LCDLimits

Consider a function

f(x)= (x2-a2)/(x-a)

        The function f(x) is defined at all points except at x = a, because at x = 1, f(x)= 0/0 because at x = a, f(x)= 0/0 which is indeterminate in the language of mathematics, hence we want to know what value does f(x) approach as x approaches a.

 Let y = f(x) be a given function defined in the neighbourhood of x=a, but not necessarily at the point x = a. The limiting behaviour of the function in the neighbourhood of x = a when |x - a| is small, is called the limit if the function when x approaches 'z' and we write this as limx→a f(x).

Let limx→a f(x) = l. It would simply mean that when we approach the point x = a from the values which are just greater than or just smaller than x = a, f(x) would have a tendency to move closer to take value 'l' This is same as saying, "difference between f(x) and l can be made as small as we like by suitably choosing x in the neighbourhood of x = a". Mathematically, we write this as, limx→af(x) = l, which is equivalent to saying that |f(x)-l|< ε for all x such that 0 < |x - a| < δ and ε depends on δ where ε and δ are sufficiently small positive numbers.

It should be clear that the limit of f(x) at x=a would exist if and only if, f(x) is well defined in the neighbourhood of x = a (not necessarily at x = a) and has a unique behaviour in the neighbourhood of x = a.

  Note: Normally students have the perception that limit should be a finite number. But it is not really so. It is quite possible that f(x) had infinite limit as x → a. If limx→a f(x) = ∞, it would simply mean that function has tendency to assume very large positive values in the neighbourhood of x = a, for example limx→0 1/|x|= ∞.

Illustration:

        limx→0 x4/x2      (0/0 form)

        limx→0 x2 = 0

Illustration:

        limx→0 x4/x2       (0/0 form)

       = limx→0 1/x2

       = a

Thus, we observe that value of 0/0 form could be zero or infinity i.e. its value depends upon the problem at hand. Hence 0/0 form is indeterminate form.

It is important to know the various forms of indeterminacy.

If the value of a function at and point acquires one of the forms 0/0; 0 × ∞; ∞/∞; ∞-∞;00;1∞,∞0 then the function is said to be indeterminate at that point. (The first of these is said to be standard indeterminate form.)

        It is again important to note that one is not allowed to manipulate the function i.e. we cannot write (x2-a2)/(x-a) = (x+a)(x-a)/((x-a)) = (x+a) because this we can do only if (x-a) ≠ 0 or x ≠ a. At x=a this manipulation is not valid due to such restriction, as the function, locating a break in the value of f(x) at x = a is almost impossible. So we can interpret that value of the function at point infinitesimally close to a exists and is finite.

  

Illustration:

        Graph of y = (x2-a2)/(x-a)(x≠a) is shown in figure given below circle at point (a, 2a) means that the point is excluded from the graph of the function.

It is for this reason that we define limits and also get rid of the manipulation or simplification constraint of function. So the moment we write

        limx→a (x2-a2)/(x-a) we can write limx→a (x+a)(x-a))/(x-a) = limx→a(x+a) = 2a

(Remember limx→a  x-a≠0)

Now continuing with same argument and considering the same example we can infer whether the limit exists at a point where the function is not defined  

The condition for the existence of limit if a function at x = a is Limit of f(x) as x approaches 'a' From Left

= Limits of f(x) as x approaches 'a' From Right

i.e. Left Hand Limit (LHL) at x = a= Right Hand Limit (RHL) at x = a or mathematically

        limx→a-  f(x) = limx→a

+  f(x)

=>        For figure given below Limit exist, L is the value of limit whether x approaches 'a' from left or right, even though f(a)  is not defined

 

 

For figure given below Limit exist, L is limit when x approaches 'a' from left and L2 is limit as x approaches 'a' from right.

 

Note:         Graphically one can infer that id there is no sudden change or break of value of f(x) slightly left and slightly right of point x=a, it means existence of limit of f(x) at x=a.

