Introduction
The property of continuity exists around various aspects of nature. The water flow in the rivers is continuous. The flow of time in human life is continuous, as we are getting older continuously. Similarly, in mathematics, we have the notion of the continuity of a function. If a function have a property that their graphs can be traced with a pencil without lifting the pencil from the page, then the functions is called continuous.
Intuitively, a function is continuous at a particular point if there is no break in its graph at that point.
let’s consider some examples to realize the intuitive notion of what it means to be continuous at a point.
(a):\( f(x)=\frac{x^21}{x1}\)  (b):\( \small {f(x)= \begin{cases} x+1 & \text{for } x ≤ 2 \\ x+2 & \text{for } x > 2 \end{cases}} \)  (c): \( \small { f(x)= \begin{cases} x+1 & \text{for } x \ne 2 \\ 4 & \text{for } x = 2 \end{cases}} \) 
 In Figure (a). We see that the graph of f(x) has a hole at x=1. In fact, f(1) is undefined. So f(x) is NOT contnuous at x=1
 In Figure (b), f(2) is defined, but the function has a jump at x=2. So f(x) is NOT contnuous at x=2.
 In Figure (c), f(2) is defined, but the function has a gap at x=2. So f(x) is NOT contnuous at x=2.
Intuitive Definition of Continuity
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, defining "a" itself
Then we say that
function f(x) is continuous at x=a
if for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
f(x)f(a)<\(\epsilon\) whenever xa<\(\delta\)
In this definition
 Given a number a, we choose δneighbourhood of a, δ is positive AND can be small enough as we like such that x  a< δ
 Now, we will try to find, εneighbourhood of f(a), ε is positive AND can be small enough as satisfied such that f(x)  f(a)< ε
 If a small change in δ implies a small change ε, then the limit exists at a.
 If a small change in δ implies a LARGE change ε, or vice versa, then the limit does NOT exist at a.
Example: Continuous at x=a
How to use the applet
 Given a number a, adjust δneighbourhood of a (drag the point a or the slider δ) , so that x  a< δ , where x is any point inside the δneighbourhood
 Try to find εneighbourhood of f(a) (largest distance from f(a) , such that f(x)  f(a)< ε , where f(x) is any point inside the εneighbourhood and the result is valid for all x  a< δ
सबै x  a< δ को लागी f(x)  f(a)< ε हुने गरि εzone बानउन सकिन्छ भने continuous exist हुन्छ ।
Once a ε is found, any higher ε is always accepted.
Once a δ is satisfied, any smaller δ is always accepted.
More Explanation
The intuitive definition says that determine a number δ>0
 take any x in the region, i.e. between a+δ and a−δ, then this x will be closer to a, that is xa<δ
 identify the point on the graph that our choice of x gives, then this point on the graph will lie in the intersection of the ε region. This means that this function value f(x) will be closer to f(a) , that is f(x)f(a)<ε
Means
if we take any value of x in the δ region then the graph for those values of x will lie in the ε region.  Once a δ is found, any smaller delta is acceptable, so there are an infinite number of possible δ's that we can choose.
 the function is continuous at given x
Example: NOT Continuous at x=1
Empirical Definition of Limit
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, defining "a" itself
Then we say that function f(x) is continuous at x=a if
all of the following three conditions hold
 f(a) is defined
 \( \displaystyle \lim_{x\to a^{}}f(x)\) exist = LHL
 \( \displaystyle \lim_{x\to a^{+}}f(x) \) exist = RHL
 LHL = RHL = f(a)

A function \( f(x) = x^2 + 1 \) is continuous at 2, since
\( \displaystyle \lim_{x \rightarrow 2} f(x) =5 =f(2)\) 
A function \( f(x) = \sqrt{4x^2} \) is NOT continuous at 3
since f(3) is not defined. 
A function \( f(x)=\frac{1}{x2}\)
is not continuous at 2 because f(2) is not defined.
 The constant function, f (x) = c, \( \forall x \in R\) is continuous on R.
 The identity function, f (x) = x, \( \forall x \in R\) is continuous on R.
 The function f (x) = x n, \( x \in N\) is continuous on R.
 The function f (x) = sinx is continuous.
Example 1
Test the continuity of a function \( f(x)= \begin{cases} x^2 +2 &\text { for } x \ne 2 \\ 1 &\text { for } x=2 \end{cases} \) at x=2.
Solution
The solution is as follows
At \( x=2^{}\), the left hand limit is
\( \displaystyle \lim_{x \to 2^{}} f(x)= \lim_{x \to 2^{} } x^2+2 =2\)
LHL
At \( x=2^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 2^{+}} f(x)= \lim_{x \to 2^{+}} = x^2+2 =2\)
RHL
At x=2, the value of the function is
f(2) = 1
Functional Value
Since, limiting value and functional value are not equal, f(x) is not continuous at x=2.
