Which of the following functions is not continuous at the origin?

Concept:

• A function is said to continuous at a given value if the function does not have a sudden change at that value.
• If two functions are individually continuous then it’s summation difference, the product will be also continuous.
• A function is continuous at x = a
  If L.H.L = R.H.L = f(a)
  i.e. \(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = f\left( a \right)\) 
• If f(x) and g(x) are individually continuous then it’s f(x) ± g(x), f(x) ⋅ g(x) and \(\frac{{f\left( x \right)}}{{g\left( x \right)}}\;;g\left( x \right) \ne 0\) will be continuous.

Calculation:

Option (A):

Let f(x) = sin x

From graph, at origin i.e. x = 0, sin x is continuous.

Option (B):

Let g(x) = x

From graph at origin i.e. x = 0

g(x) = x is continuous.

Option (C):

Let h(x) = x sin x

h(x) = g(x) ⋅ f(x)

Since f(x) and g(x) is individually continuous at origin, so h(x) will be also continuous.

Option (D):

Let \(h\left( x \right) = \frac{{\sin x}}{x} = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) 

Since sin x is continuous at origin and x is also continuous at origin and values are zero.

i.e. \(h\left( x \right) = \frac{{\sin x}}{x}\) 

At origin:

\(h\left( {x = 0} \right) = \frac{{\sin \left( 0 \right)}}{0} = \frac{0}{0}\)  is a indeterminate form)

Therefore, we can say that.

h(x) will be discontinuous at origin as at x = 0, h(x) become undefined.

So, option (D) is correct as \(\frac{{\sin x}}{x}\) is discontinuous at origin.