Question

The value of \(k\) for which the function \(f(x) = \begin{cases x^2 \sin \frac{1}{x}, & x \neq 0 \\ k(x + 1), & x = 0 \end{cases}\) is a continuous function, is :}

• A. \(\frac{1}{4}\)
• B. \(2\)
• C. \(\frac{1}{2}\)
• D. \(0\)

Hint:

Remember that \(\lim_{x \to 0} x^n \sin\left(\frac{1}{x}\right) = 0\) for any power \(n > 0\). Equating the limiting value \(0\) directly to the definition at the given point \(f(0)\) immediately yields the required constant.

Answer 1

The Correct Option is D

Concept:
• A function \(f(x)\) is continuous at a point \(x = a\) if and only if \(\lim_{x \to a} f(x) = f(a)\).
• The sine function is a bounded function such that \(-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\) for all \(x \neq 0\).
• The limit of the product of a function approaching zero and a bounded function is zero: \(\lim_{x \to 0} [x^2 \cdot (\text{bounded function})] = 0\).

Step 1: Evaluate the limit of \(f(x)\) as \(x \to 0\)
For \(x \neq 0\), the function is defined as \(f(x) = x^2 \sin \left(\frac{1}{x}\right)\). Taking the limit as \(x\) approaches \(0\): \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin \left(\frac{1}{x}\right) \]

Step 2: Use the Sandwich (Squeeze) Theorem
Since \(-1 \leq \sin \left(\frac{1}{x}\right) \leq 1\) for all \(x \neq 0\), multiplying throughout by \(x^2\) (which is strictly positive for \(x \neq 0\)) gives: \[ -x^2 \leq x^2 \sin \left(\frac{1}{x}\right) \leq x^2 \] Taking the limit as \(x \to 0\) on both sides: \[ \lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} (x^2) = 0 \] By the Sandwich Theorem: \[ \lim_{x \to 0} x^2 \sin \left(\frac{1}{x}\right) = 0 \]

Step 3: Evaluate \(f(0)\) and equate it to the limit
From the definition of the piecewise function: \[ f(0) = k(0 + 1) = k \] For \(f(x)\) to be continuous at \(x = 0\): \[ \lim_{x \to 0} f(x) = f(0) \] \[ 0 = k \] \[ k = 0 \]