Continuity of a function at a point 'a' means that the function has no jumps, breaks, or holes at that point. It implies that the function is well-behaved and doesn't have any sudden changes in its values. To check continuity at a point 'a', three conditions must be satisfied:
The function f(a) must be defined.
The limit of the function as it approaches 'a' from the left, i.e., $\lim_{x \rightarrow a^-} f(x)$, should exist and be equal to f(a).
The limit of the function as it approaches 'a' from the right, i.e., $\lim_{x \rightarrow a^+} f(x)$, should exist and be equal to f(a).
If f is not continuous at a, we say f is discontinuous at a and a is called a point of discontinuity of f
Example
Check the continuity of the function f given by f (x) = x + 2 at x = 0. Solution
First note that the function is defined at the given point x = 0 and its value f(0)=0 + 2= 2.
Lets find the limit of the function at x = 0.
$\lim_{x \rightarrow 0} f(x) =\lim_{x \rightarrow 0} (x+2)= 0 + 2=2$
Thus $\lim_{x \rightarrow 0} f(x)= f(0)=2$
Hence, f is continuous at x = 0
Continuity in an Interval
A function is continuous in an interval if it is continuous at every point in that interval.
if f function defined on a closed interval [a, b], then for f to be continuous, it needs to be continuous at every
point in [a, b] including the end points a and b. Continuity of f at a means
$\lim_{x \rightarrow a^+} f(x) =f(a)$
$\lim_{x \rightarrow b^-} f(x) =f(b)$
Continuity of Real Function
A real function f is said to be continuous if it is continuous at every point in the domain of f
Examples
Identity functions f(x) =x
To prove that the identity function \(f(x) = x\) is continuous, we need to show that it satisfies the definition of continuity at every point in its domain, which is the set of all real numbers \(\mathbb{R}\).
The definition of continuity at a point \(a\) is as follows:
A function \(f(x)\) is continuous at a point \(a\) if and only if:
1. \(f(a)\) is defined.
2. \(\lim_{{x \to a}} f(x)\) exists.
3. \(\lim_{{x \to a}} f(x) = f(a)\).
Now, let's prove that the identity function \(f(x) = x\) satisfies these conditions for all real numbers \(a\).
1. \(f(a)\) is defined: For any real number \(a\), \(f(a) = a\), which is always defined because it's just the value of \(a\) itself.
2. \(\lim_{{x \to a}} f(x)\) exists: To find the limit as \(x\) approaches \(a\), we can simply evaluate it:
\(\lim_{{x \to a}} f(x) = \lim_{{x \to a}} x = a\).
The limit exists and is equal to \(a\).
3. \(\lim_{{x \to a}} f(x) = f(a)\): We've already found that \(\lim_{{x \to a}} f(x) = a\), and we know that \(f(a) = a\). Since \(a = a\), this condition is also satisfied.
Therefore, for any real number \(a\), the identity function \(f(x) = x\) satisfies all three conditions for continuity at \(a\). Hence, the identity function is continuous for all real numbers.
Modulus Function f(x)=|x|
To prove that the identity function \(f(x) = |x|\) is continuous, we need to show that it satisfies the definition of continuity at every point in its domain, which is the set of all real numbers \(\mathbb{R}\).
Now this function can be written as
$f(x)=\begin{cases}
& -x, \text{ if } \; x < 0 \\
& x,\text{ if } \; x \geq 0
\end{cases}
$
Let a be a real number such that a < 0. Then f (a) = -a
Also
$\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} -x=-a$
Since $\lim_{x \rightarrow a} f(x)= f(a)$, Function is continous for all the negative real numbers
Let a be a real number such that a = 0. Then f (a) = 0
Also
$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} x=-0$
Since $\lim_{x \rightarrow a} f(x)= f(a)$, Function is continous at x=0
Now
Let a be a real number such that a > 0. Then f (a) = a
Also
$\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} x=a$
Since $\lim_{x \rightarrow a} f(x)= f(a)$, Function is continous for all the positive real numbers
Hence, f is continuous at all points.
Polynomial functions
Lets take an example
$f(x) = x^3+ x^2+ 1$
This function is clearly defined for any real number a
$f(a)=a^3 + a^2 +1$
Also
$\lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} x^3+ x^2+ 1=a^3 + a^2 +1$
Since $\lim_{x \rightarrow a} f(x)= f(a)$, Function is continous for all real numbers
Now
Lets defined a generic polynomials function
$p(x) = a_0 + a_1x + ... + a_nx^n$ for some natural number n, $n \ne 0$ and $a_i \in R$.
