MATHEMATICS - CONTINUITY AND DIFFERENTIABILITY - CLASS 12

MATHEMATICS 

CONTINUITY AND DIFFERENTIABILITY

CLASS 12

Calculus is one of the most important topics in Math. Especially class 11 and class 12 students need to have a clear understanding about the concepts present in calculus as it has a widespread application in many fields.  The two main topics required for better understanding of Calculus are Algebra and Trigonometry.

This is about the continuity of a function. The continuity of all the six trigonometric ratios have been discussed here. Continuity of other functions like modulus, polynomials have also been explained in detail.

Some important NCERT sums on continuity have also been included. Class 12 students can go through this to get an idea of how to prove continuity of a function.

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

 lim  f(x)  =  f(c) 

 x→c

That is, if lim f(x)   =  f(c)  =  lim  f(x)

                x→c⁻                      x→c⁺

exist and equal to each other, then f is said to be continuous at x = c.

Prove the following:

1. The sine function is everywhere continuous.

Let f(x) = sin x  and let a be any real number.

lim    f(x)  =   lim    f ( a + h ) 

x →a⁺            h→0

                 =   lim  sin ( a + h )

                       h→0

                 =   lim   ( sin a cosh  +  cos a sin h )

                       h→0

                 = sin a lim cos h  + cos a lim sin h 

                            h→0                     h→0

                 = sin a  x  1    +     cos a  x 0

                 = sin a

   lim  f(x) =  lim   f ( a - h )

 x→a⁻            h→0

                  =  lim  sin (a - h )

                      h→0

                  =  lim  (sin a cos h  -  cos a sin h)

                      h→0

                  =  sin a lim cos h   -  cos a lim sin h

                              h→0             h→0

                  =  sin a  x  1  -  cos a  x  0

                  =  sin a

Functional value:    f(a)  =  sin a

Therefore lim f(x)   =   f(a)   =  lim f(x)

                x →a⁻                  x→a⁺

Hence sine function is everywhere continuous.

2. The cosine function is everywhere continuous.

      Let  f (x )  =  cos x and let a be any real number.

      lim  f(x)   =  lim  f(a+h)

     x→ a+          h→0

                      =  lim cos (a+h)

                          h→0

                     =  lim (cos a cos h - sin a sin h)

                          h→0

                     = cos a lim cos h - sin a lim sin h

                                  h→0                 h→0

                     =  cos a  x  1   -   sin a  x  0

                     =  cos a

Similarly lim  f(x)  =  cos a

              x →a⁻

Functional value:  f(a)  =  cos a

Therefore    lim f(x)  =  f(a)  =  lim f(x)

                   x→a⁻                     x→a⁺

Hence cosine function is everywhere continuous.

3. The tangent function is continuous in its domain.

  f(x)  =  tan x = sin x  /  cos x

This function is defined for all real numbers such that cos x is not equal to zero.

  That is x is not equal to (2n+1) π/2 ,  n ∊  Z.

It is enough that we prove for sin x and cos x (proved above) and say that the tangent function being a quotient of two continuous functions is continuous wherever it is defined.

f(x) =  tan x   is continuous for all 

                          x ∈  R - { (2n+1) π /2 : n ∊ Z}

Hence the tangent function is continuous in its domain.

4. The cosecant function is continuous in its domain.

  cosec x = 1 / sin x

It is enough that we prove for the continuity of sine function (proved above) and mention the following:

If a function f(x) is continuous on D (common domain), then the function 1/f (x)  is continuous on D - {x: f(x) is not equal to zero}.

cosec x is continuous except for x = n π , n ∊ Z.

Hence the cosecant function is continuous in its domain.

5. The secant function is continuous in its domain.

    sec x  = 1 / cos x

It is enough that we prove for the continuity of cosine function (proved above) and mention the following:

If a function f(x) is continuous on D (common domain), then the function 1 / f(x) is continuous on D - {x f(x) is not equal to zero}.

Sec x is continuous except for x = (2n+1)π/2, n ∊ Z

Hence the secant function is continuous in its domain.

6. The cotangent function is continuous in its domain.

  cot x = 1 / tan x

It is enough that we prove for the continuity of  cosine and sine functions (proved above) and say that cot x being a quotient of two continuous functions is continuous wherever it is defined.

cot x is continuous except for x = nπ  , n ∊ Z.

Hence the cotangent function is continuous in its domain.

7. Discuss the continuity of the following:

(i) f (x) = sin x + cos x

It is enough to prove the continuity of sin x and cos x (proved above) and then mention that if two functions are continuous, then their sum is also continuous.

Hence f(x) = sin x + cos x is a continuous function.

(ii) f(x) = sin x - cos x

It is enough to prove the continuity of sin x and cos x (proved again)  and then mention that if two functions are continuous, then their difference is also continuous.

Hence f(x) = sin x - cos x is a continuous function.

(iii)f (x) = sin x . cos x

It is enough to prove the continuity of sin x and cos x (proved above) and then mention that if two functions are continuous ,  then their product is also continuous.

Hence f(x) = sinx . cosx is a continuous function.

