Question

Let f: R → R be defined by \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {x\sin \left( {\frac{1}{x}} \right)}&{if\;x > 0}\\ 0&{x \le 0} \end{array}} \right.\) Then
1. f is neither continuous nor differentiable at x = 0
2. f is continuous nor differentiable at x = 0
3. f is continuous but not differentiable at x = 0
4. f is not continuous but differentiable at x = 0

Answer 1

Correct Answer - Option 3 : f is continuous but not differentiable at x = 0

CONCEPT:

A function is continuous If \(\mathop {\lim }\limits_{x \to 0^-} F(x) = F(0) = \mathop {\lim }\limits_{x \to 0^+} F(0)\)

A function is differentiable if left hand derivative = right hand derivative

CALCULATION:

Given: \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {x\sin \left( {\frac{1}{x}} \right)}&{if\;x > 0}\\ 0&{x \le 0} \end{array}} \right.\)

⇒\(\mathop {\lim }\limits_{x \to 0^-} f(x) = 0\), \(\mathop {\lim }\limits_{x \to 0^+} xsin\frac{1}{x} = 0\)  and f(0) = 0

As we know that, a function is continuous If \(\mathop {\lim }\limits_{x \to 0^-} F(x) = F(0) = \mathop {\lim }\limits_{x \to 0^+} F(0)\)

So the given function f(x) is continuous at x = 0

Let's calculate the left hand derivative : 

\(\mathop {\lim }\limits_{x \to 0^-} \frac{d}{{dx}}f(0) = 0\)

Similarly, let's calculate the right hand derivative:

 ⇒ \(\mathop {\lim }\limits_{x \to 0^+} \frac{d}{{dx}}[xsin\frac{1}{x}]\) = \(\mathop {\lim }\limits_{x \to 0^+} [xsin\frac{1}{x} - \frac{1}{x}\cos \frac{1}{x}] = - \infty \)

As we can see that, left hand derivative \( \ne \) right hand derivative

So, the given function is not differentiable at x = 0

Hence, option C is correct.