Clearly this function is defined for every real number. For a fixed real number a, we have
$\lim_{x \rightarrow a} f(x)= f(a)$
By definition, p is continuous at a. Since a is any real number, p is continuous at every real number and hence p is a continuous function
Algebra of continuous functions
Theorem I
Suppose f and g be two real functions continuous at a real number c, then
(a) Function f+g is continous
(2) f – g is continuous at x = c.
(3) f . g is continuous at x = c.
(4) f/g is continuous at x = c, (provided $g (c) \ne 0$). Proof
(i)$\lim_{x \rightarrow c} f +g = \lim_{x \rightarrow c} f + \lim_{x \rightarrow c} g $
Now as f and g are contnious at c
$\lim_{x \rightarrow c} f(c)= f(c)$ and $\lim_{x \rightarrow c} g(x)= g(c)$, Therefore
=f(c) + g(c)
=(F+g) (c)
Hence, f + g is continuous at x = c
(ii)$\lim_{x \rightarrow c} f - g = \lim_{x \rightarrow c} f - \lim_{x \rightarrow c} g $
Now as f and g are contnious at c
$\lim_{x \rightarrow c} f(c)= f(c)$ and $\lim_{x \rightarrow c} g(x)= g(c)$, Therefore
=f(c) - g(c)
=(F-g) (c)
Hence, f - g is continuous at x = c
(iii)$\lim_{x \rightarrow c} f . g = \lim_{x \rightarrow c} f . \lim_{x \rightarrow c} g $
Now as f and g are contnious at c
$\lim_{x \rightarrow c} f(c)= f(c)$ and $\lim_{x \rightarrow c} g(x)= g(c)$, Therefore
=f(c) . g(c)
=(F.g) (c)
Hence, f . g is continuous at x = c
(iv)$\lim_{x \rightarrow c} f / g = \lim_{x \rightarrow c} f / \lim_{x \rightarrow c} g $
Now as f and g are contnious at c
$\lim_{x \rightarrow c} f(c)= f(c)$ and $\lim_{x \rightarrow c} g(x)= g(c)$, Therefore
=f(c) / g(c)
=(F/g) (c)
Hence, f / g is continuous at x = c
Theorem II
Suppose f and g are real valued functions such that (f o g) is defined at c.
If g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
Solved Examples
Question 1
Determine if the function \( f(x) = \begin{cases} x^2 - 1 & \text{if } x < 2 \\ 3x - 5 & \text{if } x \geq 2 \end{cases} \) is continuous at \( x = 2 \). Solution
Check if the left-hand limit (LHL) at \( x = 2 \), the right-hand limit (RHL) at \( x = 2 \), and the value of \( f(2) \) are equal.
- LHL at \( x = 2 \) is \( \lim_{x \to 2^-} (x^2 - 1) = 3 \).
- RHL at \( x = 2 \) is \( \lim_{x \to 2^+} (3x - 5) = 1 \).
- \( f(2) = 3 \times 2 - 5 = 1 \).
Since LHL ≠ RHL, the function is not continuous at \( x = 2 \).
Question 2
Prove that the function \( f(x) = x^3 \) is continuous at \( x = 1 \). Solution
- LHL = \( \lim_{x \to 1^-} x^3 = 1 \).
- RHL = \( \lim_{x \to 1^+} x^3 = 1 \).
- \( f(1) = 1^3 = 1 \).
Since LHL = RHL = \( f(1) \), the function is continuous at \( x = 1 \).
Question 3
Determine if \( f(x) = \sqrt{x - 2} \) is continuous at \( x = 2 \). Solution
- The domain of \( f(x) \) is \( x \geq 2 \).
- \( \lim_{x \to 2^+} \sqrt{x - 2} = \sqrt{2 - 2} = 0 \).
- \( f(2) = \sqrt{2 - 2} = 0 \).
Therefore, \( f(x) \) is continuous at \( x = 2 \).
Question 4
Find if the function \( f(x) = \begin{cases} 2x + 5 & \text{if } x \leq 3 \\ 3x + 2 & \text{if } x > 3 \end{cases} \) is continuous at \( x = 3 \). Solution
- LHL at \( x = 3 \) is \( \lim_{x \to 3^-} (2x + 5) = 11 \).
- RHL at \( x = 3 \) is \( \lim_{x \to 3^+} (3x + 2) = 11 \).
- \( f(3) = 2 \times 3 + 5 = 11 \).
Since LHL =RHL= \( f(3) \) ,the function is continuous at \( x = 3 \).
