$$\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{{{\sin }^2}{\text{x}}}}{{\text{x}}}$$ is equal to A. 0 B. $$\infty $$ C. 1 D. -1

The correct answer is $\boxed{1}$.

The limit $\lim_{x\to 0}\frac{\sin^2(x)}{x}$ can be evaluated using the following steps:

1. Substitute $x=0$ into the expression $\frac{\sin^2(x)}{x}$. This results in the indeterminate form $\frac{0}{0}$.
2. Use L’Hôpital’s rule, which states that if $\lim_{x\to a}\frac{f(x)}{g(x)}$ is equal to the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x\to a}\frac{f'(x)}{g'(x)}$ exists and is equal to the limit $\lim_{x\to a}\frac{f(x)}{g(x)}$.
3. Applying L’Hôpital’s rule to the limit $\lim_{x\to 0}\frac{\sin^2(x)}{x}$, we get $\lim_{x\to 0}\frac{2\sin(x)\cos(x)}{1}$.
4. Simplifying the expression $\lim_{x\to 0}\frac{2\sin(x)\cos(x)}{1}$, we get $\lim_{x\to 0}\sin(2x)$.
5. Using the substitution $u=2x$, we get $\lim_{x\to 0}\sin(2x)=\lim_{u\to 0}\sin(u)$.
6. The value of $\lim_{u\to 0}\sin(u)$ is equal to $1$.
7. Therefore, the value of $\lim_{x\to 0}\frac{\sin^2(x)}{x}$ is also equal to $1$.

The other options are incorrect because they do not represent the value of the limit $\lim_{x\to 0}\frac{\sin^2(x)}{x}$.