Find the Intervals on Which Each Function is Continuous :
Here we are going to how to examine the continuity of the function when the interval is not given.
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
sin x / x^{2}
Solution :
Let f(x) = sin x / x^{2}
(i) From the given function, we know that both "x" and "sin x " are defined for all real numbers. Here "x" is in the denominator, so it should not be 0.
(ii) lim_{ x-> x0} f(x) = lim_{ x-> x0 }sin x / x^{2}
By applying the limit, we get
= sin x_{0} / (x_{0})^{2} -------(1)
(iii) f(x_{0}) = sin x_{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 x ∈ R - {0}.
Question 2 :
Examine the continuity of the following
(x^{2 }- 16) / (x + 4)
Solution :
Let f(x) = (x^{2 }- 16) / (x + 4)
If we apply x = -4, then the denominator will become zero. So the entire value will become infinity.
Hence it is continuous for all x ∈ R - {-4}.
Question 3 :
Examine the continuity of the following
|x + 2| + |x - 1|
Solution :
Since we have a absolute value function, we have to split into piece wise function and check its continuity.
x + 2 = 0 x = -2 |
x - 1 = 0 x = 1 |
f(x) = -x - 2 - x + 1 = -2x - 1 If x < -2
f(x) = x + 2 - x + 1 = 3 If -2 ≤ x < 1
f(x) = x + 2 + x - 1 = 2x + 1 If x ≥ 1
From the above piece wise unction, we have to check if it is continuous at x = -2 and x = 1
lim x_{->-2}- f(x) = -2(-2) - 1 = 4 - 1 = 3 ---(1) |
lim x_{->-2}+ f(x) = 3 ---(2) |
Since left hand limit and right hand limit are equal for -2, it is continuous at x = -2.
lim x_{->1}- f(x) = 3 |
lim x_{->1}+ f(x) = 2x + 1 = 2(1) + 1 = 3 |
Since left hand limit and right hand limit are equal for 1, it is continuous at x = 1.
Hence the function is continuous for x ∈ R
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