Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.11

And (^) x 0lim f(x) lim|x| lim( x) 0.→ − =x 0→ − =x 0→ −− =
∴ limf(x) 0.x 0→ =
Example 30 : Find analytically x 3lim x 3,→ − if it exists.
xlim x 3 0,→ 3 + − = but xlim x 3→ 3 − − does not exist. As when x → 3 – 0 is definitely less than 3, whereas x – 3 is
negative and square root of negative is never real but an imaginary.
∴ the limit does not exist.
Alternatively : Put x = 3 + h, as x → 3, h → 0
x 3lim x 3 lim 3 h 3 lim h→ − =h 0→ + − =h 0→
hlim h 0, but lim h→ + 0 = =h 0→ − does not exist (as the h near to but less than zero corresponds no real value of h).
∴ xlim x 3→ 3 − does not exist.
Note : At x = 3, f (x) = x 3 3 3 0− = − = ∴ f(3) exists.
Example 31 : Do the following limits exists? If so find the values
Solution:
(i) xlim→ − (^2) x 2+^1 (ii) limx→ (^01) x (iii) ( )

2 (^2 )

x 1

x 1

lim x 1→ x 1

(^) − (^)
− + −
(^)
(i) xlim→ − (^2) x 2+^1 =limh 0→ − + +2 h 2^1 =h 0lim→ h^1
Now xlim→ + (^0) h^1 = +ve; limh 0→ −h^1 = −ve. As the two values are not same, so the limit does not exist.
(ii) xlim→ + (^0) x^1 = +∞, limh 0→ −x^1 = − ∞.The limit doesn’t exist as the two values are unequal.
(iii) Expr. ( )

( )( )

( )

2

x 1

lim x 1 x 1 x 1

→ x 1

(^) − + − − (^)
(^) − (^)
= {( ) ( )}
2
lim x 1 x 1 ,x 1→ − + − = as (x – 1) = 0 ;
= lim x x 2 1 1 2 0.x 1→(^2 + − )=(+ − )= On putting the limiting value of x, the value of the function exists and its
value is 0.
Distinction between lim f xx a→ ( ) and f(a). By x alim f x→ ( ) we mean the value of f(x) when x has any arbitary value
near a but not a. The quantity f (a) is the value of f(x), when x is exactly equal to a.
Note : The following cases may arise :