Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.13

Example 33 : Evaluate

2

x 1 2

limx 3x 2.

→ x 4x 3

− +

− +

Reqd. limit =

( )( )

x 1( ) ( ) x 1

lim x 1 x 2 limx 2 1 2 1.

→ x 1 x 3 →x 3 1 3 2

− − = − = − =

− − − −

If we put x = 1, the function becomes^00 which is undefined. Further limit of the denominator is zero, we
cannot apply theorem 2, hence cancelling the common factor (x – 1) which is ≠ 0 as x →1, we may obtain
the above result.
Problem when variable tends of infinity :

Example 34 : Find the limit of xlim 2x 5x 2→ ∞(^2 − + )

Solution:
Now 2x^2 – 5x + 2 = x^2

∴xlim 2x 5x 2 limx 2→ ∞(^2 − + )=x→ ∞^2 −5 2x+x 2 =limx lim 2x→ ∞^2 ×x→ ∞ −x5 2+x 2

Now limx ; lim2 2 ; lim 0; lim 0x→ ∞^2 = ∞ x→ ∞ = x→ ∞^5 x= x→ ∞x^22 =

∴lim 2x 5x 2x→ ∞(^2 − + )= ∞ × = ∞2.

Example 35 : Find

5 3

x^84

lim4x 2x 5

→ ∞7x x 2

+ −

+ +

Solution:
Here the highest power of x in the denominator is 8. We now divide both the numberator and the

denominator by x^8 to avoid the undefined form ∞∞. So we get

3 5 8 3 5 8

x u 0^48

4 8

4 2 5

lim x x x lim4u 2u 5u

7 1 2 7 u 2u

x x

→ ∞ →

+ − + −

= + +

+ + (put

(^1) u,
x= so as x → ∞ , u → 0)
u 0(^358 )
4 8
u 0
lim 4u 2u 5u
lim(7 u 2u )
→
→

+ −

= + + (as denominator ≠ 0)

3 5 8

u 0 u 0 4 u 0 8

u 0 u 0

4limu 2lim u 5limu 0 0 0 0

→7 lim u 2limu→ → 7 0 0 7 0

→ →

+ − + −

= + + = + + = =