  

Illustration:

Find limx→2 (3x+1)

Solution:  

Let f(x)= (3x+1)

f(1.9) = 3 × 1.9 + 1 = 6.7

f(1.99) = 3 (1.99) + 1 = 6.97

f(1.999) = 3 (1.999) + 1 = 6.997

Thus we observe that as x→2-, f(x)→7

Consider,   f(2.1)=7.3

                f(2.01) = 3 (2.01) + 1

                          = 7.03

                f(2.001) = 7.003

i.e. as x → 2+, f(x) → 7+

         we can write

                limx→2 (3x+1) = 3(2)+1 = 7

We, observe that, as long as we do not get one of the indeterminate forms, we can put the value of x in the function in the limiting case as well.

Now we define limit mathematically

        'l' is said to be limit of f(x), at x = a, if for an arbitrary small number ∈ > 0 there exist δ > 0 such that

|f(x)-l|<0 for all 0<|x-a|<δ i.e. ∈ is independent and δ is dependent on ∈.

i.e. Difference between 'f(x)' and 'l' become negligibly small when 'x' is sufficiently close to 'a', |x-a|<δ is called the neighbourhood of 'a' means x can take any value from the interval (a δ,a+δ).

 

Mathematically we write as limx→a  f(x) = l

Left and Right Limit

Let y = f(x) be a given function, and x = a is the point under consideration. Left tendency of f(x) at x = a is called it's left limit and right tendency is called it's right limit.

Left tendency (left limit) is denoted by f(a - 0) or f(a-) and right tendency (right limit) is denoted by f(a + 0) or f(a+) and are written as

where 'h' is a small positive number.

Thus for the existence of the limit of f(x) at x = a, it is necessary and sufficient that f(a-0)  = f(a + 0), if these are finite or f(a-0) and f(a+0) both should be either + ∞ or -∞.

 

Remark:    For the existence of the limit at x = a, f(x) need not be defined at x = a. However if f(a) exists, limit need not exist or even if it exists then if need not be equal to f(a).

  

Illustration:      For what values of m does the limx→2 f(x) exist when

                       

Solution:   limx→2-  f(x) =  limx→2

-  f(mx-3) = 2m - 3;

                 limx→2+  f(x) =  limx→2

+  x/2 = 2/m

                 limx→2 f(x) exists when  limx→2- f(x) =  limx→2

+ f(x)

                ⇒ 2m - 3= 2/m ⇒ 2m2 - 3m -2 = 0 ⇒ m = -1/2, 2.

 Algebra of limit

Suppose limx→a f(x) = α and limx→a g(x)=β then we can define the following rules:

  

Caution:    The above rules are applicable only when both limx→a f(x)

              And limx→a g(x) exist separately

Let us try to prove some of these Rules:

 

(i)     limx→a  f(x) + g(x) = limx→a f(x) + limx→a g(x)

        To prove the above result we choose ∈, then by definition of limit, there exist δ1 and δ2 both greater than zero such that

       

 

Let δ= min (δ1, δ2) then by algebra, we have |f(x) + g(x0 -α-β| < |f(x)g(x) - αβ| < ∈ whenever 0 < |x - α| < δ

Hence limx→a |f(x)g(x) | = limx→a f(x) limx→a g(x)

We have to prove that for ∀∈>0, there exist some δ>0 such that             |f(x)g(x)-αβ |<∈whenever 0<|x-a|<δ

Now |f(x)g(x) - αβ| = |f(x) g(x) - αg(x) + αg(x) - αβ| < |f(x) g(x) - αβ(x)| + |αβ(x) - αβ| = ||f(x) - α|| g(x) - β|

From definition ∀ ∈ >0,∃δ_1,δ_2>0 such that |f(x)-α |<∈/2A.

Whenever 0<|x-a|<δ1

Where A is the upper bound of g(x) in the neighbourhood of x - a. and       |g(x)-β | < ∈/2|α| whenever 0<|x-α|<δ (proved)

  

Illustration:      Find limx→3  (2x3-3x2-x-1)

Solution:

                

 

Enquiry:    Can a function have two distinct limits as x approaches 'a'?

If limx→a f(x) exists then it is unique i.e. there cannot be two distinct number L1 and L2 such that when x ®  a the function tends to both L1 and L2. Hence if limx→a f(x)=L1 and limx→a f(x)L2 then L1=L2.