8. Find all points of discontinuity of f, where

                f(x) = sin x / x,  if x < 0

                            x+1 ,    if x  ≥ 0

  lim   f(x)  =  lim  f(0-h)

x→0⁻             h→0

                  = lim f(-h)

                     h→0

                   = lim  sin (- h ) / (-h)

                      h→0

                  = lim  (-sin h) / -h

                    h→0

                  = lim sinh / h

                     h→0                          

                  = 1

lim f(x)      =  lim  f(0+h)

x→0 ⁺              h→0

                   =  lim  f(h)

                       h→0

                   =  lim (h+1)

                       h→0

                   = 1

Functional value: f(x=0) = 0 + 1 = 1

Therefore L.H.L = Functional value = R.H.L                    

Hence f is continuous.

There is no point of discontinuity.

(Note: L.H.L  - Left Hand Limit

          R.H.L  - Right Hand Limit)

9. Determine if f defined by

     f(x) = x^2 sin (1/x ) , if x ≠ 0

                0 ,     if x = 0     is a continuous function?

lim  f(x)  = lim  f(0-h)

x→0⁻         h→0

               = lim f(-h)

                  h→0

               = lim (-h)^2 sin (1/-h)

                  h→0

               = lim  -h^2 sin (1/h)

                  h→0

              =  0

lim  f(x)  = lim f(0+h)

x →0⁺       h→0

                = lim f(h)

                  h→0

               = lim h^2 sin(1/h)

                  h→0

               = 0

functional value f(x=0) = 0

Therefore L.H.L = Functional value  = R.H.L

Hence f is a continuous function.

10. Examine the continuity of f, where f is defined by

            f(x)  =  sin x - cos x,    if x ≠ 0

                         -1     ,  if x = 0

lim  f(x)  = lim f (0-h)

x→0⁻         h→0

               = lim f(-h)

                  h→0  

               = lim  (sin (-h)  -  cos(-h))

                  h→0

               =lim (-sin h  -  cos h)

                  h→0

               = -0-1  = -1

lim  f(x)  = lim f(0+h)

x→0⁺       h→0

               = lim  f(h)

                  h→0

               = lim (sin h  -  cos h)

                  h→0

               = 0 - 1  =  -1

Functional value: f(x=0)  = -1

Therefore L.H.L  = Functional value =  R.H.L

Hence f is continuous.

11. Show that the function defined by f(x) = cos(x^2) is a continuous function.

This can be proved in two ways:

Either by taking a real number a and then determining the L.H.L , R.H.L and the functional value which will be cos a^2 if proved by the method given above.

OR

By using the theorem of continuity of composite functions.

The given function f(x)=cosx^2 is defined for every real number.

Consider f as a composition goh of two functions g and h such that g(x) = cosx  and h(x) = x^2

  goh (x) = g(h(x))  = g(x^2)  =  cos x^2

We know that cosx is a continuous function and x^2 being a polynomial function is also continuous.  

If two functions g and h are continuous, then the composition of functions goh is also continuous.

Hence f(x)=cosx^2 is a continuous function.

12. Show that the function defined by f(x)=|cosx| is a continuous function.

Let a be any real number.

lim f(x) = lim f(a-h)

x→a⁻       h→0

             = lim |cos(a-h)|

                h→0

             = |cosa|

lim f(x) = lim f(a+h)

x→a⁺      h→0

             = lim |cos(a+h)|

                 h→0

             = |cosa|

functional value: f(x=a) = f(a) = |cosa|

Therefore L.H.L  = Functional value  = R.H.L

Hence f(x)=|cosx| is a continuous function.

13. Examine that sin|x| is a continuous function.

Let f(x) = sin|x| and let a be any real number.

lim  f(x)  =  lim f(a-h) 

x→a⁻          h→0

               = lim sin |a-h|

                 h→0

               = sin |a|

lim f(x)  =  lim f(a+h)

x→a⁺        h→0

              =  lim  sin |a+h|

                  h→0

              =  sin |a|

Functional value:  f(x=a) = sin|a|

Therefore L.H.L = Functional value = R.H.L

Hence f(x)=sin|x| is a continuous function.

14. Find all the points of discontinuity of f defined by                   f(x)= |x| - |x+1|

By the definition of Modulus function, we have

f(x)  = -x+x+1  , x<-1

           -x - (x+1)  ,  -1 ≤ x < 0

            x - (x+1)  ,  x ≥ 0

On simplifying, we get 

f(x)  = 1 ,   x < -1

          -2x -1 ,     -1  ≤  x < 0

          -1 ,    x  ≥ 0           

It is clear that f(x) is continuous for all x satisfying x<-1, -1<x<0 and x>0.  So, we need to find whether f is continuous at x=-1 and x=0.

Let us discuss continuity of f at x=-1

lim f(x)  = lim 1 = 1

x→- 1⁻     x→-1⁻

lim f(x)  =  lim     (-2x-1)  = -2(-1) -1  = 2 -1  = 1

x→-1⁺       x→-1⁺

functional value: f(x=-1)  = -2(-1) -1 = 2-1 = 1

Therefore L.H.L = Functional value = R.H.L

So f(x) is continuous at x = -1.

Continuity at x=0 :

lim f(x)  = lim (-2x-1)  = -2(0) -1 = -1

x→0⁻        x→0⁻   

lim f(x)  = lim -1  =  -1

x→0⁺       x→0⁺

functional value f(x=0)  = -1

Therefore L.H.L  = Functional value = R.H.L

So f(x) is continuous at x=0

Hence f(x) is everywhere continuous.