Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.15

Example 39 : Evaluate (i) x 0
lim4x |x|
→ 3x |x|

+

+ (ii) x 2

lim|x 2|

→ x 2

−

−

Solution:

(i) x 0lim→ +4x x3x x++ =x 0lim→ 5x 54x 4= , as x ≠ 0 and also for x > 0, | x | = x

x 0 x 0

lim 4x x lim 3x 3

→ − 3x x → −2x 2

− = =

− as x ≠ 0, and for x < 0, | x | = – x
∴ the given limit doesn’t exist, as the above two limiting values are unequal.

(ii) x 2 x 2
lim x 2 limx 2 1.
→ +x 2 → +x 2

− = − =

− −

for | x -– 2 | > 0 i.e., for x > 2, | x – 2 | = x – 2

( )

x 2 x 2 ( )

lim x 2 lim x 2 1

→ − x 2 → − x 2

− = − − = −

− −

for | x – 2 | < 0 i.e., for x < 2, | x – 2 | = – (x – 2)
As the two limiting values are unequal, so the given expression does not exist.
Some Useful Limits :

(A) lim 1 x ex 2→ (+ )1/x= (B) x 0lim log 1 x 1→^1 x e(+ )=

(C)

x

x 0

lime 1 1

→ x

− = (D) x

x 0 e

lima 1 log a

→ x

− =

(E)

n n n

x a

limx a an 1

→ x a

− = −

− (F)

( )n

x 0

lim 1 x 1 n.

→ x

+ − =

Example 40 :

3 x

x 0

lime 1 3.

→ x

− = Put 3x = u, as x → 0 u → 0.

( )

3x u u

x 0 u 0 u 0

lime 1 e 1lim 3 lim. 3 1 by C 3.e 1

→ x → u/ 3 → u

− = − = × − = × =

Example 41 : x 0lim→log 1 6x(x+ )=6.

Solution:
Put 6x = u, as x → u, u → 0

Expression u 0 ( ) u 0 ( )
limlog 1 u 6 limlog 1 u
→ u/6 → u

= + = × +

= 6 × 1 (by B) = 6.