1. We need to evaluate

$$\lim_{x\to 0^+}\left(\sin(\sin kx)+\cos x+x\right)^{\frac{2}{x}}=e^6.$$

Let

$$F(x)=\sin(\sin kx)+\cos x+x.$$

As $x\to 0^+$,

$$\sin(\sin kx)\to 0,\quad \cos x\to 1,\quad x\to 0,$$

so

$$F(x)\to 1.$$

Thus the limit is of the standard form $1^{\infty}$.

2. Use the standard result:

$$\lim_{x\to 0^+}F(x)^{\frac{2}{x}}=e^{\lim\limits_{x\to 0^+}\frac{2}{x}(F(x)-1)}$$

provided the latter limit exists.

So we compute

$$F(x)-1=\sin(\sin kx)+\cos x+x-1.$$

3. Expand each term near $x=0$:

• Since $\sin t\sim t$ as $t\to 0$,

$$\sin(\sin kx)\sim \sin kx\sim kx.$$

• Also,

$$\cos x=1-\frac{x^2}{2}+o(x^2).$$

Hence,

$$F(x)-1 = kx + \left(1-\frac{x^2}{2}+o(x^2)\right)+x-1 = (k+1)x-\frac{x^2}{2}+o(x^2).$$

Therefore,

$$\frac{2}{x}(F(x)-1)=\frac{2}{x}\left((k+1)x-\frac{x^2}{2}+o(x^2)\right) =2(k+1)-x+o(x).$$

So,

$$\lim_{x\to 0^+}\frac{2}{x}(F(x)-1)=2(k+1).$$

4. Therefore,

$$\lim_{x\to 0^+}\left(\sin(\sin kx)+\cos x+x\right)^{\frac{2}{x}}=e^{2(k+1)}.$$

Given this equals $e^6$, we get

$$e^{2(k+1)}=e^6 \implies 2(k+1)=6.$$

Thus,

$$k+1=3 \implies k=2.$$

5. Check options:

• A: $1$ ❌
• B: $2$ ✅
• C: $3$ ❌
• D: $4$ ❌

So the correct option is B.