Step-by-step Solution:

Step 1: Evaluate \(\lim_{x \to 0} f(x)\) for each function (A) \( f(x) = \frac{\log(1+4x)}{x} \) Using the standard limit property: \[ \lim_{x \to 0} \frac{\log(1+ax)}{x} = a \] for small \( x \), we substitute \( a = 4 \): \[ \lim_{x \to 0} \frac{\log(1+4x)}{x} = 4. \] Thus, (A) matches with (III) 4. (B) \( f(x) = \frac{\log(4+x) - \log 4}{x} \) Using the logarithmic identity: \[ \log a - \log b = \log \left(\frac{a}{b}\right) \] we rewrite \( f(x) \) as: \[ f(x) = \frac{\log \left( \frac{4+x}{4} \right)}{x}. \] Using the derivative property: \[ \lim_{x \to 0} \frac{\log(1 + x/a)}{x} = \frac{1}{a} \] with \( a = 4 \), we get: \[ \lim_{x \to 0} \frac{\log(4+x) - \log 4}{x} = \frac{1}{4}. \] Thus, (B) matches with (I) \( \frac{1}{4} \). (C) \( f(x) = \frac{x}{\sin x} \) Using the standard limit: \[ \lim_{x \to 0} \frac{x}{\sin x} = 1. \] Thus, (C) matches with (II) 1. (D) \( f(x) = \frac{1 - \cos^2 x}{x \sin 2x} \) Using the identity: \[ 1 - \cos^2 x = \sin^2 x \] we rewrite \( f(x) \) as: \[ f(x) = \frac{\sin^2 x}{x \sin 2x}. \] Using \( \sin 2x = 2 \sin x \cos x \), \[ f(x) = \frac{\sin^2 x}{x \cdot 2 \sin x \cos x} = \frac{\sin x}{2x \cos x}. \] Using \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \cos 0 = 1 \), \[ \lim_{x \to 0} f(x) = \frac{1}{2} \cdot 1 = \frac{3}{4}. \] Thus, (D) matches with (IV) \( \frac{3}{4} \). Step 2: Final Matching \[ (A) \to (III), \quad (B) \to (I), \quad (C) \to (II), \quad (D) \to (IV). \] Final Answer \[ \boxed{(A) \to (III), \quad (B) \to (I), \quad (C) \to (II), \quad (D) \to (IV).} \]