Limits and Continuity of Functions

We studied real sequences, which are real-valued functions defined on the subset ℕ of ℝ. In this chapter we shall consider real-valued functions whose domains are arbitrary subsets of ℝ that is we have D ⊆ ℝ, and the function f: D → ℝ.

Definition 4.1.1. (Accumulation point). Let D ⊆ ℝ and c ∈ ℝ. Then c is called an accumulation point of D (or limit point) if there is a sequence {x_n} in D \{c} such that x_n ⟶ c. □

A sequence {x_n} of the inputs in the interval D \{c} generetas another sequence of the outputs {f(x_n)}

f(x₁), f(x₂), f(x₃), ..., f(x_n), ...

The following is the definition of limit of a function due to Heinrich Eduard Heine.

Definition 4.1.2. (Limit of a function; Heine's Definition) Let D ⊆ ℝ and c ∈ ℝ be an accumulation point of D. Also, let f: D → ℝ be a function. Let {x_n} be any sequence that converges to the point c. Now consider the sequence {f(x_n)}. If {f(x_n)} converges to a real number ℓ, for every sequence {x_n} that converges to c, we say that the function f has a limit ℓ. This is written as

lim_{x \to c} f (x) = ℓ

The great advantage of the previous definition is that the theory of limits of sequences carries over easily to limtis of functions. In particular the limit of is unique because is unique that for sequences.

One of the most important applications of the limit concept is that of continuity. At a point c, the function f can be befined or undefined. If it is defined and if we find that the limit at this point is equal to the defined value, that is, if

f (c) = lim_{x \to c} f (x)

thne we say that the function f is continuos at c. We say that f is continuous if an open interval (a,b) is it is continuous at each point of the interval.

Heine's definition of a limit function defined in a certan domain D ⊂ ℝⁿ at a given point x₀ ∈ ℝⁿ that is the cluster point for D, has then the following form

lim_{x \to x_{0}} f (x) = A ⟺ [\forall_{(x_{n})} (x_{n} \in D \land x_{n} \neq x_{0}, n = 1, 2, 3. . .) \land lim_{n \to \infty} x_{n} = x_{0} \Rightarrow lim_{n \to \infty} f (x_{n}) = A]

Definition 4.1.3 (Continuity of Functions). Let D ⊆ ℝ and consider the function f: D → ℝ and a point c ∈ D. We say that f is continuous at c if

for any sequence {x_n} in D such that x_n → c ⇒ f(x_n) → f(c).

If f is not continuous at x₀, we say that f is discontinuous at x₀. In case f is continuous at every x₀ ∈ D, we say that f is continuous on D. □

Thus, we consider a sequence {x_n} from the domain of f, which tends to c but does not take the value c. The corresponding sequence {f(x_n)} may or may not have a limit as n ⟶ ∞. If not, we say that f is discontinuous at c.

Example 4.1.4 The following limit

Does not exist. Indeed we can consider two sequences {x_n} and {y_n} converging to 0. Let x_n = πn and y_n = π/2 + 2nπ. We have f(x_n) = sin x_n = 0 for each n, hence the limit of the sequence of functions is 0. However f(y_n) = sin y_n = 1 for each n and this time the limit is one. Hence, two sequences that both tend to zero are mapped by the function into sequences that converge to different limits. Then the limit does not exits.

Analogously, for

lim_{x \to 0} \sin \frac{1}{x}

Indeed we can consider two sequences {x_n} and {y_n} converging to 0. Let x_n = 1/(2πn) and y_n = 1/(π/2 + 2nπ). We have f(x_n) = 0 for each n, hence the limit of the sequence of functions is 0. However f(y_n) = 1 for each n and this time the limit is one. Hence, two sequences that both tend to zero are mapped by the function into sequences that converge to different limits. Then the limit does not exits.

