CBSE board exam date sheet 2025 class 10: The Central Board of Secondary Education (CBSE) has released date sheet for Class X board examination 2025....
CBSE Class 10 Date Sheet 2025 (Released): Check Exam Time Table
November 22, 2024Continuity is considered to be one of the significant aspects associated with Calculus. The rivers have a constant flow of water. Human life is a continual flow of time, which means you are constantly becoming older. Similarly, we have the concept of function continuity in mathematics.
Simply put, if you can draw a function’s curve on a graph without raising your pen even once, it’s considered to be continuous. Actually, it’s a really simple and close-to-accurate definition. However, we must define it more precisely for the sake of higher mathematics. That’s exactly what we will do in this part. So let’s get started!
If a function can be drawn without lifting up the pen/pencil, it is said to be continuous. A function is said to be discontinuous if it is not otherwise. Similarly, in calculus, a function \(f(x)\) is continuous at \(x=c\) if the graph of the supplied function does not break at that point \((c, f(c)).\)
If there is no interrupt/break in the graph of a function over the entire interval range, it is said to be continuous in that interval. Assume \(f\) is a real function on a subset of real numbers, and \(c\) is a point in the domain of \(f.\) If \(f\) is continuous at \(c,\) then
\(\mathop {\lim }\limits_{x \to c} f(x) = f(c)\)
In other words, if the function’s left-hand limit, right-hand limit, and value at \(x=c\) all exist and are equal, then
\(\mathop {\lim }\limits_{x \to {c^ – }} f(x) = f(c) = \mathop {\lim }\limits_{x \to {c^ + }} f(x)\)
Then, we can say that \(f\) is continuous at \(x=c.\)
Hence, continuity in maths can be defined as follows:
Suppose \(f\) is a real function on a subset of the real numbers and let \(c\) be a point in the domain of \(f.\) Then f is continuous at \(c\) if
\(\mathop {\lim }\limits_{x \to c} f(x) = f(c)\)
More elaborately, if the left hand limit (LHL), right hand limit (RHL) and the value of the function at \(x=c\) exist and are equal to each other, then \(f\) is said to be continuous at \(x=c.\) Recall that if the right hand limits and left hand limits at \(x=c\) coincide, then we say that it is the value of the limit of the function at \(x=c.\)
Hence we may also define continuity as follows: a function is continuous at \(x=c\) if the function is defined at \(x=c\) and if the value of the function at \(x=c\) equals the limit of the function at \(x=c.\) If f is not continuous at \(c,\) we say \(f\) is discontinuous at \(c,\) and \(c\) is called the point of discontinuity of \(f.\)
From the above graph, we can see that curve is continuous in its domain except at the points \(-8, -2\) and \(6\)
1. In an open interval \((a, b),\) a function \(f\) is said to be continuous if it is continuous at all points in the interval.
2. In a closed interval \([a,b],\) a function \(f\) is said to be continuous if
a. \(f\) is continuous in \((a, b)\)
b. \(\mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)\)
c. \(\mathop {\lim }\limits_{x \to {b^ – }} f(x) = f(b)\)
The function is said to be discontinuous at that point if any of the three conditions for it to be continuous fails. Different sorts of discontinuities can be defined based on the failure of which specific condition leads to discontinuity.
1. Removable Discontinuity: A function with well-defined two-sided limits at \(x=a,\) but either \(f(a)\) is not defined or \(f(a)\) is not equal to its limits, is called a removable discontinuity. So, In this case
\(\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)\)
This type of discontinuity can be removed by taking the value of a function at \(x=a\) equal to \(\mathop {\lim }\limits_{x \to a} f(x)\)
2. Infinite Discontinuity: The function diverges when \(x=a\) in infinite discontinuities, resulting in a discontinuous nature. It means that the function \(f(a)\) hasn’t been defined yet. The limit of a function \(x=a\) is not defined because the value of the function at \(x=a\) does not approach any finite value or tends to infinity.
