Paper 4: Fundamentals of Business Mathematics & Statistic

3.10 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Calculus

Solution:

Now

2

2

( 1)(2 3) 2 5 3

( 2)(3 4) 3 10 8

+ + = + +

+ + + +

x x x x

x x x x

2

2

2 5 3

3 10 8

+ +

=

+ +

x x

x x

[Dividing by x^2 ]

∴

Lt (x 1)(2x 3) 2

→∞( 2)(3 4) 3

+ + =

x x x+ +

Or

Lt (x 1)(2x 3)

→∞( 2)(3 4)

+ +

x x x+ +

1 1 2 3

Lt 1 2 2

→∞ 1 2 3 4 1 3 3

(^) + (^) + (^)
=^ = ⋅ =
(^) + (^) + ⋅
(^)
x
x x
x x

SOLVED EXAMPLES

Example 28 : limx 2→ x 2−−^4 =4.

x^2

For if x = 2 + h, whether h be positive or negative,

( )( )

( )

x 4^2 x 2 x 2 h 4 h( ) 4 h

x 2 x 2 h

− = − + = + = +

− −

and by taking h numerically small, the difference of

x 4^2

x 2

−

− and 4 can be made as small as we like. It may

be noted here that however small h may be, as h ≠ 0, one can cancel the factor

(x – 2) i.e., h between numerator and denominator here. Hence

2

x 2

limx 4 4.

→ x 4

− =

− But for x = 2, the function

x 4^2

x 2

−

− is undefined as we cannot cancel the factor x – 2, which is equal to zero.

Now writing

2

x 2

f(x) x 4, limf(x) 4

x 2 →

= − =

− whereas f(2) does not exist.

Example 29 : For f (x) = | x |, find, x 0lim f(x).→

Solution:

f(x) = | x | means

x, for x 0

x, for x 0

≥

− <

Now (^) x 0lim f(x) lim|x| lim x 0→ + =x 0→ + =x 0→ + =