  

Some important theorems:

1.           Let f < g in an open interval containing 'a' then limx→a f(x) < limx→a g(x)

2.           Sandwich Theorem: Let f, g, h be three continuous function such that f<g<h in an open interval containing 'a' and suppose limx→a f(x) = limx→a h(x) = l. Then limx→a g(x) also exists and is equal to 'l'

Illustration:

As shown in figure given below, we have f ≤ g ≤ h

In (x1, x2) and as x approaches a.

        limx→a f(x) = limx→a  h(x) limx→a g(x) = 1.

3.           If 0 ≤ f≤ g in an open interval containing a and limx→a g(x)=0 then limx→a f(x) exists and is equal to zero.

4.           limx→a f(x)=0 if and only if  limx→a f(x)

 

Some Important Results on Limits:

        

Algebra of limit

Frequently Used Series Expansions

  Following are some of the frequently used series expansions:

 

    

Illustration:

Using the formulae evaluate the followings

  

Solution:   

(i)      Method 1:

limx→0 sin3x/x = limx→0 3 sinx - 4 sin3 x/x

= limx→0 3(sinx/x) - limx→0 4(sinx/x) (sinx)2

= 3×1-4×1×0=3

 

 Method 2:

 limx→0 sin3x/x

= lim3x→0 (sin3x/3x)3

= 1×3=3

 

(ii)   limx→0 (1-x)1/x

= limx→0  (((1+(-x))1/-x)-1

= 1/e

 

  Note: we know that

                   √(x2 ) = |x| = x ∀ x ≥0

                             = -x ∀ x <0

 

(iii)    limx→0+  |x|/x = limx→0

+  x/x = limx→0+ 1 = 1

 

(iv)  ∴   limx→0- |x|/x = limx→0

- (-x)/x = limx→0- -1=-1

       ∴   limx→0+ |x|/x ≠ limx→0

- |-x|/x

      ∴   limx→0 |x|/x does not exist

         does not exist

 

Algebra of limit

(v)    limx→5+ y = limx→5

+ f(x)                   let x=5+h, where h>0

                =limh→0 f(5+h)(as x→5+,h→0)

                =limh→0 (5+h)2

                =(5+0)2=25                                             ......(i)

limx→5- y = limx→5

- f(x) = limh→0 f(5-h)         let x = 5-h, h>0

        = limx→0 (4(5-h)+5) = (4(5-0)+5) = 25

                                                As x→5-. h→0             ......(ii)

From (i) and (ii) we conclude

∴ limx→5 y=25

 

Illustration:

Evaluate x → limx→2 (x2-3x+2)/(x2-x-2)

Let f(x) = (x2-3x+2)/(x2-x-2)

  

Solution:  

Method 1:

(i)     Let x→ 2+

        i.e. x = 2+h, h→ 0, h>0

        limx→2+  f(x)

        = limx→0  f(2+h)

        = limh→0 ((2+h)2 - 3(2+h)+2)/((2+h)2 - (2+h)-2)

        = limh→0  h(1+h)/h(3+h) (since h≠ 0, we cancel h)

        = limh→0 (1+h)/(3+h) = (1+0)/(3+0) = 1/3

(ii)    Let x→ 2- i.e. x=2-h, h→ 0, h>0

      limx→2-  f(x) = limh→0 f(2-h)

      = limh→0  ((2-h)2 - 3(2-h)+2)/((2-h)2-(2-h)-2)

     = limh→0 ((h2-h)/(h2-3h)) = limh→0 (h-1)/(h-3) = (0-1)/(0-3) = 1/3

     ∴ limx→2+ f(x) = limx→2

- f(x) = 1/3

 

Method 2:

           limx→2 (x2-3x+2)/(x2-x-2)

       = limx→2 (x-1)(x-2)/(x+1)(x-2)

        = limx→2 ((x-1))/((x+1) )=1/3

 

Limits at infinity and infinite limits

        When the term limits at infinity is used, it obviously means functions limits to be evaluated as x→ ∞

For example limx→∞ 1/x2 =0

The term infinite limit means that when x tends to a particular value 'a'. then the limit of the function tends to infinity i.e.

limx→2  f(x) = ∞

Evaluation of the limit at infinity in a problem is solved by changing the expression f(x) to g(1/x) form. Here, we have to find the value of the function f(x) as x→∞.