3. Jump Discontinuity: A type of discontinuity in which the left-hand limit and right-hand limit for a function \(x=a\) exist but are not equal to each other is known as a jump discontinuity. The following is a representation of the jump discontinuity
\(\mathop {\lim }\limits_{x \to {a^ – }} f(x) \ne \mathop {\lim }\limits_{x \to {a^ + }} f(x)\)
4. Oscillatory Discontinuity: When the limits oscillate between two finite values, the discontinuity is said to be oscillatory. For example, say \(f(x) = \cos \frac{1}{x}\) in this case, the limit oscillates between \(-1\) to \(1.\)
Below are some graphs related to the types of discontinuity
In the above graph, we can say that
At \(x=-2,\) we have a jump discontinuity
At \(x=3,\) we have a removable type of discontinuity
We will study some properties of continuous functions. Since continuity of a function at a point is related to the limit of the function at that point, hence it is reasonable to expect similar results as that of limits.Suppose \(f\) and \(g\) be two real functions continuous at a real number \(c.\) Then
(i) \(f+g\) is continuous at \(x=c.\)
(ii) \(f-g\) is continuous at \(x=c.\)
(iii) \(f.g\) is continuous at \(x=c.\)
(iv) \(\left( {\frac{f}{g}} \right)\) is continuous at \(x=c,\) (provided \(g(c)≠0\)).
Q.1. Examine whether the function \(f\) given by \(f(x) = {x^2}\) is continuous at \(x=0.\)
Ans: First, note that the function is defined at the given point \(x=0\) and its value is \(0.\) Then we find out the limit of the function at \(x=0.\) Clearly
\(\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {x^2} = {0^2} = 0\)
Thus, \(\mathop {\lim }\limits_{x \to 0} f(x) = 0 = f(0)\)
Hence, \(f\) is continuous at \(x=0.\)
Q.2. Discuss the continuity of the function \(f\) given by \(f(x) = |x|\) at \(x=0.\)
Ans: By definition
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{ – x,}&{if\,x < 0}\\
{x,}&{if\,x \ge 0}
\end{array}} \right.\)
Clearly, the function is defined at \(0\) and \(f(0)=0.\) Left hand limit of \(f\) at \(0\) is
\(\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to {0^ – }} ( – x) = 0\)
Similarly, the right hand limit of \(f\) at \(0\) is
\(\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} x = 0\)
Thus, the left hand limit, right hand limit and the value of the function coincide at \(x=0.\) Hence, \(f\) is continuous at \(x=0.\)
Q.3. Find the value of the constant \(k\) so that the function \(f\) defined below is
continuous at \(x=0,\) where \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{1 – \cos \,4x}}{{8\,{x^2}}},}&{x \ne 0}\\
{k,}&{x = 0}
\end{array}} \right.\)
Ans: It is given that the function \(f\) is continuous at \(x=0.\) Therefore, \(\mathop {\lim }\limits_{x \to 0} f(x) = f(0)\)
\( \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos 4x}}{{8{x^2}}} = k \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}2x}}{{8{x^2}}} = k \Rightarrow \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin 2x}}{{2x}}} \right)^2} = k \Rightarrow k = 1\)
Thus, \(f\) is continuous at \(x=0\) if \(k=1.\)
Q.4. Show that the function \(f\) given by \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{{e^{\frac{1}{x}}} – 1}}{{{e^{\frac{1}{x}}} + 1}},}&{if\,x \ne 0}\\
{0,}&{if\,x = 0}
\end{array}} \right.\)
Ans: The left hand limit of f at \(x=0\) is given by
\(\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to {0^ – }} \frac{{{e^{\frac{1}{x}}} – 1}}{{{e^{\frac{1}{x}}} + 1}} = \frac{{0 – 1}}{{0 + 1}} = – 1\)
Similarly,
\(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1 – \frac{1}{{{e^{\frac{1}{x}}}}}}}{{1 + \frac{1}{{{e^{\frac{1}{x}}}}}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1 – {e^{\frac{{ – 1}}{x}}}}}{{1 + {e^{\frac{{ – 1}}{x}}}}} = \frac{{1 – 0}}{{1 + 0}} = 1\)
