**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 = 4 - 1 = 3 ---(1) |
lim x |

Since left hand limit and right hand limit are equal for -2, it is continuous at x = -2.

lim x |
lim x = 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

After having gone through the stuff given above, we hope that the students would have understood, "Find the Intervals on Which Each Function is Continuous"

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