfies the definition of limit. The reversed statement is not another way of saying the same thing, but a much stronger claim.

Proposed problem 5A limit at only one point

The following function equals $x$ at the rationals and $0$ at the irrationals. Prove that its limit at $0$ is $0$ , and that it has no limit at any point $a \neq 0$ .

h(x) = \begin{cases} x & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}

Hint

At $0$ , observe that $\abs{h(x)} \leq \abs{x}$ in all cases. At $a \neq 0$ , propose the challenge $\varepsilon = \frac{\abs{a}}{3}$ and use the fact that every interval contains rationals and irrationals, as was proved in the lesson on the real line: a nearby irrational forces $L$ to be close to $0$ , and a nearby rational forces it to be close to $a$ .

Answer

At $0$ , $\delta = \varepsilon$ responds. At $a \neq 0$ , if $L$ were the limit, an irrational $y$ and a rational $x$ of the punctured interval of radius $\min\left(\delta, \frac{\abs{a}}{3}\right)$ would give at once the following two inequalities, which are incompatible:

\abs{L} < \frac{\abs{a}}{3}, \qquad \abs{L} > \abs{x} - \frac{\abs{a}}{3} > \frac{\abs{a}}{3}

Worked solution

Solution

$0 < \abs{x} < \varepsilon \rightarrow \abs{h(x) - 0} \leq \abs{x} < \varepsilon$
At $0$ : at the rationals $\abs{h(x)} = \abs{x}$ , and at the irrationals $\abs{h(x)} = 0$ ; in both cases, the distance from $h(x)$ to $0$ does not exceed that from $x$ to $0$ , and $\delta = \varepsilon$ responds.
$\delta' = \min\left(\delta, \frac{\abs{a}}{3}\right)$
Let $a \neq 0$ and suppose that $L$ is the limit at $a$ . With $\varepsilon = \frac{\abs{a}}{3}$ there is a $\delta$ that responds, and we reduce it, if necessary, to $\frac{\abs{a}}{3}$ .
$\abs{0 - L} < \frac{\abs{a}}{3} \rightarrow \dato{i}{\abs{L} < \frac{\abs{a}}{3}}$
In the punctured interval of radius $\delta'$ there is an irrational $y$ , at which $h$ equals $0$ : the limit lies close to $0$ .
$\abs{x - L} < \frac{\abs{a}}{3}, \qquad \dato{r}{\abs{x} > \abs{a} - \frac{\abs{a}}{3} = \frac{2\abs{a}}{3}}$
There is also a rational $x$ , at which $h$ equals $x$ . Since the distance from $x$ to $a$ is less than $\frac{\abs{a}}{3}$ , its absolute value is greater than $\frac{2\abs{a}}{3}$ .
$\abs{L} \geq \abs{x} - \abs{x - L} > \frac{2\abs{a}}{3} - \frac{\abs{a}}{3} = \resaltar{\frac{\abs{a}}{3}}$
By the triangle inequality, $\abs{L}$ would be at once less than and greater than $\frac{\abs{a}}{3}$ : a contradiction. There is no limit at $a$ .