Three requirements have to be satisfied for the continuity of a function y = f(x) at x = x0 :
(i) f(x) must be defined in a neighbourhood of x0 (i.e., f(x0) exists);
(ii) lim x->x0 f(x) exists.
(iii) f(x0) = lim x -> x0 f(x)
(1) Constant function is continuous at each point of R (R stands for real numbers)
(2) Power functions with positive integer exponents are continuous at every point of R
(3) Polynomial functions, p(x) are continuous at every point of R.
(4) Quotients of polynomials namely rational functions of the form
R(x) = p(x) / q(x), are continuous at every point where q(x) ≠ 0, and
(5) The circular functions sin x and cos x are continuous at every point of their domain R = (- ∞, ∞) since lim x-> x0 sin x = sin x0, lim x-> x0 cos x = cos x0
(6) The nth root functions, f (x) = x1/n are continuous in their proper domain since lim x -> x0 (x1/n) = x0 1/n
(7) The reciprocal function f (x) = 1/x is not defined at 0 and hence it is not continuous at 0. It is continuous at each point of R − {0} .
(8) The exponential function f(x) = ex is continuous on R.
(9) The logarithmic function f(x) = log x (x > 0) in continuous in (0, ∞)
(10) The modulus function
f(x) = |x|
-x if x < 0
0 if x = 0
x if x > 0
is continuous at all points of the real line R .
Question 1 :
Prove that f(x) = 2x2 + 3x - 5 is continuous at all points in R.
Solution :
Since the given function f(x) is a polynomial, it is continuous at all points in R.
Question 2 :
Examine the continuity of the following
x + sin x
Solution :
Let f(x) = x + sin x
(i) From the given function, we come to know that both "x" and "sin x" are defined for all real numbers.
(ii) lim x-> x0 f(x) = lim x-> x0 x + sin x
By applying the limit, we get
= x0 + sin x0 -------(1)
(iii) f(x0) = x0 + sin x0 -------(2)
From (1) and (2)
lim x-> x0 f(x) = f(x0)
Hence the given function is continuous for all real values.
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