3.14 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS
Calculus
Example 36 : Find
2
x^2
lim 5 2x.
→ ∞3x 5x
−
+
Solution:
Expression
2 2
x u 0
(^52)
limx lim5u 2
(^35) 3u 5
x
→ ∞ →
− −
= = +
+
(^1) u, asx ,u 0
x
(^) = → ∞ →
(^)
2
u 0
u 0
5 lim u 2 0 2 2
3 lim u 5 0 5 5→.
→
− −
= + = + = −
Problems regarding rationalisation :
Example 37 : Find the value of : hlim→ 0 x h x+h−
( )( )
h 0 h 0 ( ) h 0 ( )
x h x x h x x h x x h x
lim h lim lim
→ → h x h x → h x h x
+ − + − + + + −
= =
+ + + +
h (^0) ( ) h 0
lim h lim^1 ,
→ h x h x → x h x
= =
- (h ≠ 0)
(^11).
x 0 x 2 x
- (h ≠ 0)
= =
+ +
Problem regarding both sides limits :
Example 38 : A function is defined as
f(x) = x^2 , for x > 1
= 4.1, for x = 1
= 2x, for x < 1.
Does lim f(x)x 1→ exist?
Solution:
Let us find x 1lim f(x)→ + and x 1lim f(x).→ −
Now lim f(x) lim x 1 1,x 1→ + =x 1→ + 2 2= = as f(x) = x^2 for x > 1
Again x 1lim f(x) lim 2x 2.1 2,→ − =x 1→ − = = as f(x) = 2x for x < 1
Since the two limiting values are not equal, so lim f xx 1→ ( ) doesn’t exist.