  

Illustration:

        limx→ ∞ ((3x-1)(4x-2))/((x-8)(x-1))

        limx→ ∞ ((3-1/x)(4-2/x))/((1-8/x)(1-1/x))

Solution:  

          Given expression has been changed from f(x) to g(1/x) form.

       

 

Note:  When infinite limit is encountered i.e. '0' in the denominator as x→a, then either (x-a) get cancelled from numerator and denominator both and if it is not possible then the limit's value is put as infinity. Such limits are called improper limits i.e. limx→∞  1/x2 =∞..

  

Illustration:

        limx→∞  (x2+3x+1)/(x3-2x+1)

 

Solution:  

Since, the highest power of x is '3', we divide both numerator and denominator by x3

  

 Limits using Sandwich Theorem

Limits using Sandwich Theorem

 Let f(x).g(x) and h(x) be there real numbers having a common domain D such that h (x) ≤ f(x) ≤ g(x) ∀ x ∈ D. If limx→a h(x) - limx→a g(x)=l. This is known as Sandwich Theorem.

 

Illustration:      Examine limx→0 √(1-cos x )/x

Solution:

               

 

Illustration:      Evaluate the following limits, if these exist;

                        (i) limx→0  (ax-1)/(√(1+x)-1)

                        (ii) limx→0 (sin x )x

 

Solution:  

 

Clearly in this case (sin x)x in not defined towards the left of x=0.

Hence the given limit will not exist.

 

Illustration:   Prove that limn→∞ x5 [1/x3 ] = 0, where [.] denotes the greatest

                   integer function.

Solution:           We know that x - 1 <|x| ≤ x ∀ x ∈ R

                    So, 1/x3 -1 < [1/x3 ] ≤ 1/x2

                    ⇒ x5 (1/x3 -1) < x5 [1/x3 ] ≤ x5 (1/x3 ) for x > 0

                   And limx→0  x2 - x5 = limx→0) x2 = 0

                    ⇒ limx→0+  x5 [1/x3 ] = 0

                    Also, x5 (1/x3 ) ≤ x5 [1/x3 ] < x5 (1/x3 -1) for x<0

                  ⇒ limx→0-  x5 [1/x3 ] = 0   => limx→0  x5 [1/x3 ] = 0

 

Illustration: Evaluate limn→∞ ([x]+[2x]+[3x]+ ...........=[nx]) / n2 ,

                     where [.] denotes the greatest integer function.

Solution: We know that x - 1 < x

     ⇒ x + 2x + .......... + nx -n < ∑r=1n [rx] ≤ x + 2x +  ......... + nx

     ⇒ (x.n (n+1))/2-n  < ∑r=1n [rx] ≤ (x.n (n+1))/2

     ⇒ x/2 (1+1/n)-1/n < 1/n2 ∑r=1n [rx] ≤ x/2 (1+1/n)

   Now, limn→∞ x/2 (1+1/n) = x/2 and limn→∞ x/2 (1+1/n)-1/n = x/2.

   Using Sandwich theorem we find that limn→∞ ([x]+[2x]+ ..........+[nx])/n2 = x/2

 

Alternative solution:

We know that [rx] = rx = rx -{xr}

for r=1, 2, 3, ........ n and 0 ≤{x_r }<1 for each r.

Also ∑r=1n [rx] = ∑r=1

n ((rx)-{xr } )

                     = x∑r=1n r - ∑r=1

n {xr} = x (n(n+1))/2-k

where k < n (since each {xr}<1).

 Hence limn→∞  1/n2∑r=1n  [rx] = limn→∞ ∑r=1

n [x/2 (1+1/n)-k/n2 ] = x/2

 

Illustration:      Evaluate the following limits, if these exist. Here {x} denotes the fractional part and [.] the greatest integer part.

