Continuity

Intuitive Concept of Continuity

Intuitively, a function is said to be continuous at a point if the graph of the function around that point can be drawn without lifting the pen from the plane of the paper. This idea is only a visual guide, but it is useful in building intuition.

Consider the function $$f(x) = \begin{cases} 1, & x \le 0 \\ 2, & x > 0 \end{cases}$$ At $x=0$, the left hand limit is $$\lim_{x \to 0^-} f(x) = 1,$$ and the right hand limit is $$\lim_{x \to 0^+} f(x) = 2.$$ Since these two one-sided limits are different, the limit of $f(x)$ at $x=0$ does not exist. Therefore, the function is not continuous at $x=0$.

Now consider another function $$f(x) = \begin{cases} 1, & x \ne 0 \\ 2, & x = 0 \end{cases}$$ Here, $$\lim_{x \to 0^-} f(x) = 1 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 1,$$ so the limit exists and is equal to 1. However, $$f(0) = 2.$$ Since the limit exists but is not equal to the value of the function at the point, this function is also not continuous at $x=0$.

Thus, continuity requires more than just the existence of the function value or the limit separately; they must agree.

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Formal Definition of Continuity at a Point

Suppose $f$ is a real-valued function defined on a subset of the real numbers, and let $c$ be a point in the domain of $f$. Then $f$ is said to be continuous at $c$ if $$\lim_{x \to c} f(x) = f(c).$$

More explicitly, a function is continuous at $x=c$ if all three of the following conditions hold:

1. $f(c)$ is defined,
2. $\lim_{x \to c} f(x)$ exists,
3. $\lim_{x \to c} f(x) = f(c)$.

Equivalently, for interior points of the domain, continuity at $c$ means $$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c).$$

If any one of these conditions fails, the function is said to be discontinuous at $c$, and $c$ is called a point of discontinuity.

If the domain of $f$ consists of only one point, then $f$ is continuous on that domain by definition, because there is no nearby point in the domain at which continuity can fail.

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Continuity of a Function

A real function $f$ is said to be continuous if it is continuous at every point in its domain.

If a function is defined on a closed interval $[a,b]$, then continuity on $[a,b]$ means:

• $f$ is continuous at every interior point of $(a,b)$,
• $f$ is right continuous at $a$, that is, $$\lim_{x \to a^+} f(x) = f(a),$$
• $f$ is left continuous at $b$, that is, $$\lim_{x \to b^-} f(x) = f(b).$$

This endpoint condition is important because at the endpoints of a closed interval, only one-sided limits are relevant.

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Algebra of Continuous Functions

The algebra of continuous functions closely parallels the algebra of limits.

Suppose $f$ and $g$ are two real-valued functions continuous at a real number $c$. Then:

1. $f+g$ is continuous at $c$,
2. $f-g$ is continuous at $c$,
3. $fg$ is continuous at $c$,
4. $\dfrac{f}{g}$ is continuous at $c$, provided $g(c) \ne 0$.

These facts follow from the corresponding laws of limits.

Special consequences:

• If $\lambda$ is a constant and $g$ is continuous at $c$, then $\lambda g$ is continuous at $c$.
• If $f$ is continuous at $c$, then $-f$ is continuous at $c$.
• If $g$ is continuous at $c$ and $g(c) \ne 0$, then $$\frac{1}{g}$$ is continuous at $c$.

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Continuity of Composite Functions

Let $f$ and $g$ be real-valued functions such that the composite function $(f \circ g)$ is defined at $c$.

If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then the composite function $$(f \circ g)(x) = f(g(x))$$ is continuous at $c$.

This theorem is extremely useful. It allows us to conclude continuity of complicated expressions by recognizing them as compositions of simpler continuous functions.

Solved Examples

Example 1: Checking Continuity at a Point

Check the continuity of the function $f(x) = 2x + 3$ at $x = 1$.

Solution: Step 1: Find the value of the function at the point: $$f(1) = 2(1) + 3 = 5.$$

Step 2: Find the limit of the function as $x \to 1$: $$\lim_{x \to 1} (2x+3) = 2(1)+3 = 5.$$

Step 3: Compare the limit with the function value: $$\lim_{x \to 1} f(x) = 5 = f(1).$$

Hence, $f$ is continuous at $x=1$.

In fact, since $2x+3$ is a linear polynomial, it is continuous for every real number.

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Example 2: Continuity of Modulus Function

Examine whether the function $f(x) = |x|$ is continuous at $x = 0$.

Solution: Recall that $$|x| = \begin{cases} -x, & x<0 \\ x, & x \ge 0 \end{cases}$$

Step 1: Find the function value at 0: $$f(0) = |0| = 0.$$

Step 2: Find the left hand limit: $$\lim_{x \to 0^-} |x| = \lim_{x \to 0^-} (-x) = 0.$$

Step 3: Find the right hand limit: $$\lim_{x \to 0^+} |x| = \lim_{x \to 0^+} x = 0.$$

Step 4: Compare all three values: $$\lim_{x \to 0^-} |x| = \lim_{x \to 0^+} |x| = f(0) = 0.$$

Therefore, $f(x) = |x|$ is continuous at $x=0$.

Since $|x|$ is also continuous for all $x<0$ and all $x>0$, it is continuous for every real number.

