Question

The value of \(k\) for which the function \[ f(x)= \begin{cases} x^2\sin\!\left(\dfrac{1}{x}\right), & x\ne0,\\[6pt] k(x+1), & x=0, \end{cases} \] is continuous, is \[ \_\_\_\_. \]

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

Hint:

Any function composed of \( x^n \times (\text{bounded function}) \) as \( x \to 0 \) will always have a limit of 0 as long as \( n > 0 \). Consequently, simply evaluate the other branch at 0 and set it equal to 0.

Answer 1

The Correct Option is D

Concept: For a function \( f(x) \) to be continuous at a specific point \( x = a \), the limit of the function as \( x \) approaches \( a \) must exist and be equal to the value of the function at that point: \[ \lim_{x \to a} f(x) = f(a) \]

Step 1: Evaluate the limit of \( f(x) \) as \( x \to 0 \).
We need to find: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right) \] We can evaluate this limit using the Sandwich (Squeeze) Theorem. We know that the sine function is bounded between \(-1\) and \(1\) for all real arguments: \[ -1 \le \sin \frac{1}{x} \le 1 \]

Step 2: Apply the inequality bounded limits.
Multiply the entire inequality chain by the non-negative term \( x^2 \) (since \( x^2 \ge 0 \) for all real \( x \)): \[ -x^2 \le x^2 \sin \frac{1}{x} \le x^2 \] Taking limits as \( x \to 0 \): \[ \lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} (x^2) = 0 \] Since the lower and upper bounds both approach 0, by the Squeeze Theorem: \[ \lim_{x \to 0} \left( x^2 \sin \frac{1}{x} \right) = 0 \]

Step 3: Equate the limit to the value of the function at \( x = 0 \).
From the function definition, the value at \( x = 0 \) is: \[ f(0) = k(0 + 1) = k \] For continuity at \( x = 0 \): \[ \lim_{x \to 0} f(x) = f(0) \quad \Rightarrow \quad 0 = k \] Thus, \( k = 0 \), which matches option (D).