(i)                  limx→0 (x tan 2x - 2x tan x )/(1-cos 2x )2

(ii)                 limx→∞ (1+1/(a+bx))c+dx  (a, b, c, d are positive)

(iii)                 limx→3 ([x]-3)/((x-3))

Solution:          

    

    (iii) limx→3 ([x]-3)/(x-3)

           Towards the right of x = 3, [x]=3

         ⇒[x] -3=0, in the right neighbourhood of x=3

         ⇒limx→3+0 ([x]-3)/(x-3)=0

             Towards the left of x=3, [x]=2

         ⇒[x] -3=-1, in the left neighborhood of x =3

         ⇒limx→3+0 ([x]-3)/(x-3) = limx→3+0 (-1)/(x-3) = ∞.

           Thus limx→3 ([x]-3)/(x-3) does not exist.

 

 L’ Hospital’s rule

L' Hospital's Rule

We have dealt with problems which had indeterminate from either 0/0 or ∞/∞ .

The other indeterminate forms are ∞-∞,0,∞,00,∞0,1∞

We state below a rule, called L' Hospital's Rule, meant for problems on limit of the form 0/0 or ∞/∞  .

Let f(x) and g(x) be functions differentiable in the neighbourhood of the point a, except may be at the point a itself. If  limx→a f(x) = 0 = limx→a g(x) or limx→a f(x)= ∞ = ∞  g(x), then limx→a f(x)/g(x) = limx→a f' (x)/g(x) = limx→a f' (x)/g'(x)    provided that the limit on the right either exists as a finite number or is ± ∞ .

 

Illustration:

        Evaluate  limx→1 (1-x+lnx)/(1+cos π x )

Solution:

  limx→1 (1-x+lnx)/(1+cos π x ) (of the form 0/0)

        = limx→1 (1-1/x)/(-π sin π  x) (still of the form 0/0)

       =  limx→1 (x-1)/(πx sin π x) (algebraic simplification)

       =  limx→1 1/(πx sin π x + π2 x cos π x ) (L' Hospital's rule again)

       = - 1/π2

 

Illustration:     

Evaluate limx→y  (xy-yx)/(xx-yy )

Solution:            

          limx→y  (xy-yx)/(xx-yy );   [0/0] = limx→y (yxy-1 - yx log y)/(xx log(ex) )

                                                     = (1-log y)/log(ey)

 

Continuity

A function f(x) is said to be continuous at x= a if limx→a- f(x)= limx→a

+ f(x)=f(a)

i.e. L.H.L. = R.H.L. = f(a)= value of the function at a i.e. limx→a f(x) = f(a)

If f(x) is not continuous at x= a, we say that f(x) is discontinuous at x=a.

For the function to be continuous at any point x=a, the function must be defined at that point and limiting values of f(x) when x approaches a, is equal to f(a).

        Continuity means the function should not have any break or sudden jump at any point in the given domain.

(i)

Continuous function in [a, b]

(ii)

Discontinuous function at x=c

(iii)

Discontinuous at x=c

So the condition for continuity if function at x=a can be defined as L.H.L.=f(a)= R.H.L. i.e.

       limx→a-  f(x) = limx→a

+ f(x) = f(a)

i.e. left hand limit is equal to the value of the function at that point and is equal to the right hand limit of the function at that point.

 

f(x) will be discontinuous at x=a in any of the following cases:

  limx→a- f(x) and limx→a

+ f(x) exist but are not equal.

  limx→a- f(x) and limx→a

+ f(x) exist and are equal but not equal to f(a).

  f(a) is not defined

  At least one of the limits does not exist.

   

Hence f(x) is continuous at x=2

Note: Another way is that if you given a function and you are needed to comment on the continuity of the function then simply finding points on domain where function is not defined may solve the problem.

 

Illustration:

Comment on the continuity of

                y = logc (x2 + x - 12)

Solution:

        The function can be written as

                y = loge ((x-3) (x + 4)

        And since log of non positive number is not defined so we have region of discontinuity as -4 < x < 3

        For all other value of x the function is continuous (∵ log function is a continuous function)

        A few examples of continuous function are xn, sing x, log x, ex, cos x etc. in the domain x ∈|-∞,∞| except for log x where x ∈|0,∞|

 

Properties of Continuous Functions

Let f(x) and g(x) be functions, both continuous at x=a. Then

 cf (x) is continuous at x =a where c is any constant

 f(x) + g(x) is continuous at x= a.

 f(x).g(x) is continuous at x= a.

 f(x)/g(x) is continuous at x= a, provided g(a) ≠ 0.

