Key Concepts and Formulas

• Limit of the form $1^\infty$: A limit of the form $\mathop {\lim }\limits_{x \to a} {f(x)^{g(x)}}$ where $\mathop {\lim }\limits_{x \to a} f(x) = 1$ and $\mathop {\lim }\limits_{x \to a} g(x) = \infty$ can be evaluated using the formula: $\mathop {\lim }\limits_{x \to a} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x \to a} {g(x)(f(x)-1)}}$
• Algebraic Manipulation of Limits: Techniques like dividing numerator and denominator by the highest power of $x$ are crucial for evaluating limits at infinity.
• Standard Limit: The standard limit $\mathop {\lim }\limits_{y \to \infty} {\left( {1 + {1 \over y}} \right)^y} = e$ or its variant $\mathop {\lim }\limits_{x \to a} {\left( {1 + h(x)} \right)^{k(x)}} = e^{\mathop {\lim }\limits_{x \to a} {h(x)k(x)}}$ where $h(x) \to 0$ and $k(x) \to \infty$ as $x \to a$.

Step-by-Step Solution

Step 1: Identify the form of the limit. We are asked to evaluate $\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}$. As $x \to \infty$, the base $\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)$ approaches $\mathop {\lim }\limits_{x \to \infty } {{{x^2}(1 + {5/x} + {3/x^2})} \over {{x^2}(1 + {1/x} + {2/x^2})}} = {{1+0+0} \over {1+0+0}} = 1$. The exponent $x$ approaches $\infty$. Thus, the limit is of the indeterminate form $1^\infty$.

Step 2: Rewrite the base in the form $1 + h(x)$. We rewrite the base $\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)$ as $1$ plus a term that goes to zero as $x \to \infty$. $\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 2}} = \frac{{({x^2} + x + 2) + (4x + 1)}}{{{x^2} + x + 2}} = 1 + \frac{{4x + 1}}{{{x^2} + x + 2}}$ So, the limit becomes: $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {{4x + 1} \over {{x^2} + x + 2 }}} \right)^x}$

Step 3: Apply the formula for the $1^\infty$ indeterminate form. Using the formula $\mathop {\lim }\limits_{x \to a} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x \to a} {g(x)(f(x)-1)}}$, where $f(x) = 1 + \frac{{4x + 1}}{{{x^2} + x + 2}}$ and $g(x) = x$. The exponent of $e$ will be: $\mathop {\lim }\limits_{x \to \infty } {x \left( {\left( {1 + {{4x + 1} \over {{x^2} + x + 2 }}} \right) - 1} \right)}$ $= \mathop {\lim }\limits_{x \to \infty } {x \left( {{{4x + 1} \over {{x^2} + x + 2 }}} \right)}$ $= \mathop {\lim }\limits_{x \to \infty } {{{4x^2 + x} \over {{x^2} + x + 2 }}}$

Step 4: Evaluate the limit of the exponent. To evaluate $\mathop {\lim }\limits_{x \to \infty } {{{4x^2 + x} \over {{x^2} + x + 2 }}}$, we divide the numerator and the denominator by the highest power of $x$ in the denominator, which is $x^2$. \mathop {\lim }\limits_{x \to \infty } {{{4x^2/x^2 + x/x^2} \over {{x^2/x^2} + x/x^2 + 2/x^2}}}} $= \mathop {\lim }\limits_{x \to \infty } {{{4 + {1/x}} \over {1 + {1/x} + {2/x^2}}}}$ As $x \to \infty$, the terms $1/x$, $2/x^2$, and $1/x$ all approach $0$. $= {{4 + 0} \over {1 + 0 + 0}} = 4$

Step 5: Combine the result with the base $e$. The original limit is $e$ raised to the power of the limit we just calculated. $\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x} = e^4$

Common Mistakes & Tips

• Incorrectly Identifying the Form: Always check if the limit is indeed of the $1^\infty$ form before applying the special formula. If it's $0^0$, $\infty^0$, or $0^\infty$, different techniques apply.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when subtracting 1 from the base. A small error in calculating the numerator of $f(x)-1$ can lead to a wrong answer.
• Direct Application of $e^{\lambda}$: The formula $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \lambda x} \right)^{{1 \over x}}} = {e^\lambda }$ is specific. The general form $\mathop {\lim }\limits_{x \to a} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x \to a} {g(x)(f(x)-1)}}$ is more universally applicable for $1^\infty$ forms. The provided solution used a slightly different notation in the final step which can be confusing if not fully understood. The core idea is to evaluate the limit of the exponent.

Summary

The given limit is of the indeterminate form $1^\infty$. To evaluate it, we first rewrite the base of the expression in the form $1 + h(x)$, where $h(x) \to 0$ as $x \to \infty$. Then, we use the standard formula for $1^\infty$ limits, which states that $\mathop {\lim }\limits_{x \to \infty} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x \to \infty} {g(x)(f(x)-1)}}$. We calculated the limit of the exponent $g(x)(f(x)-1)$ by simplifying the expression and dividing the numerator and denominator by the highest power of $x$. The resulting limit of the exponent was 4, leading to the final answer of $e^4$.

The final answer is $\boxed{e^4}$. This corresponds to option (A).