Question 3

The limit is of the form $1^\infty$. Let $L = \lim_{x \to 0^+} (\sin(\sin kx) + \cos x + x)^{\frac{2}{x}}$. Then $L = e^P$ where $P = \lim_{x \to 0^+} \frac{2}{x} (\sin(\sin kx) + \cos x + x - 1)$. Using Taylor expansions for $x \to 0$: $\sin(\sin kx) = kx + O(x^3)$ $\cos x = 1 - \frac{x^2}{2} + O(x^4)$ Substituting these into $P$: $P = \lim_{x \to 0^+} \frac{2}{x} \left( kx + \left(1 - \frac{x^2}{2}\right) + x - 1 \right) = \lim_{x \to 0^+} \frac{2((k+1)x - x^2/2)}{x}$ $P = \lim_{x \to 0^+} 2(k + 1 - x/2) = 2(k + 1)$ Given $e^{2(k+1)} = e^6$, we have $2(k+1) = 6 \Rightarrow k+1 = 3 \Rightarrow k = 2$.