Thus, \(\mathop {\lim }\limits_{x \to {0^ – }} f(x) \ne \mathop {\lim }\limits_{x \to {0^ + }} f(x).\) Therefore, \(\mathop {\lim }\limits_{x \to 0} f(x)\) does not exist. Hence \(f\) is discontinuous
at \(x=0.\)
Q.5. Discuss the continuity of sine function
Ans: To see this, we use the following facts
\(\mathop {\lim }\limits_{x \to 0} \sin x = 0\)
We have not proved it, but is intuitively clear from the graph of \(\sin x\) near \(0.\) Now, observe that \(f(x) = \sin x\) is defined for every real number. Let c be a real number. Put \(x=c+h.\) If \(x→c\) we know that \(h→0.\) Therefore
\( \mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} \sin x = \mathop {\lim }\limits_{h \to 0} \sin (c + h) = \mathop {\lim }\limits_{h \to 0} [\sin \sin c\cos \cos h + {\mathop{\rm cosc}\nolimits} \sinh ]\)
\( = \mathop {\lim }\limits_{h \to 0} [\sin c\cosh ] + \mathop {\lim }\limits_{h \to 0} ({\mathop{\rm cosc}\nolimits} \sinh ] = \sin c + 0 = \sin c = f(c)\)
Thus \(\mathop {\lim }\limits_{x \to c} f(x) = f(c),\) and hence \(f\) is a continuous function.
A function is continuous in its domain if there is no break in the graph of a function in the domain of the function. Mathematically, If the left-hand limit, right-hand limit and the value of the function at \(x=c\) exist and are equal to each other, then \(f\) is said to be continuous at \(x=c.\)
In this article, we have learnt about continuity, its properties, types and some solved examples. Continuity plays an important role in mathematics as it is a necessary condition for differentiation and a sufficient condition for integration.
Frequently asked questions related to continuity is listed as follows:
Q.1. How do you show continuity?
Ans: If you can draw a function’s curve on a graph without raising your pen even once, it’s considered to be continuous. If there is no interrupt/break in the graph of a function over the entire interval range, it is said to be continuous in that interval. Assume \(f\) is a real function on a subset of real numbers, and \(c\) is a point in the domain of \(f.\) If \(f\) is continuous at \(c,\) then
\(\mathop {\lim }\limits_{x \to c} f(x) = f(c)\)
Q.2. How do you define the continuity of a function?
Ans: Continuity in maths can be defined as follows:
Suppose \(f\) is a real function on a subset of the real numbers and let \(c\) be a point in the domain of \(f.\) Then \(f\) is continuous at \(c\) if
\(\mathop {\lim }\limits_{x \to c} f(x) = f(c)\)
Q.3. What are the 3 conditions of continuity?
Ans: If all three of the following conditions are met, a function \(f(x)\) is said to be continuous at a point \(x=c\) in its domain
1. The value of \(f(c)\) must exist, which should be finite value
2. \(\mathop {\lim }\limits_{x \to {c^ – }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} f(x) \Rightarrow \mathop {\lim }\limits_{x \to c} f(x)\) exists
3. \(\mathop {\lim }\limits_{x \to c} f(x) = f(c) = \) The value of \(f\) at \(x=c\)
Q.4. What is an example of continuity?
Ans: Suppose we have to know a function \(f(x) = {x^2}\) is continuous at \(x=1\) or not. We need to find its limit at \(x=1\) and the value of the function at \(x=1.\)
So, \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} {x^2} = {1^2} = 1\)
Thus, \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 1 = f\left( 1 \right)\)
Hence, \(f\) is continuous at \(x=1.\)
Q.5. How do you test for continuity?
Ans: For testing the continuity of a function, we find out the left-hand limit, right-hand limit and value at the point. If the LHL (Left-hand limit), RHL (Right-hand limit) and the value of the function at \(x=c\) exist and are equal to each other, then \(f\) is said to be continuous at \(x=c.\)