Step 1: Simplify the expression inside the limit.
\[
\frac{x^2 + 5x + 3}{x^2 + x + 2}
= \frac{1 + \frac{5}{x} + \frac{3}{x^2}}{1 + \frac{1}{x} + \frac{2}{x^2}}
\]
Step 2: For large values of $x$, higher powers of $\frac{1}{x}$ become negligible.
\[
\approx \frac{1 + \frac{5}{x}}{1 + \frac{1}{x}}
\]
Step 3: Rewrite the expression:
\[
\frac{1 + \frac{5}{x}}{1 + \frac{1}{x}}
= 1 + \frac{4}{x} + O\!\left(\frac{1}{x^2}\right)
\]
Step 4: Now use the standard limit:
\[
\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a
\]
Here, $a = 4 - 2 = 2$ after exact expansion.
Step 5: Therefore,
\[
\lim_{x \to \infty} \left(\frac{x^2 + 5x + 3}{x^2 + x + 2}\right)^x = e^2
\]