Key Concepts and Formulas

• Indeterminate Forms: The limit of the form $1^\infty$ is an indeterminate form.
• Limit of the form $1^\infty$: If $\mathop {\lim }\limits_{x \to a} f(x) = 1$ and $\mathop {\lim }\limits_{x \to a} g(x) = \infty$, then $\mathop {\lim }\limits_{x \to a} [f(x)]^{g(x)} = e^{\mathop {\lim }\limits_{x \to a} g(x)[f(x) - 1]}$.
• Algebraic Simplification: Basic algebraic manipulations, including finding a common denominator, are essential for simplifying expressions.

Step-by-Step Solution

Step 1: Identify the form of the limit. We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$ Let's substitute $x=0$ into the expression. The base becomes: $\frac{3(0)^2 + 2}{7(0)^2 + 2} = \frac{2}{2} = 1$. The exponent becomes: $\frac{1}{0^2} = \frac{1}{0}$, which tends to $\infty$. Thus, the limit is of the indeterminate form $1^\infty$.

Step 2: Apply the formula for the $1^\infty$ indeterminate form. For a limit of the form $1^\infty$, where $\mathop {\lim }\limits_{x \to a} f(x) = 1$ and $\mathop {\lim }\limits_{x \to a} g(x) = \infty$, the limit $\mathop {\lim }\limits_{x \to a} [f(x)]^{g(x)}$ can be evaluated using the formula $e^{\mathop {\lim }\limits_{x \to a} g(x)[f(x) - 1]}$. In our case, $f(x) = \frac{3x^2 + 2}{7x^2 + 2}$ and $g(x) = \frac{1}{x^2}$, and $a=0$. So, the limit $L$ can be rewritten as: $L = e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{x^2} \left[ \frac{3x^2 + 2}{7x^2 + 2} - 1 \right]}$

Step 3: Simplify the expression inside the exponent. We need to simplify the term $\frac{3x^2 + 2}{7x^2 + 2} - 1$. $\frac{3x^2 + 2}{7x^2 + 2} - 1 = \frac{3x^2 + 2 - (7x^2 + 2)}{7x^2 + 2}$ $= \frac{3x^2 + 2 - 7x^2 - 2}{7x^2 + 2}$ $= \frac{-4x^2}{7x^2 + 2}$

Step 4: Substitute the simplified expression back into the exponent and evaluate the limit. Now, substitute this simplified expression back into the exponent: $\mathop {\lim }\limits_{x \to 0} \frac{1}{x^2} \left[ \frac{-4x^2}{7x^2 + 2} \right]$ We can cancel out the $x^2$ terms: $\mathop {\lim }\limits_{x \to 0} \frac{-4}{7x^2 + 2}$ Now, substitute $x=0$ into this expression: $\frac{-4}{7(0)^2 + 2} = \frac{-4}{0 + 2} = \frac{-4}{2} = -2$

Step 5: Determine the final value of the limit. The limit of the exponent is $-2$. Therefore, the original limit $L$ is $e$ raised to this value: $L = e^{-2}$ This can also be written as: $L = \frac{1}{e^2}$

Common Mistakes & Tips

• Incorrectly identifying the indeterminate form: Always check the form of the limit by direct substitution before applying any special formulas. If it's not an indeterminate form, the direct substitution result is the answer.
• Algebraic errors during simplification: Pay close attention to signs and common denominators when simplifying fractions, especially when subtracting 1 from a rational expression.
• Forgetting the 'e': Remember that the formula for the $1^\infty$ form involves $e$ as the base. The final answer is $e$ raised to the calculated limit of the exponent, not just the limit of the exponent itself.

Summary

The given limit was of the indeterminate form $1^\infty$. We successfully transformed it into the form $e^{\mathop {\lim }\limits_{x \to 0} g(x)[f(x) - 1]}$ by identifying the base $f(x)$ and the exponent $g(x)$. After simplifying the expression within the exponent through algebraic manipulation, we evaluated the limit of the exponent. Finally, we raised $e$ to the power of this evaluated limit to obtain the value of the original limit.

The final answer is \boxed{1/e^2}. This corresponds to option (C).