**How to Examine the Continuity of a Function :**

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

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, ∞).

After having gone through the stuff given above, we hope that the students would have understood, "How to Examine the Continuity of a Function"

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