Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.19

Example 45: Discuss the continuity of f (x) at x = 4, where
f(x) = 2x + 1, x ≠ 4
= 8 x = 4.

Herex 4lim f x , lim 2x 1 2.4 1 9→ + ( ) x 4→ +( +)= + =

x 4lim f x lim 2x 1 2.4 1 9→ − ( )=x 4→ −( +)= + =
f (4) = 8, which is different from the previous value.
∴ f (x) is not continuous at x = 4.
SOLVED EXAMPLES
Example 46: Show that f(X) = 2x + 3 is continuous at X = 1.
Solution :

x 1lim f x lim 2x 3 2.1 3 5.→ +( )=x 1→ +( + )= + =

x 1lim f x lim 2x 3 2.1 3 5.→ −( )=x 1→ −( + )= + = and at x = 1, f(1) = 2.1 + 3 = 5

∴ x 1lim f x lim f x f 1 5 ;→ +( )=x 1→ − ( )= ()= ∴ f(x) is continuous at x = 1.

Example 47: Discuss the continuity of f (x) = | x | at x = 0.
F (x)= | x | means f(x) = x for x > 0
= 0 for x = 0
= – x for x < 0.

Solution : x 0lim f x lim x 0→ +( )=x 0→ + =

x 0lim f x lim x 0→ −( )=x 0→ −(− )= and at x = 0, f(0) = 0

∴x 0lim f x lim f x f 0 0 ;→ + ( )=x 0→ −( )= ( )= ∴ f(x) is continous at x = 0.

Example 48: Show that f x( )=x 1^1 − is discontinuous at x = 1.

Solution :

f 1( )=1 1 01 1− = , undefined.

Now x 1lim f x lim→ + ( )=h 0→ 1 h 1 h+ −^1 =limh 0→^1 = + ∞ (Putting x = 1 + h, as x → 1, h → 0)

And x 1lim f x lim→ −( )=h 0→ 1 h 1− −^1 =limh 0→ −^1 h= − ∞ (Putting x = 1 – h, as x → 1, h → 0)

We find right hand limit is not equal to left hand limit.
∴ at x = 1, f (x) is discontinuous.