Question

Let \( k \in \mathbb{R} \). If \[ \lim_{x \to 0^+} \frac{\left(\sin(\sin(kx)) + \cos x + x\right)^2}{x} = e^6, \] then the value of \( k \) is:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( 4 \)

Hint:

For limits involving small \( x \), use Taylor expansions to approximate trigonometric and logarithmic terms.

Answer 1

The Correct Option is B

Let \[ y = \lim_{x \to 0^+} \frac{\left(\sin(\sin(kx)) + \cos x + x\right)^2}{x}. \] Taking the natural logarithm: \[ \ln y = \lim_{x \to 0^+} \frac{\ln\left(\sin(\sin(kx)) + \cos x + x\right)}{x}. \] We know \( y = e^6 \), so: \[ \begin{aligned} \lim_{x \to 0^+} 2 \times \frac{k \cos(\sin(kx)) \cos(kx) - \sin(x) + 1}{\sin(\sin(kx)) + \cos x + x} &= \lim_{x \to 0^+} 2 \times \left( \frac{k+1}{1} \right) \\ \therefore \quad y &= e^{2k+2} = e^6 \implies k = 2. \end{aligned} \] From this, solving gives \( k = 2 \).