Properties of continuous functions

i. If, f, g ate two continuous functions at a point a of their common domain D, then f ± g. fg are continuous at a point, then f/g is also continuous at a.

ii. If f is continuous at a and f(a) ≠ 0, then there exists an open interval (a - δ, a + δ) such that for all x ϵ (a - δ, a + δ), f(x) has the same sign as f(a).

iii. If a function f is continuous on a closed interval [a, b], then it is bounded on [a, b] i.e. there exist real numbers K and K such that K ≤ f(x) ≤ K for all x ϵ [a, b].

iv. If f is a continuous function defied on [a, b] such that f(a) and f(b) are of opposite signs, then there exists at least one solution of the equation f (x) = 0 in the open interval (a, b).

v. If f is continuous on [a, b], then f assumes at least once, every value between minimum and maximum values of f (x), then there exist at least solution of the equation f (x) = K in open interval (a, b).

vi. If f is continuous on [a, b] and maps [a, b] into [a, b], then for some x ϵ [a, b] we have f(x) = x.

vii. If g is continuous at a and f is continuous at g(a). The fog is continuous at a.

Evaluate: Show that the function  is continuous at x = 0

Solution: we have,

(LHL at x = 0)

= f(x)

= f(0 - h)

= f(- h)

=  -h sin (1/-h)

= h sin (1/h)

= 0 x (An oscillating number between - 1 and 1)

= 0

(RHL at x = 0)

=  f(x)

= f(0 + h)

= f(h)

= h sin (1/h)

= 0 x (An oscillating number between - 1 and 1)

= 0

And,

f(0) = 0.

Thus, .

Hence, f(x) is continuous at x = 0.