Let $k\in <!--\; \in \; -->\mathbb{R}$. If $\underset{x\to <!--\; \to \; -->0^{+}}{lim}(\sin <!--\; \; -->(\sin <!--\; \; -->kx)+\cos <!--\; \; -->x+x{)}^{\frac{2}{x}}=e^6$, then the value of k is

The correct answer is 2.

We are given that:
$\underset{x\to 0^{+}}{lim}(\sin (\sin kx)+\cos x+x{)}^{\frac{2}{x}}=e^6$

Let the limit be denoted by L. As $x\to 0^{+}$, the base expression is:
$\sin (\sin (k·0))+\cos (0)+0=0+1+0=1$
and the exponent is:
$\frac{2}{x}\to \infty$
Thus, the limit is of the indeterminate form $1^{\infty }$.

For a limit of the form $\underset{x\to a}{lim}[f(x){]}^{g(x)}$ where $f(x)\to 1$ and $g(x)\to \infty$, the limit is evaluated as:
$L=e^{\underset{x\to a}{lim}(f(x)-1)g(x)}$

Applying this rule to our limit:
$L=e^{\underset{x\to 0^{+}}{lim}(\sin (\sin kx)+\cos x+x-1)·\frac{2}{x}}$

Let us evaluate the limit in the exponent:
$E=\underset{x\to 0^{+}}{lim}2·[\frac{\sin (\sin kx)}{x}+\frac{\cos x-1}{x}+\frac{x}{x}]$

We can analyze each term inside the brackets as $x\to 0^{+}$:
1) First term:
$\underset{x\to 0^{+}}{lim}\frac{\sin (\sin kx)}{x}=\underset{x\to 0^{+}}{lim}[\frac{\sin (\sin kx)}{\sin kx}·\frac{\sin kx}{kx}·k]=1·1·k=k$
2) Second term:
$\underset{x\to 0^{+}}{lim}\frac{\cos x-1}{x}=\underset{x\to 0^{+}}{lim}-x·\left(\frac{1-\cos x}{x^2}\right)=-0·\frac{1}{2}=0$
3) Third term:
$\underset{x\to 0^{+}}{lim}\frac{x}{x}=1$

Substituting these values back into the exponent expression, we obtain:
$E=2·(k+0+1)=2(k+1)$

Therefore, the limit is:
$L=e^{2(k+1)}$

We are given that $L=e^6$. Comparing the exponents:
$2(k+1)=6$
$k+1=3$
$k=2$

Thus, the value of k is 2.