Continuity in an Interval

A real function 'f' is said to be continuous in an open interval (a, b) if it is continuous at every point in the interval (a, b). Function 'f' is said to be continuous for closed interval [a, b] if it is continuous ∀ x ∈ (a, b) and

        limn→a+ f(x) = f(a) and  limn→b

-  f(x) = f(b)

        Here the limits of end points of interval are one sided limit while solving problems one does not evaluate continuity condition at all points in the interval but uses his elementary knowledge of the function to find points of discontinuity and if none exists then function is continuous in the entire interval.

Let us define a function h(x) given by

        h(x) = f(x) for a < x < b

                = g(x) for b < x < c

Where f(x) and g(x) are continuous in their respective intervals. Then continuity if h(x) is checked only at x = b, as it is the only point where h(x) changes from f(x) to g(x) and hence is a likely point of discontinuity.

 

  Important

1. The necessary and sufficient condition for continuity of f at x = a ('a' being finite) is that both limx→a

- f(x) and limn→a+ f(x) should exist and be equal to f(a)

2. If a function is continuous on a closed interval [a, b], a and b are necessarily finite, it is bounded on it. It may not be always true in case of open interval (a, b).

eg     f(x)= 1/x ∀ 0 < x ≤ 1

The functions is continuous but not bounded in the interval (0. 1)

f(x) is said to be continuous in an open interval (a, b) if it is continuous at every point in this interval. f(x) is said to be continuous in the closed interval [a, b] if

f(x) is continuous in (a, b)

limx→a+ f(x)=f(a)

limx→a- f(x)=f(a)

Illustration:

       

Discuss the continuity of f(x)

Solution:

If we look at x2, 2x-1 or x+3, these functions are continuous in their respective intervals (Because every polynomial function is a continuous function). Hence we will check the continuity of f(x) at x=1 and x=2, because these are the points where the function is changing values. Consider x=1

        limx→1+  f(x) = limh→0 (1-h)2 = 1

        limx→1+  f(x) = limh→0  2 (1+h) - 1 = 1

         f(1) = 2 - 1 - 1 = 1

Both limits equal to f(1), hence f(x) is continues at x=1

Consider x=2

        limx→2-  f(x) = limh→0  2(2-h)-1 = 3

        limx→2+  f(x) = limh→0 (2+h) + 3 - 5

        f(2) = 2 + 3 = 5

Since LHL is not equal to f(2) hence f(x) is discontinuous at x = 2

 

Continuity of some of the common functions

Functions f(x) Interval in which f(x) is continuous

Constant C (-∞,∞)

bn, n is an integer > 0 (-∞,∞)

|x-a| (-∞,∞)

x-n, n is a positive integer. (-∞,∞)-{0}

a0xn + a1xn-1 +........ + an-1x + an (-∞,∞)

p(x)/q(x), p(x) and q(x) are polynomials in x

R - {x:q(x)=0}

sin x R

cos xR

R-{(2n-1π)/2:n=0,±1,........}

tan x R-{nπ:n=0,±1,........}

cot x R-{(2n-1)π/2:n=0,±1,± 2,........}

sec x R-{(2n-1)π/2:n=0,±1,± 2,........}

ex R

Ln x (0, ∞)

Illustration:     

Let f(x)=  

For what value of k is f(x) continuous at x-0 ?

 

Solution:

    

                = e × e = e2

   Since f(x) is continuous at x = 0, limx→a  f(x) = f(0)

      ⇒ a2 = k

    Hence f(x) is continuous at x = 0 when k = e2

 

Illustration:     

Discuss the continuity of

                    f(x)=

Solution:

We rewrite f(x) as f(x) =

                        As we can see, f(x) is defined as a polynomial function in each of the intervals (-∞, -2). (-2, 0), (0, 3), and (3, ∞). Therefore it is continuous in each of these four intervals.

       At the point x = -2

      limx→-2-  f(x)= limx→-2(-x-1)=1, and limx→-2

+ f(x) = limx→-2+(2x+3)=-1

     Therefore, limx→-2 f(x) does not exist.