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Example 3: Finding Points of Discontinuity

Show that the function $f$ given by $f(x) = x^3 + 3$ if $x \ne 0$, and $f(x) = 1$ if $x = 0$ is not continuous at $x = 0$.

Solution: Step 1: Find the function value at the point: $$f(0) = 1.$$

Step 2: For $x \ne 0$, the function is $x^3+3$, so $$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^3+3) = 0^3 + 3 = 3.$$

Step 3: Compare the limit with the function value: $$\lim_{x \to 0} f(x) = 3 \ne 1 = f(0).$$

Hence the function is not continuous at $x=0$.

Also, for every $x \ne 0$, the function behaves like the polynomial $x^3+3$, which is continuous. Therefore, the only point of discontinuity is $x=0$.

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Example 4: Continuity of Constant and Polynomial Functions

Prove that every polynomial function is continuous.

Solution: Let $$p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$$ be a polynomial.

Each constant function is continuous. The function $x$ is continuous. Therefore, powers such as $x^2, x^3, \dots, x^n$ are continuous because products of continuous functions are continuous.

Now each term $$a_k x^k$$ is continuous, since it is a constant multiple of a continuous function.

Finally, a polynomial is a finite sum of such continuous terms. Since sums of continuous functions are continuous, $p(x)$ is continuous.

Hence every polynomial function is continuous at every real number.

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Example 5: Continuity of Greatest Integer Function

Find all the points of discontinuity of the greatest integer function defined by $f(x) = [x]$.

Solution: The greatest integer function $[x]$ gives the greatest integer less than or equal to $x$.

We consider two cases.

Case 1: $c$ is not an integer If $c$ is not an integer, then for all $x$ sufficiently close to $c$, the value of $[x]$ remains equal to $[c]$. Hence, $$\lim_{x \to c} [x] = [c] = f(c).$$ So the function is continuous at every non-integer point.

Case 2: $c$ is an integer Let $c=n$, where $n$ is an integer. Then:

• just to the left of $n$, we have $[x] = n-1$,
• just to the right of $n$, we have $[x] = n$.

Therefore, $$\lim_{x \to n^-} [x] = n-1,$$ $$\lim_{x \to n^+} [x] = n.$$ Since these one-sided limits are different, the limit at $x=n$ does not exist. Hence the function is discontinuous at every integer.

So the greatest integer function is continuous at all non-integers and discontinuous at all integers.

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Example 6: Discussing Continuity of Piecewise Functions

Discuss the continuity of the function defined by $f(x) = x + 2$ if $x \le 1$, and $f(x) = x - 2$ if $x > 1$.

Solution: The function is defined for all real numbers. We check continuity in three cases.

Case 1: $c < 1$ Then near $c$, the function is simply $$f(x) = x+2.$$ So, $$\lim_{x \to c} f(x) = \lim_{x \to c} (x+2) = c+2 = f(c).$$ Hence the function is continuous for all $x<1$.

Case 2: $c > 1$ Then near $c$, the function is $$f(x) = x-2.$$ So, $$\lim_{x \to c} f(x) = \lim_{x \to c} (x-2) = c-2 = f(c).$$ Hence the function is continuous for all $x>1$.

Case 3: $c = 1$ Left hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+2) = 3.$$ Right hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-2) = -1.$$ Since $$\lim_{x \to 1^-} f(x) \ne \lim_{x \to 1^+} f(x),$$ the limit at $x=1$ does not exist. Therefore the function is not continuous at $x=1$.

Hence the function is continuous everywhere except at $x=1$, which is the only point of discontinuity.

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Example 7: Continuity of Trigonometric Functions

Discuss the continuity of the sine function.

Solution: Let $$f(x) = \sin x.$$ We want to check continuity at an arbitrary real number $c$.

Step 1: Write $x = c+h$, where $h \to 0$ as $x \to c$. Then, $$\lim_{x \to c} \sin x = \lim_{h \to 0} \sin(c+h).$$

Step 2: Use the angle addition formula: $$\sin(c+h) = \sin c \cos h + \cos c \sin h.$$ So, $$\lim_{x \to c} \sin x = \lim_{h \to 0} \big(\sin c \cos h + \cos c \sin h\big).$$

Step 3: Use the standard limits $$\lim_{h \to 0} \cos h = 1, \qquad \lim_{h \to 0} \sin h = 0.$$ Therefore, $$\lim_{x \to c} \sin x = \sin c \cdot 1 + \cos c \cdot 0 = \sin c.$$

Since $$\lim_{x \to c} \sin x = \sin c = f(c),$$ $\sin x$ is continuous at $x=c$.

As $c$ was arbitrary, the sine function is continuous for all real numbers.

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Example 8: Continuity of Composite Functions

Show that the function defined by $f(x) = \sin(x^2)$ is a continuous function.

Solution: Write $$f(x) = \sin(x^2)$$ as a composition of two functions: $$h(x) = x^2, \qquad g(x) = \sin x.$$ Then $$f(x) = (g \circ h)(x).$$

Now:

• $h(x) = x^2$ is a polynomial, so it is continuous for all real numbers.
• $g(x) = \sin x$ is continuous for all real numbers.

By the theorem on continuity of composite functions, since $h$ is continuous at every real number and $g$ is continuous at every value of $h(x)$, the composite function $$g(h(x)) = \sin(x^2)$$ is continuous for all real numbers.

Hence $f(x)=\sin(x^2)$ is a continuous function.