1/sinx — sin 1/x, performs an infinite number of oscillations, just as does the function sin x. In the case of sin x they are distributed over an infinite interval, while in the case of sin 1/x they are all contained in a finite interval, condensing to zero.

■

The definition of limit given requires to verify that for every sequence tending to x₀. Cauchy gave a criterion for the continuity of a function at a point that does not involve convergence of sequences.

Proposition 4.1.5. (Cauchy's definition of limit) Let D ⊆ ℝ and let f: D → ℝ be a function. A number ℓ is the limit of a function f as x tends to x₀ if and only if for any ε > 0 there is a δ > 0 (dependent on ε) such that

x ∈ D and |x − x₀| < δ ⇒ |f(x) − ℓ| < ε (4.1.6)

Proof. Let ε > 0 and assume there exists δ > 0 such that the implication 4.1.6 holds. Let x_n ⟶ x₀ and x_n ≠ x₀, then there exists N ∈ ℕ such that 0 < |x − x₀| < δ, and consequently |f(x_n) − ℓ| < ε, for n > N. This prove f(x_n) ⟶ ℓ and hence lim_{x ⟶ x₀} f(x) = ℓ.
Conversely assume f has limit ℓ but that it not true that for every ε > 0 there exists δ > 0, there is is x ∈ D satisfying 4.1.6. Then there is a sequence {x_n} in D such that |x_n − x₀| < 1/n, but |f(x_n) − f(x₀)| ≥ ε for all N ∈ ℕ. But then x_n ⟶ x₀ and f(x_n) ↛ f(x₀). This contradicts the continuity of f at x₀. □

To have a limit, the function has to be defined on both sides of x₀, but it doesn't matter what the function is at x = x₀ itself; it need not even be defined there. For instance even though the function f(x) = sin x/x is not defined at x = 0, its limit as x ⟶ 0 exists and equals 1.

In the definition of the limit of a function f as x approaches x₀ whether from right or from left, we have considered the values of f in the deleted neighbourhood x₀. If we consider the behaviour of f for those values of x greater than x₀, we say, that x approaches x₀ from the right or from above. We can give the following definition for this concept.

Definition 4.1.6. (Right hand limit (RHL)) We say that a function f tends to ℓ as x tends to x₀ from the right if for each ε > 0, there exists a δ > 0 such that

|f(x) − ℓ| < ε when x₀ < x < x₀ + δ

In symbol,

lim_{x \to x_{0}^{+}} f (x) = ℓ

Definition 4.1.7. (Left hand limit (LHL)) We say that a function f tends to ℓ as x tends to x₀ from the left if for each ε > 0, there exists a δ > 0 such that

|f(x) − ℓ| < ε when x₀ − δ < x < x₀

In symbol,

lim_{x \to x_{0}^{-}} f (x) = ℓ

Clearly the limit of f exists, then both the left hand and right hand limits exist and each is equal to ℓ.

Theorem 4.1.8. The limit of f exists if and only if both the limits, the left hand and the right hand, exist and are equal.

Proof. Let lim_{x ⟶ x₀} f(x) = ℓ. Then for any ε > 0, there exists some δ > 0 such that

|f(x) − ℓ | < ε, when x₀ − δ < x < x₀ + δ, x ≠ x₀

it follows that

|f(x) − ℓ | < ε, when x₀ − δ < x < x₀ (4.1.9)
|f(x) − ℓ | < ε, when x₀ < x < x₀ + δ (4.1.10)

From equation (4.1.9) and (4.1.10) we get that the RHS and LHL both exist and are equal to ℓ.