     Thus f(x) is discontinuous at x=-2. At a point x = 0

     limx→0- f(x) = limx→0

+ f(x) = f(0)=3,

     Therefore f(x) is continuous at x=0. At a point x = 3

     limx→3- f(x) = limx→3

+ f(x) = f(3)=12

     Therefore, f(a) is continuous at x=3.

     So, we conclude that f(x) is continuous at all points in R except at x=-2.

 Illustration:     

Let f(x) be a continuous function and g(x) be a discontinuous function. Prove that f(x) + g(x) is a discontinuous function.

Solution:            

Suppose that h(x) = f(x)+g(x) is continuous. Then, in view of the fact that f(x) is continuous, g(x)=h(x)-f(x), a difference of continuous functions, is continuous.

   But this is a contraction since g(x) is given as a discontinuous function.

                        Hence h(x) = f(x) + g(x) is discontinuous.

 

Continuity of Composite Functions

If the function u = f(x) is continuous at the point x=a, and the function y=g(u) is continuous at the point u = f(a), then the composite function y=(gof)(x)=g(f(x)) is continuous at the point x=a.

Illustration:

        Find the points of discontinuity of y = 1/(u2+u-2) where u= 1/(x-1)

Solution:  

The function u = f(x)= 1/(x-1) is discontinuous at the point x=1.

 The function y = g(u)= 1/(u2+u-2) = 1/((u+2)(u-1)) is discontinuous at

u = -2 and u =1.

                ⇒1/(x-1)=-2                 ⇒ x=1/2

     When u = 1

                ⇒1/(x-1)=1                    ⇒x=2

Hence the composite function y = g(f(x)) is discontinuous at three  points

  x = 1/2, x = 1 and x =2.

Types of Discontinuities

Types of Discontinuities

1.     Discontinuity of first kind (or) ordinary discontinuity)

        limx→a+ f(x) = L1 limx→a

+  f(x) = L2

And L1 ≠ L2 where L1 and L2 are finite real number.

 

Illustration :

Consider f(x) =[x], where [x] is greatest integer < x.

Solution:

Consider x =1. Where I is some integer

RHL = limx→1+ f(x) = 1 and

LHL = limx→a-  f(x) = 1-1

Hence both RHL and LHL are finite real numbers but they are not equal.

 

2.     Discontinuity of second kind: This kind of continuity exists when neither limx→a+ f(x) nor

limx→a- f(x) exist.

  

Illustration:

       

Consider x = a where 'a' is a real number.

Solution:

        Here limx→a+ f(x) can have value 1 or 0 depending upon whether a+ is rational or irrational.

Likewise limx→a- f(x) does not have a definite value.

Hence both RHL and LHL do not exist.

 

3.     Mixed discontinuity: If any one of limx→a+ f(x) or limx→a

- f(x) exists and the other does not.

 

 

Types of Discontinuities

Illustration:

        f(x)=log x

Solution:

        Here limx→0+ f(x) exists but limx→0

- f(x) does not, as log x is not defined for negative values of x.

 

4.     Removable discontinuity

If limx→a f(x) exists but is not equal to f(a), then f(x) has removable discontinuity at x = a and it can be removed by redefining f(x) for x=a.

 

Illustration:

Redefine the function f(x)=[x] + [-x] in such a way that it becomes continuous for x ∈ (0, 2).

 

Solution:

          Here limx→1 f(x) = -1 but f(a)=0

                Hence, f(x) has a removable discontinuity at x= 1.

                To remove this we define f(x) as follows

                f(x)=[x]+[-x], x ∈ (0, 1) U (1, 2)

                = -1,                 x=1.

                Now, f(x) is continuous for x ∈ (0, 2)

 

5.     Non-removable discontinuity

If  limx→a f(x) does not exists, then we can remove this discontinuity so that this become a non-removable or essential discontinuity.

e.g. f(x) = [x+3] has essential discontinuity at any x ∊ 1.

  

6.     Jump discontinuity: A function f(x) is said to have a jump discontinuity at a point x=a if, limx→a

-  f(x) ≠ limx→a+ f(x) and f(x) and may be equal to either of previous limits.

 

Illustration:

        f(x)=[x]:[ ] denotes greatest integer has jump discontinuity at all integer values.