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->x₀  f(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 x₀ -------(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 

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 + x²

Solution :

Let f(x)  =  e^2x + 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} e^2x + x²

By applying the limit, we get

 =  e^2x₀ + (x₀)² -------(1)

(iii) f(x₀)  =   e^2x₀ + (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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