Let $$k \in \mathbb{R}$$. If $$\lim_{x \to 0^+} \left(\sin(\sin kx) + \cos x + x\right)^{\frac{2}{x}} = e^6$$, then the value of $$k$$ is

Let

$$F(x)=\left(\sin(\sin kx)+\cos x+x\right)^{\frac{2}{x}}$$

and put

$$L=\lim_{x\to 0^{+}}F(x).$$

We know the standard limit property: if $$u(x)\to 0$$ and $$v(x)\to\infty$$ such that $$u(x)\,v(x)\to \ell,$$ then

$$\lim_{x\to 0}\bigl(1+u(x)\bigr)^{v(x)}=e^{\ell}.$$

Hence we must rewrite the base inside the bracket in the form $$1+u(x).$$

Expand each elementary function near $$x=0$$ (retain terms up to $$x^{2}$$ because the final coefficient of $$x$$ decides the limit):

1. $$\cos x = 1-\dfrac{x^{2}}{2}+O(x^{4}).$$

2. $$\sin kx = kx-\dfrac{k^{3}x^{3}}{6}+O(x^{5}).$$

3. For small $$y$$, $$\sin y = y-\dfrac{y^{3}}{6}+O(y^{5}).$$ Put $$y=\sin kx$$ in this expansion:

$$\sin(\sin kx)=\left(kx-\dfrac{k^{3}x^{3}}{6}\right)-\dfrac{(kx)^{3}}{6}+O(x^{5}) =kx-\dfrac{k^{3}x^{3}}{6}-\dfrac{k^{3}x^{3}}{6}+O(x^{5}) =kx-\dfrac{k^{3}x^{3}}{3}+O(x^{5}).$$

Now add the three series along with the term $$x$$:

$$\sin(\sin kx)+\cos x+x =\left(kx-\dfrac{k^{3}x^{3}}{3}\right)+\left(1-\dfrac{x^{2}}{2}\right)+x+O(x^{3}).$$

Combine like powers of $$x$$:

$$=1+\bigl(k+1\bigr)x-\dfrac{x^{2}}{2}+O(x^{3}).$$

Thus the bracket equals $$1+u(x)$$ where

$$u(x)=\bigl(k+1\bigr)x-\dfrac{x^{2}}{2}+O(x^{3})$$

and indeed $$u(x)\to 0$$ as $$x\to 0^{+}.$$

Take natural logarithm to prepare the exponent:

$$\ln F(x)=\frac{2}{x}\,\ln\!\bigl(1+u(x)\bigr).$$

For small $$u(x)$$, $$\ln(1+u)=u-\dfrac{u^{2}}{2}+O(u^{3}).$$ Using only the linear part (higher powers vanish after multiplying by $$\dfrac{2}{x}$$ and then letting $$x\to 0$$):

$$\ln\!\bigl(1+u(x)\bigr)=u(x)+O\!\bigl(u(x)^{2}\bigr) =\bigl(k+1\bigr)x+O(x^{2}).$$

Therefore

$$\ln F(x)=\frac{2}{x}\,\Bigl[\bigl(k+1\bigr)x+O(x^{2})\Bigr] =2\bigl(k+1\bigr)+O(x).$$

Sending $$x\to 0^{+}$$ yields

$$\lim_{x\to 0^{+}}\ln F(x)=2\bigl(k+1\bigr).$$

Consequently

$$L=\exp\!\Bigl[2\bigl(k+1\bigr)\Bigr] =e^{2(k+1)}.$$

The problem states that $$L=e^{6},$$ so

$$2(k+1)=6\quad\Longrightarrow\quad k+1=3\quad\Longrightarrow\quad k=2.$$

Hence the required value of $$k$$ is $$2$$.

Option B which is: $$2$$