Home Topics in Analysis Analysis Elementary Linear Algebra Linear Algebra Analysis of Functions of Many Variables Linear Algebra and Analysis Complex Analysis Calculus of One And Many Variables Engineering Math One Variable Advanced Calculus Interrogation of Hiroshi Oshima

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⇐ Suppose the sequence condition holds at x. Why is f continuous at x? If this werenot the case, then there would exist ε > 0 and xn with |xn − x|< 1/n but | f (xn)− f (x)| ≥ε . However, xn → x and so f (xn) → f (x) so for large enough n, | f (x)− f (xn)| < ε , acontradiction.

Because of this theorem, I will use either of the equivalent definitions without commentin what follows.

The other thing to notice is that the concept of continuity as described in the definitionis a point property. That is to say it is a property which a function may or may not have ata single point. Here is an example.

Example 4.0.3 Let f (x) =

{x if x is rational

0 if x is irrational. This function is continuous at x = 0

and nowhere else.

Let xn → 0. Then | f (xn)| ≤ |xn| and so f (xn) → 0. Thus it is continuous at 0. Ifx ̸= 0, then if x is irrational, you could pick a sequence of rational numbers xn → x. Thenf (xn)→ x ̸= 0 but f (0) = 0. If x is rational, then let xn → x where xn is irrational. Thenf (xn) = 0 → 0 but f (x) = x ̸= 0 so this is not continuous at any other point than 0.

Here is another example.

Example 4.0.4 Show the function f (x) = 3x+10 is continuous at x = −3.

Let xn → −3. Then by the limit theorems for sequences, 3xn + 10 → 3(−3)+ 10 =f (−3) so this is continuous.

Here is another example.

Example 4.0.5 Show the function f (x) =√

x is continuous at x = 5.

Note f (5) =√

5 and so | f (x)− f (5)| =∣∣∣√x−

√5∣∣∣ . For x positive,

∣∣∣√x−√

5∣∣∣ ≤

|x−5|√x+

√5≤ |x−5|√

5. Now let xn → 5. Eventually xn is positive and so

∣∣∣√xn −√

5∣∣∣ ≤ |xn−5|√

5and the expression on the right converges to 0 as n → ∞.

The following is a useful theorem which can remove the need to constantly use the ε,δdefinition given above.

Theorem 4.0.6 The following assertions are valid

1. The function a f +bg is continuous at x when f , g are continuous at x ∈ D( f )∩D(g)and a,b ∈ R.

2. If and f and g are each continuous at x in the domains of both f and g, then f g iscontinuous at x. If, in addition to this, g(x) ̸= 0, then f/g is continuous at x.

3. If f is continuous at x, f (x) ∈ D(g) ⊆ R, and g is continuous at f (x) ,then g ◦ f iscontinuous at x.

4. The function f : R→ R, given by f (x) = |x| is continuous.

99< Suppose the sequence condition holds at x. Why is f continuous at x? If this werenot the case, then there would exist € > 0 and x, with |x, —x| < 1/n but |f (x,) — f (x)| >€. However, x, + x and so f (x,) + f(x) so for large enough n,|f (x) — f (xn)| < €, acontradiction. JBecause of this theorem, I will use either of the equivalent definitions without commentin what follows.The other thing to notice is that the concept of continuity as described in the definitionis a point property. That is to say it is a property which a function may or may not have ata single point. Here is an example.x if x is rational . oo, .Example 4.0.3 Let f (x) = . This function is continuous at x = 00 if x is irrationaland nowhere else.Let x, > 0. Then |f(xn)| < |xn| and so f(x,) > 0. Thus it is continuous at 0. Ifx #0, then if x is irrational, you could pick a sequence of rational numbers x, + x. Thenf Xn) > x £0 but f (0) = 0. If x is rational, then let x, — x where x, is irrational. Thenf (%n) =0— 0 but f (x) = x £0 so this is not continuous at any other point than 0.Here is another example.Example 4.0.4 Show the function f (x) = 3x + 10 is continuous at x = —3.Let x, — —3. Then by the limit theorems for sequences, 3x, + 10 + 3(—3) +10 =f (—3) so this is continuous.Here is another example.Example 4.0.5 Show the function f (x) = \/x is continuous at x = 5.Note f (5) = V5 and so |f (x) — f (5)| = ve- v5. For x positive,|x—5| < [x=5)Vxtv5 — V5and the expression on the right converges to 0 as n + ©,The following is a useful theorem which can remove the need to constantly use the €,6definition given above.vz~v3| <Now let x, — 5. Eventually x, is positive and so | Vn ~/5 | < PalTheorem 4.0.6 The following assertions are valid1. The function af + bg is continuous at x when f, g are continuous at x € D(f) ND (g)anda,beER.2. If and f and g are each continuous at x in the domains of both f and g, then fg iscontinuous at x. If, in addition to this, g(x) 40, then f/g is continuous at x.3. If f is continuous at x, f (x) € D(g) CR, and g is continuous at f (x) ,then go f iscontinuous at x.4. The function f :R — R, given by f (x) = |x| is continuous.