FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.17(ii) x 0limf x→ ( ) when f(x) = 1, x ³ 0.
= – 1, x < 0
[Ans. doesn’t exist, in graph (0, – 1) will be excluded]- A function f(x) is defined as follow :
f(x)= x^2 , for x < 1.
= 2. 5, for x = 1
= x^2 + 2, for x > 1.
Does limf xx 1→ ( ) exist? [Ans. no] - Given f(x) = x^2 , for x > 1
= 2. 1, for x = 1
= x, for x < 1.
Find limf x .x 1→ ( ) [Ans. 1]
Verify the following limits :
- (i)
3
x^32
lim 4x 5x 1 2.
→ ∞6x 7x 4 3+ − =
+ + (ii)2
x^2
lim 6 5x^1.
→ ∞4x 15x 3− = −
+
(iii)5
x^7
lim^4 x x 7 0.
→ ∞2x 3x 2− + =
+ + (iv)9 2
x^5
lim2x 5x 2
→ ∞9x 2x 12− + = ∞
+ +
- (i)
x 1^2
x 0
lim .e
x−
→ [Ans. 0] (ii) x^1
limlogx
→ x 1− [Ans. 1]3.3 CONTINUITYIntroduction :
A function is said to be continuous if its graph is a continuous curve without any break. If, however, there is
any break in the graph, then function is not continuous at that point.
If for a value of k, the limit of f(x) does not exist i.e., if on the curve of f(x) a point is absent, the graph will be
discontinuous i.e., not continuous.
A function f(x) is said to be continuous at x = a, x alim f(x)→ if exists in finite and is equal to f(a)
i.e., x alim f(x) lim f x f a→ + =x a→ −( )= ( )
i.e., f (a + 0) = f (a – 0) = f (a) briefly or limf(a h) f(a)x 0→ + =
Thus we are to find following three values :
(i) x alim f(x),→ + (ii)x alim f(x)→ − (iii) f(a)