Question

Find the value of \(k\) if the function \(f(x) = \dfrac{k\sin x}{x}\) is continuous at \(x = 0\) and \(f(0)=3\).

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(6\)

Hint:

Whenever continuity at \(x=0\) involves expressions like \( \frac{\sin x}{x} \), remember the standard limit: \[ \lim_{x\to0}\frac{\sin x}{x}=1 \] This limit is very commonly used in calculus problems.

Answer 1

The Correct Option is C

Concept: A function is continuous at a point \(x=a\) if the following condition holds: \[ \lim_{x \to a} f(x) = f(a) \] In this problem, the function must be continuous at \(x=0\). Therefore, \[ \lim_{x \to 0} f(x) = f(0) \] We also use the important trigonometric limit: \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Step 1: Apply the continuity condition. Given \[ f(x) = \frac{k\sin x}{x} \] For continuity at \(x=0\): \[ \lim_{x\to0} \frac{k\sin x}{x} = f(0) \]

Step 2: Evaluate the limit. \[ \lim_{x\to0} \frac{k\sin x}{x} = k \lim_{x\to0} \frac{\sin x}{x} \] Using the standard limit: \[ \lim_{x\to0} \frac{\sin x}{x} = 1 \] Thus, \[ \lim_{x\to0} f(x) = k \]

Step 3: Use the given value of the function. Since \(f(0)=3\), \[ k = 3 \] \[ \boxed{k = 3} \]