The value of the function \[{\text{f}}\left( {\text{x}} \right) = \mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{{{\text{x}}^3} + {{\text{x}}^2}}}{{2{{\text{x}}^3} – 7{{\text{x}}^2}}}\] is A. 0 B. \[ – \frac{1}{7}\] C. \[\frac{1}{7}\] D. \[\infty \]

The correct answer is $\boxed{\text{D}. \infty}$.

To solve this, we can use direct substitution. However, when we substitute $x=0$, we get the indeterminate form $\frac{0}{0}$. This means that we cannot evaluate the limit directly.

To solve this, we can use L’Hôpital’s rule. L’Hôpital’s rule states that if $\lim_{x\to a} \frac{f(x)}{g(x)}$ is indeterminate, then $\lim_{x\to a} \frac{f'(x)}{g'(x)}$ also exists and is equal to the limit of the original expression.

In this case, we have:

$$\begin{align}
\lim_{x\to 0} \frac{{x^3} + {x^2}}{{2x^3} – 7x^2} &= \lim_{x\to 0} \frac{{3x^2} + 2x}{{6x^2} – 14x} \
&= \lim_{x\to 0} \frac{{3x^2} + 2x}{{6x^2} – 14x} \cdot \frac{{x^2}}{{x^2}} \
&= \lim_{x\to 0} \frac{{3x} + 2}{{6} – 14} \
&= \lim_{x\to 0} \frac{{3x} + 2}{{-8}} \
&= \infty
\end{align}$$

Therefore, the value of the function $\text{f}(x) = \lim_{x\to 0} \frac{{x^3} + {x^2}}{{2x^3} – 7x^2} = \infty$.