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