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: L'Hopital for the first limit:
Both numerator and denominator vanish at $x = k$. Differentiate: $\dfrac{3x^2}{2x} \to \dfrac{3k^2}{2k} = \dfrac{3k}{2}$.

Step 2: Standard limit for the second:
$\dfrac{1-\cos 2x}{x \sin x} = \dfrac{2\sin^2 x}{x\sin x} = 2\cdot\dfrac{\sin x}{x} \to 2$.

Step 3: Equate:
$\dfrac{3k}{2} = 2$, so $k = \dfrac43$.

Step 4: Check:
With $k = \frac43$, the left limit is $\frac{3}{2}\cdot\frac43 = 2$, matching the right limit.

Final Answer:
$k = \dfrac{4}{3}$, option (A). \[ \boxed{\frac{4}{3} \text{ (A)}} \]