Find Parameter k in Exponential Limit Equation

To find the value of $k \in \mathbb{R}$, we evaluate the given limit: $\lim_{x \to 0+} \left( \sin(\sin kx) + \cos x + x \right)^{\frac{2}{x}} = e^6$

First, let us examine the form of the limit as $x \to 0+$: $\lim_{x \to 0+} \left( \sin(\sin kx) + \cos x + x \right) = 0 + 1 + 0 = 1$ and the exponent satisfies: $\lim_{x \to 0+} \frac{2}{x} = \infty$

Thus, the limit is of the indeterminate form $1^\infty$. For any limit of the form $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)(f(x) - 1)}$ when $f(x) \to 1$ and $g(x) \to \infty$, we can evaluate the exponent limit $P$: $P = \lim_{x \to 0+} \frac{2}{x} \left( \sin(\sin kx) + \cos x + x - 1 \right)$

We can rewrite $P$ by expanding the terms using Taylor series expansions around $x = 0$:

1. $\sin(\sin kx) = \sin(kx + O(x^3)) = kx + O(x^3)$
2. $\cos x = 1 - \frac{x^2}{2} + O(x^4)$

Substituting these expansions back into the expression: $\sin(\sin kx) + \cos x + x - 1 = \left( kx + O(x^3) \right) + \left( 1 - \frac{x^2}{2} + O(x^4) \right) + x - 1$ $= (k + 1)x - \frac{x^2}{2} + O(x^3)$

Now, substitute this result into the limit $P$: $P = \lim_{x \to 0+} \frac{2}{x} \left( (k + 1)x - \frac{x^2}{2} + O(x^3) \right)$ $P = \lim_{x \to 0+} \left( 2(k + 1) - x + O(x^2) \right) = 2(k + 1)$

Therefore, the original limit evaluates to: $e^P = e^{2(k + 1)}$

We are given that this limit equals $e^6$: $e^{2(k + 1)} = e^6$

Equating the exponents gives: $2(k + 1) = 6$ $k + 1 = 3$ $k = 2$

Thus, the correct option is (B).