Mathematics Revision Notes of Continuity of functions for NDA
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Revision Notes of Continuity of functions for NDA

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Introduction :

- The graph of a function is said to be continuous at x = a if while travelling along the graph of the function and in crossing over the point at `x = a` either from Left to Right or from Right to Left one does not have to lift his pen.

- ln case one has to lift his pen the graph of the function is said to have a break or discontinuous at `x = a`. Different type of situations which may come up at `x = a` along the graph can be :

Revision of Continuity For NDA :

`"(A) Continuity at a point :"`

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

`lim_(x-> c) f(x) = f(c)`

`lim_(x-> c^+) f(x) = lim_(x-> c^(-)) f(x) = f(c)`

or `f(c^+)= f(c^(-)) = f(c)`

or `lim_(h->0) f(c+h) =lim_(h-> 0) f (c-h) = f(c)`


`"(B) Continuity in an interval : "`

- A real function `f` is said to be continuous if it is continuous at every point in the domain of `f`.

- Suppose `f` is a 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-> a^+) f(x) =f(a)`

and continuity of `f` at `b` means `lim_(x-> b^(-)) f(x) =f(b)`

Observe that only `lim_(x-> a^(-))` and ` lim_(x-> b^(+))` do not make sense. As a consequence of this definition, if f is defined only at one point, it is continuous there, i.e., if the domain of `f` is a singleton, `f` is a continuous function.

Some important points about continuity

(i) Continuity at `x = a =>` existence of limit at `x = a`, but not the converse

(ii) Continuity at `x = a =>` f is well defined at `x = a`, but not the converse

(iii) Discontinuity at `x = a` is meaningful to talk if in the immediate neighborhood of `x = a`, i.e. the function has a graph in the immediate neighborhood of `x = a`, not necessarily at `x = a`.

(iv) Continuity is always talked in the domain of function and hence `f(x) = 1/(x-1) , 1/x tan x` are all continuous functions but if you want to talk of discontinuity then we can say `1/(x-1)` is discontinuous at `x=1 , 1/x` is discontinuous at `x = 0`.

Note that all rational functions are continuous. Because continuity is always talked in the domain of `f(x)`.

Types Of Discontinuity :

`"1. Removable Discontinuity :"`

Here `lim_(x->a) f(x)` necessarily exists, but is either not equal to `f(a)` or `f(a)` is not defined. In this case, therefore it is possible to redefine the function in such a manner that `Lim_(x-.a) f(x) =f(a)` and thus making the function continuous. These discontinuities can be further classified as :

(A) Missing point discontinuity :

Here `Lim_(x->a) f(x)` exists. But `f(a)` is not defined.

(a) `f(x) =( ((x-1) (9-x^2) )/(x-1) ) x ne 1`

at `x = 1, f(1)` is not defined. Hence `f(x)` has missing point of discontinuity at `x = 1`.

(b) `f(x) =(sin x)/x , x ne 0`
`f(0)` is not defined. `f(x)` has missing point of discontinuity at `x = 0`.

(B) Isolated point discontinuity :

Here `Lim_(x->a) f(x) ` exists, also `f(a)` is defined but `Lim_(x->a) f(x) ne f(a)`


(a) `f(x) = [x] + [-x] = [ tt ( (0, text(if) x in I) , (-1 , text(if) x notin I) )`

has isolated point of discontinuity at all integeral points.

`"2 . Non -Removable Discontinuity :"`

- Here `Lim_(x->a) f(x)` does not exists and therefore it is not possible to redefine the function in any manner to
make it continuous. Such discontinuities can be further classified into 3 fold.

(a) Finite type (both limits finite and unequal) :

(i) `Lim_(x->0) tan^(-1) (1/x)= [tt( (f(0^+) =pi/2) , ( f(0^-) = - pi/2) )` ; jump `= pi`

(ii) `Lim_(x->0) (|sin x |/x)= [tt( (f(0^+) =1) , ( f(0^-) = - 1) )` ; jump `= 2`

(b) Infinite type (at least one of the two limit are infinity) : (i) `f(x) = (x/(1-x))` at `x=1 ,[tt( (f(1^+) =-oo) , ( f(1^-) = + oo) )`

(i) `f(x) = 2^(tanx)` at `x=pi/2 ,[tt( (f((pi/2)^+) =0) , ( f((pi/2)^-) = oo) )`

(c) Oscillatory (limits oscillate between two finite quantities) : (i) ` tt ( (f(x) =sin( 1/x) ) ,( text (or)) ,( f(x) =cos (1/x)) ) ]` at `x=0` oscillates between `- 1` & `1`
Q 2605056868









Column IColumn II
(A) `f(x) =1/(x-1)` (P) Removable discontinuity
(B) `f(x) =(x^3-x)/(x^2-1)` (Q) Non-removable discontinuity
(C) `f(x)=(|x-1|)/(x-1)` (R) Jump of discontinuity
(D) `f(x)= sin (1/(x-1))` (S) Discontinuity due to vertical asymptote
(T) Missing point discontinuity
(T) Oscillating discontinuity


(A)

`A-> (S),(Q); quad B-> (T),(P); quad C->(R),(Q), quad D->(U),(Q)`

(B)

`A-> (S),(T); quad B-> (S),(P); quad C->(R),(U), quad D->(Q),(Q)`

(C)

`A-> (S),(Q); quad B-> (R),(Q) ; quad C->(T),(P), quad D->(U),(Q)`

(D)

`A-> (S),(Q); quad B-> (R),(Q) ; quad C-> (U),(Q), quad D->(T),(P)`

Solution:

Correct Answer is `=>` (A) `A-> (S),(Q); quad B-> (T),(P); quad C->(R),(Q), quad D->(U),(Q)`

Theorems on Continuity :

Sum, difference, product and quotient of two continuous functions is always a continuous function.

However `h(x) = (f(x))/(g(x))` is continuous at `x = a` only if `g(a) ne 0`.

Following Important Notes Should be Remembered :

(a) lf `f(x)` is continuous and `g (x)` is discontinuous then `f(x) + g (x)` is a discontinuous function.

(b) If `f(x)` is continuous & `g(x)` is discontinuous at `x = a` then the product function `phi (x) = f(x) * g(x)` is not
necessarily be discontinuous at `x = a`.

e.g. `f(x) = x` & `g(x) = [ tt ((sin \ 1/x , x ne 0),(0, x=0))`

Then `f(x) * g(x) = { tt ((x sin \ (1/x), x ne 0),(0, x=0)) ` is continuous at `x = 0`.

(c) If `f(x)` and `g(x)` both are discontinuous at `x = a` then the product function `phi (x) = f (x) * g(x)` is not
necessarily be discontinuous at `x = a`.

e.g., `f(x) =-g(x) = [ tt ((1 , x ge 0),(-1, x < 0))`

`:. f(x) g(x) = 1 AA x in R` which is continuous function.
 
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