Key Concepts and Formulas

• Standard Limit Form 1: The fundamental limit form $\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{k}{x}} \right)^x = e^k$. This is a crucial result for evaluating limits of the form $(1 + \text{something tending to 0})^\text{something tending to infinity}$.
• Limit of a Power: If $\mathop {\lim }\limits_{x \to c} f(x) = L$ and $\mathop {\lim }\limits_{x \to c} g(x) = M$, then $\mathop {\lim }\limits_{x \to c} [f(x)]^{g(x)}$ can often be evaluated by considering $e^{\mathop {\lim }\limits_{x \to c} g(x) \log f(x)}$.
• Algebraic Manipulation for Limits: When dealing with indeterminate forms, algebraic manipulation is often required to simplify the expression into a recognizable limit form.

Step-by-Step Solution

Step 1: Identify the Indeterminate Form As $x \to \infty$, the base of the given expression $\left( {1 + \frac{a}{x} + \frac{b}{x^2}} \right)$ approaches $(1 + 0 + 0) = 1$. The exponent $2x$ approaches $\infty$. This is an indeterminate form of the type $1^\infty$.

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}}$

Step 2: Apply the Exponential Transformation To evaluate limits of the form $1^\infty$, we can use the property $\lim_{y \to c} [f(y)]^{g(y)} = e^{\lim_{y \to c} g(y) \log f(y)}$. Here, $f(x) = 1 + \frac{a}{x} + \frac{b}{x^2}$ and $g(x) = 2x$.

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^{\mathop {\lim }\limits_{x \to \infty } 2x \log \left( {1 + {a \over x} + {b \over {{x^2}}}} \right)}}$

Step 3: Use the Taylor Expansion of $\log(1+u)$ For small values of $u$, the Taylor expansion of $\log(1+u)$ is $\log(1+u) = u - \frac{u^2}{2} + \dots$. As $x \to \infty$, $\frac{a}{x} + \frac{b}{x^2}$ approaches 0. Let $u = \frac{a}{x} + \frac{b}{x^2}$.

$\log \left( {1 + {a \over x} + {b \over {{x^2}}}} \right) = \left( {a \over x} + {b \over {{x^2}}} \right) - \frac{1}{2} \left( {a \over x} + {b \over {{x^2}}} \right)^2 + \dots$

Step 4: Simplify the Exponent Term Substitute the expansion back into the exponent:

$\mathop {\lim }\limits_{x \to \infty } 2x \log \left( {1 + {a \over x} + {b \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to \infty } 2x \left[ \left( {a \over x} + {b \over {{x^2}}} \right) - \frac{1}{2} \left( {a^2 \over {{x^2}}} + \dots \right) + \dots \right]$

Now, distribute the $2x$:

$= \mathop {\lim }\limits_{x \to \infty } \left[ 2x \left( {a \over x} \right) + 2x \left( {b \over {{x^2}}} \right) - 2x \left( \frac{1}{2} {a^2 \over {{x^2}}} \right) + \dots \right]$

$= \mathop {\lim }\limits_{x \to \infty } \left[ 2a + {2b \over x} - {a^2 \over x} + \dots \right]$

Step 5: Evaluate the Limit of the Exponent As $x \to \infty$, terms with $x$ in the denominator go to 0.

$\mathop {\lim }\limits_{x \to \infty } \left[ 2a + {2b \over x} - {a^2 \over x} + \dots \right] = 2a + 0 - 0 + \dots = 2a$

Step 6: Equate the Result with the Given Limit We are given that the original limit is $e^2$. Therefore, the exponent must be equal to 2.

$e^{2a} = e^2$

This implies:

$2a = 2$ $a = 1$

Step 7: Consider the Value of $b$ In the limit calculation of the exponent, the term involving $b$ was $\frac{2b}{x}$, which tends to 0 as $x \to \infty$. This means that the value of $b$ does not affect the leading term of the exponent, which determines the value of $a$. Therefore, $b$ can be any real number.

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^{2a}}$

Since we found $a = 1$, the limit becomes $e^{2(1)} = e^2$, which matches the given condition. The value of $b$ does not influence this outcome, as long as the expression remains well-defined.

Common Mistakes & Tips

• Incorrectly applying $\lim_{x \to \infty} (1 + k/x)^x = e^k$: This formula is for a specific form. The given problem has an additional term $b/x^2$ in the base and a different multiplier for $x$ in the exponent.
• Forgetting the $\log$ in the exponential transformation: The correct transformation is $e^{\lim g(x) \log f(x)}$, not just $\lim g(x) f(x)$.
• Approximation errors in Taylor series: Ensure you use enough terms in the Taylor expansion of $\log(1+u)$ to correctly capture the dominant terms as $x \to \infty$. For this problem, the first two terms of $u$ are sufficient.

Summary

The problem involves evaluating a limit of the indeterminate form $1^\infty$. We transformed the limit using the exponential property $\lim [f(x)]^{g(x)} = e^{\lim g(x) \log f(x)}$. By using the Taylor expansion of $\log(1+u)$ for small $u$, we simplified the exponent. Equating the resulting limit of the exponent with the given exponent of $e$ allowed us to find the value of $a$. The term involving $b$ in the exponent vanished as $x \to \infty$, indicating that $b$ can be any real number.

The final answer is $a=1$ and $b \in \mathbb{R}$. This corresponds to option (C).

The final answer is \boxed{a=1, b \in R}.