Let $k \in \mathbb{R}$. If $\lim\limits_{x \to 0^+} (\sin(\sin kx) + \cos x + x)^{2/x} = e^6$, then the value of $k$ is
Step-by-Step Solution
Key Concept: The limit of a $1^\infty$ form is $e^L$, where $L = \lim (B(x)-1)E(x)$. Using standard limits like $\lim_{x\to0} \frac{\sin x}{x} = 1$ and $\lim_{x\to0} \frac{1-\cos x}{x} = 0$.
As $x \to 0^+$, the limit is of the form $1^\infty$.
Let $L = \lim\limits_{x \to 0^+} \dfrac{2}{x} (\sin(\sin kx) + \cos x + x - 1)$:
$$L = 2 \lim_{x \to 0^+} \left( \dfrac{\sin(\sin kx)}{x} + \dfrac{\cos x - 1}{x} + \dfrac{x}{x} \right)$$
Evaluate each limit separately:
1) $\lim\limits_{x \to 0^+} \dfrac{\sin(\sin kx)}{x} = \lim\limits_{x \to 0^+} \dfrac{\sin(\sin kx)}{\sin kx} \cdot \dfrac{\sin kx}{kx} \cdot k = k$
2) $\lim\limits_{x \to 0^+} \dfrac{\cos x - 1}{x} = 0$
3) $\lim\limits_{x \to 0^+} \dfrac{x}{x} = 1$
Thus:
$$L = 2(k + 0 + 1) = 2k + 2$$
We are given the limit is $e^6$, so:
$$2k + 2 = 6 \implies 2k = 4 \implies k = 2$$
Thus, the correct option is B.
Correct Answer: B