Question

If \(\underset{x\rightarrow k}{lim}\frac{x^3-k^3}{x^2-k^2} = \underset{x\rightarrow 0}{lim}\frac{1-cos(2x)}{xsinx}\), then the value of \(k\) is \(\ldots\)

• A. \(\frac{4}{3}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{8}{3}\)
• D. \(\frac{3}{8}\)

Hint:

Evaluate each limit separately: the left one gives 3k/2 and the right one gives 2.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We evaluate both limits, equate them, and solve for \(k\).

Step 2: Left limit:
\[ \lim_{x\to k}\frac{x^3 - k^3}{x^2 - k^2} = \lim_{x\to k}\frac{(x-k)(x^2 + xk + k^2)}{(x-k)(x+k)} = \frac{3k^2}{2k} = \frac{3k}{2} \]

Step 3: Right limit:
Use \(1 - \cos 2x = 2\sin^2 x\):
\[ \lim_{x\to 0}\frac{1 - \cos 2x}{x\sin x} = \lim_{x\to 0}\frac{2\sin^2 x}{x \sin x} = \lim_{x\to 0}\frac{2\sin x}{x} = 2 \]

Step 4: Equate and solve:
\[ \frac{3k}{2} = 2 \Rightarrow k = \frac{4}{3} \]
The value \(\frac{3}{4}\) in option (B) arises from inverting the equation. Options (C) and (D) come from the wrong value of the right-hand limit.

Final Answer:
The value of \(k\) is \(\dfrac{4}{3}\), option (A). \[ \boxed{\frac{4}{3} \text{ (A)}} \]