CONTINUITY OF A ABSOLUTE VALUE FUNCTION AT A POINT

For a function to be continuous at a point,

  • It must be defined at that point
  • Its limit must exist at the point
  • The value of the function at that point must be equal to the value of the limit at that point.

If the function f(x) is continuous at a point x = a, then

lim xa-f(x)=lim xa+f(x)and so lim xaf(x) exists

Problem 1 :

Check if the function is continuous at x = 0

f(x) = 2x - |x|

Solution :

Step 1 :

Defining the given absolute value function as piece wise function.

Step 2 :

Evaluating the left hand limit :

lim x0-f(x)=lim x0- xApplying the limit, we get=0lim x0-f(x) = 0 ------(1)

Evaluating the right hand limit :

lim x0+f(x)=lim x0+ 3xApplying the limit, we get=3(0lim x0+f(x) = 0 ------(2)

Both left hand limit and right hand limit they are equal.

lim x0-f(x)=lim x0+f(x)

So, the function is continuous at x = 0.

Problem 2 :

Discuss the continuity of the function of given by

f(x) = |x - 1| + |x - 2| at x = 1 and x = 2.

Solution :

continuity-of-abs-functionq1

The condition for three pieces will be

x < 1, 1 ⩽ x < 2 and x ⩾ 2

  • Choosing one of the value in the first condition (x < 1) and applying the absolute value function, we get negative for x - 1 and for x - 2.
  • Choosing one of the value in the second condition (1 ⩽ x < 2) and applying the absolute value function, we get positive for x - 1 and negative for x - 2.
  • Choosing one of the value in the third condition (x ⩾ 2) and applying the absolute value function, we get positive for x - 1 and x - 2.
f(x)=-(x-1)-(x-2)if x <1(x-1)-(x-2)if 1x<2(x-1)+(x-2)if x 2f(x)=-x+1-x+2if x <1x-1-x+2if 1x<2x-1+x-2if x 2f(x)=-2x+3 if x <11 if 1x<22x-3 if x 2

Continuity at x = 1 :

Evaluating the left hand limit :

lim x1-f(x)=lim x1--2x+3=2(-1)+3lim x1-f(x)

Evaluating the right hand limit :

lim x1+f(x)=lim x1+lim x1+f(x)
lim x1-f(x)=lim x1+

So, the function is continuous at x = 1.

Continuity at x = 2 :

lim x2-f(x)=lim x2-lim x2+f(x)=lim x2+lim x2-f(x)lim x2+f(x)

So, the function is continuous at x = 2.

Problem 3 :

Show that 

f(x)=x-|x|2when x02when x=0

is discontinuous at x = 0.

Solution :

f(x)=x-|x|2when x02when x=0f(x)=x-(-x)2x<02x=0x-x2x>0f(x)=2x2x<02x=002x>0f(x)=xx<02x=00x>0

Checking the left hand and right hand limit :

lim x0-f(x)=lim x0-lim x0+f(x)=lim x0+lim x0+f(x)lim x0+f(x)

f(0) = 2

lim x0-f(x)=lim x0+

So, the function f(x) is discontinuous at x = 0.

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