Question

Let \(k\in R\). If \(\)\(\)\(\)\(\lim\limits_{x→0+}(\sin(\sin kx)+\cos x+x)^{\frac{2}{x}}=e^6\), then the value of k is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

Step 1: Given Limit Expression 

We are given: \[ y = \lim_{x \to 0^+} \frac{(\sin(\sin(kx)) + \cos x + x)^2}{x} \]

Step 2: Taking Natural Logarithm

Taking the natural logarithm on both sides: \[ \ln y = \lim_{x \to 0^+} \frac{\ln(\sin(\sin(kx)) + \cos x + x)}{x} \]

Step 3: Given \( y = e^6 \)

We use differentiation: \[ \lim_{x \to 0^+} 2 \times \frac{k \cos(\sin(kx)) \cos(kx) - \sin(x) + 1}{\sin(\sin(kx)) + \cos x + x} \] Simplifying: \[ \lim_{x \to 0^+} 2 \times \frac{k + 1}{1} = 6 \]

Step 4: Solving for \( k \)

\[ 2(k+1) = 6 \] \[ k+1 = 3 \] \[ k = 2 \]

Final Answer:

\[ k = 2 \]