Example 2
Test the continuity of a function \(f(x)=\frac{x}{x}\) for \(x\ne 0\) at x=0.
Solution
The solution is
At \( x=0^{}\), the left hand limit is
\( \displaystyle \lim_{x \to 0^{}} f(x)= \lim_{x \to 0^{} } \frac{x}{x}= \lim_{x \to 0^{} } \frac{x}{x}=\lim_{x \to 0^{} } 1=1\) LHL
At \( x=0^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 0^{+}} f(x)= \lim_{x \to 0^{+} } \frac{x}{x}= \lim_{x \to 0^{+} } \frac{x}{x}=\lim_{x \to 0^{+} } 1=1\) RHL
At x=0, the value of the function is
\(f(0)= \frac{x}{x}=\frac{0}{0}=\) Indeterminate f(x)
At x = 0, the value of f(0) is not defined.
So, the function is NOT continuous at x=0.
Example 3
Is\( f(x)=\frac{x+4}{x3}\) continuous at x =1? At x =3?
Solution
The solution is as follows
At x=1, the limit value is
\( \displaystyle \lim_{x \to 1} f(x)= \lim_{x \to 1}\frac{x+4}{x3}=\lim_{x \to 1} \frac{1+4}{13} =\frac{5}{2}\)
This is Limit Value
At x=1, the value of the function is
\( \displaystyle \lim_{x \to 1} f(x)= \lim_{x \to 1} \frac{1+4}{13} =\frac{5}{2}\)
This is Functional Value
Since, limiting value and functional value are equal, f(x) is continuous at x=1.
Next,
At x=3, the limit value is
At \( x=3^{}\), the left hand limit is
\( \displaystyle \lim_{x \to 3^{}} f(x)= \lim_{x \to 3^{}} \frac{x+4}{x3}=\lim_{h \to 0 } \frac{(3h)+4}{(3h)3}=\lim_{h \to 0 } \frac{7h}{h}=\infty\)
LHL
At \( x=3^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 3^{+}} f(x)= \lim_{x \to 3^{+}} \frac{x+4}{x3}=\lim_{h \to 0 } \frac{(3+h)+4}{(3+h)3}=\lim_{h \to 0 } \frac{7+h}{h}=\infty\)
RHL
At x=3, the value of the function is
\( \displaystyle \lim_{x \to 3} f(x)= \lim_{x \to 3} \frac{x+4}{x3}= \frac{3+4}{0}=\infty\)
f(x)
Since, LHL ≠RHL, f(x) is NOT continuous at x=3.
Discontinuity of a function
When a function is not continuous at a point, then we can say it is discontinuous at that point. There are several types of behaviors that lead to discontinuities. These several types can be classified into two major types of discontinuity, they are given below. Removable discontinuity
 Missing point discontinuity
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a.
Then we say that function f(x) has missing point discontinuity at x=a if
 f(a) is not defined
 LHL = RHL
 Isolated point discontinuity
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a.
Then we say that function f(x) has issolated point discontinuity at x=a if
 f(a) is defined
 f(a)≠LHL = RHL
 Missing point discontinuity
 Nonremovable discontinuity
 Finite point discontinuity
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a.
Then we say that function f(x) has finite point discontinuity at x=a if
 finite=LHL≠RHL=finite
onesided limits are both finite, yet not equal to each other
One More Example is given below.
Test the continuity of a function \( f(x)= \begin{cases} 4x3 & x < 2\\ (x3)^2 & x ≥ 2 \end{cases} \) at x=2.  finite=LHL≠RHL=finite
 Infinite discontinuity
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a.
Then we say that function f(x) has infinite discontinuity at x=a if
 At least one of LHL or RHL is infinite
One more example is given below
At what point is the function \(f(x)=\frac{x+1}{(x2)(x3)}\) (i) discontinuous (ii) continuous ?  Oscillatory discontinuity
Let 𝑓(𝑥) be a function defined at all values in an open interval containing a.
Then we say that function f(x) has oscillatory discontinuity at x=a if
the values of the function appear to be approaching two or more values simultaneously.
 consider a function \( f(x)=\sin \frac{1}{x} \)
 take \( x=\frac{1}{\frac{\pi}{2} (2n+1)};n \in Z \)
 f(x) jumps alternatively to 1 and 1 as x approaches to 0 because \( \frac{1}{x}=\frac{\pi}{2} (2n+1);n \in Z\)
 \( f(x)=\sin \frac{1}{x} \) has oscillating discontinuity at x=0
 Finite point discontinuity
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