HOW TO EXAMINE THE CONTINUITY OF A FUNCTION

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->xf(x) exists.

(iii) f(x0)  =  lim x -> x0 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 

x2 cos x

Solution :

Let f(x)  =  x2 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 x2 cos x

By applying the limit, we get

 =  x02 cos x0 -------(1)

(iii) f(x0)  =   x02 cos x-------(2)

From (1) and (2)

lim x-> x0 f(x)  =  f(x0)

Hence the given function is continuous for all real numbers.

Question 2 :

Examine the continuity of the following 

ex tan x

Solution :

Let f(x)  =  ex tan x

From the given function, we know that the exponential function is defined for all real values.But tan is not defined aπ/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 

e2x + x2

Solution :

Let f(x)  =  e2x + x2

(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 e2x + x2

By applying the limit, we get

 =  e2x0 + (x0)2 -------(1)

(iii) f(x0)  =   e2x0 + (x0)2 -------(2)

From (1) and (2)

lim x-> x0 f(x)  =  f(x0)

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 x0 from (0, ∞)

(ii)  lim x-> x0 f(x)  =  lim x-> x0  x ln x

By applying the limit, we get

 =   x0 ln x0 -------(1)

(iii) f(x0)  =   x0 ln x0 -------(2)

From (1) and (2)

lim x-> x0 f(x)  =  f(x0)

Hence the given function is continuous on (0, ∞).

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