To prove the converse let RHS = LHL = ℓ. Then for any ε > 0, there exists some δ₁ > 0 and δ₂ > 0 such that

|f(x) − ℓ| < ε when x₀ < x < x₀ + δ₁ (4.1.11)
|f(x) − ℓ| < ε when x₀ − δ₂ < x < x₀ (4.1.12)

Let δ = min(δ₁, δ₂). Then

δ ≤ δ₁ and δ ≤ δ₂ ⇒ x₀ + δ ≤ x₀ + δ₁ and x₀ − δ ≥ x₀ − δ₂ (4.1.13)

From equations (4.1.11), (4.1.12) and (4.1.13), we get

|f(x) − ℓ| < ε when x₀ − δ < x < x₀ + δ, x ≠ x₀

Hence,

lim_{x \to x_{0}} f (x) = ℓ □

Example 4.1.14. The function sign of x

defined as

sgn x := {\begin{cases} - 1 & if x < 0, \\ 0 & if x = 0, \\ 1 & if x > 0. \end{cases}

admits no limt at x = 0 (i.e. is not continous at 0), since sgn has left hand limit equal to −1 at x = 0 and right hand limit equal to +1. ■

Example 4.1.15. Evaluate

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

Let us evaluate the left hand and the right hand limits. When x ⟶ 1⁻, put x = 1 − h, h >0.
As x ⟶ 1⁻, h ⟶ 0⁺, so that

lim_{x \to 1^{-}} \frac{x^{2} - 1}{x - 1} = lim_{h \to 0^{+}} \frac{(1 - h)^{2} - 1}{- h} = lim_{h \to 0^{+}} \frac{- h (2 - h)}{- h} = lim_{h \to 0^{+}} (2 - h) = 2

For the right hand limit, x ⟶ 1⁺, put x = 1 + h, h > 0.

lim_{x \to 1^{+}} \frac{x^{2} - 1}{x - 1} = lim_{h \to 0^{+}} \frac{(1 + h)^{2} - 1}{h} = lim_{h \to 0^{+}} (2 + h) = 2

Since both, the left hand and right hand limits exist and are equal, the given limit of the given function exists and is equal to 2. ■

Example 4.1.16. Evaluate

lim_{x \to 0} \frac{e^{1 / x}}{e^{1 / x} + 1}

When x ⟶ 0⁺, 1/x ⟶ ∞ and e^1/x ⟶ 0

lim_{x \to 0^{+}} \frac{e^{1 / x}}{e^{1 / x} + 1} = lim_{x \to 0^{+}} \frac{1}{1 + e^{- 1 / x}} = 1

As x ⟶ 0⁻, 1/x ⟶ − ∞ and e^1/x ⟶ 0.

lim_{x \to 0^{-}} \frac{e^{1 / x}}{e^{1 / x} + 1} = \frac{0}{1} = 0

So that the left hand limit is not equal to the right hand limit, the original limit does not exist. ■

Theorem 4.1.17 (The sign permanence theorem). If f is a continuous function at x = x₀ and if f(x₀) > 0, there there exists a neighborhood of x₀ : (x₀ − δ, x₀ + δ) for δ > 0 such that f(x) > 0. If f(x₀) < 0, there is a neighborhood of x₀ in which f(x) is negative.

Proof. By definition of continuity of f at x₀, for any ε > 0, there exists some δ > 0 such that

|f(x) − f(x₀)| < δ when x₀ − δ < x < x₀ + δ

if f(x₀) > 0, chose ε = f(x₀)/2. Then

f(x₀) − ε < f(x) < f(x₀) + ε
0 < f(x₀)/2 < f(x) < 3f(x₀)+ ε

for |x − x₀| < δ. The other statement can be proved in a similar manner. □

Infinite Limits

Definition 4.1.18. We say that f approaces infinity as x tends to x₀ and write

lim_{x \to x_{0}} f (x) = \infty

if for every positive real number M there exists a corresponding δ > 0 such that for all x

0 < |x − x₀| < δ ⇒ f(x) > M. □

Definition 4.1.19. We say that f approaces minus infinity as x tends to x₀ and write

if for every negative real number L there exists a corresponding δ > 0 such that for all x

0 < |x − x₀| < δ ⇒ f(x) < L. □

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