Three requirements have to be satisfied for the continuity of a function y = f(x) at x = x_{0} :
(i) f(x) must be defined in a neighbourhood of x_{0} (i.e., f(x_{0}) exists);
(ii) lim _{x->}x_{0 }f(x) exists.
(iii) f(x_{0}) = lim _{x ->} x_{0} f(x)
To know the points to be remembered in order to decide whether the function is continuous at particular point or not, you may look into the page " How to check continuity of a function, if interval is not given "
Question 1 :
Examine the continuity of the following
x^{2} cos x
Solution :
Let f(x) = x^{2} cos x
(i) From the given function, we know that both "x" and "cos x" are defined for all real numbers.
(ii) lim_{ x-> x0} f(x) = lim_{ x-> x0 }x^{2} cos x
By applying the limit, we get
= x_{0}^{2} cos x_{0} -------(1)
(iii) f(x_{0}) = x_{0}^{2} cos x_{0 }-------(2)
From (1) and (2)
lim_{ x-> x0} f(x) = f(x_{0})
Hence the given function is continuous for all real numbers.
Question 2 :
Examine the continuity of the following
e^{x} tan x
Solution :
Let f(x) = e^{x} tan x
From the given function, we know that the exponential function is defined for all real values.But tan is not defined at π/2.
So, the function is continuous for all real values except (2n+1)π/2.
Hence the answer is continuous for all x ∈ R- (2n+1)π/2.
Question 3 :
Examine the continuity of the following
e^{2x} + x^{2}
Solution :
Let f(x) = e^{2x} + x^{2}
(i) From the given function, we know that both "x" and "exponential function" are defined for all real numbers.
(ii) lim_{ x-> x0} f(x) = lim_{ x-> x0 }e^{2x} + x^{2}
By applying the limit, we get
= e^{2x}_{0} + (x_{0})^{2} -------(1)
(iii) f(x_{0}) = e^{2x}_{0} + (x_{0})^{2}_{ }-------(2)
From (1) and (2)
lim_{ x-> x0} f(x) = f(x_{0})
Hence the given function is continuous for all real numbers.
Question 4 :
Examine the continuity of the following
x ln x
Solution :
Let f(x) = x ln x
(i) From the given function, we know that both "x" defined for all real values, but logarithmic function is defined only on (0, ∞)
Let us take x_{0} from (0, ∞)
(ii) lim_{ x-> x0} f(x) = lim_{ x-> x0 } x ln x
By applying the limit, we get
= x_{0} ln x_{0} -------(1)
(iii) f(x_{0}) = x_{0} ln x_{0}_{ }-------(2)
From (1) and (2)
lim_{ x-> x0} f(x) = f(x_{0})
Hence the given function is continuous on (0, ∞).
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 22, 24 06:32 AM
May 17, 24 08:12 AM
May 14, 24 08:53 AM