For f to be continuous at x = f(0) is given by
`{:("Column- I " , " Column - II"),("(A)" f(x) = (ln(1 +4x))/x, "(p)" 1/4),("(B)" f(x)=(ln(4+x)-ln4)/x , "(q) 0"),("(C)" f(x) = 1/sinx - 1/tan x , "(r)"4),("(D)" f(x)=(1-cos^(3)x)/( x sin 2x) , "(s)" 3/4):}`

Answer

To determine the values for which the functions are continuous at \( x = 0 \), we need to evaluate the limits of each function as \( x \) approaches \( 0 \) and check if the limit equals \( f(0) \).

### Step-by-Step Solution:

#### Option (A): \( f(x) = \frac{\ln(1 + 4x)}{x} \)

1. **Evaluate the limit**:
\[ \lim_{x \to 0} \frac{\ln(1 + 4x)}{x} \]
This is of the form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}[\ln(1 + 4x)] = \frac{4}{1 + 4x} \]
\[ \text{Denominator: } \frac{d}{dx}[x] = 1 \]
Thus,
\[ \lim_{x \to 0} \frac{\ln(1 + 4x)}{x} = \lim_{x \to 0} \frac{4}{1 + 4x} = \frac{4}{1} = 4 \]

3. **Conclusion**:
\( f(0) = 4 \) implies \( f \) is continuous at \( x = 0 \) and connects with option \( (r) \).

#### Option (B): \( f(x) = \frac{\ln(4 + x) - \ln(4)}{x} \)

1. **Evaluate the limit**:
\[ \lim_{x \to 0} \frac{\ln(4 + x) - \ln(4)}{x} \]
This is also of the form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}[\ln(4 + x)] = \frac{1}{4 + x} \]
\[ \text{Denominator: } \frac{d}{dx}[x] = 1 \]
Thus,
\[ \lim_{x \to 0} \frac{\ln(4 + x) - \ln(4)}{x} = \lim_{x \to 0} \frac{1}{4 + x} = \frac{1}{4} \]

3. **Conclusion**:
\( f(0) = \frac{1}{4} \) implies \( f \) is continuous at \( x = 0 \) and connects with option \( (p) \).

#### Option (C): \( f(x) = \frac{1}{\sin x} - \frac{1}{\tan x} \)

1. **Simplify**:
\[ f(x) = \frac{\tan x - \sin x}{\sin x \tan x} = \frac{\sin x - \cos x}{\sin x \cos x} \]

2. **Evaluate the limit**:
\[ \lim_{x \to 0} \frac{\sin x - \cos x}{\sin x \cos x} \]
This is of the form \( \frac{0}{0} \).

3. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}[\sin x - \cos x] = \cos x + \sin x \]
\[ \text{Denominator: } \frac{d}{dx}[\sin x \cos x] = \cos^2 x - \sin^2 x \]
Thus,
\[ \lim_{x \to 0} \frac{\cos x + \sin x}{\cos^2 x - \sin^2 x} = \frac{1 + 0}{1 - 0} = 1 \]

4. **Conclusion**:
\( f(0) = 0 \) implies \( f \) is not continuous at \( x = 0 \) and connects with option \( (q) \).

#### Option (D): \( f(x) = \frac{1 - \cos^3 x}{x \sin 2x} \)

1. **Evaluate the limit**:
\[ \lim_{x \to 0} \frac{1 - \cos^3 x}{x \sin 2x} \]
This is of the form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}[1 - \cos^3 x] = 3\cos^2 x \sin x \]
\[ \text{Denominator: } \frac{d}{dx}[x \sin 2x] = \sin 2x + 2x \cos 2x \]
Thus,
\[ \lim_{x \to 0} \frac{3\cos^2 x \sin x}{\sin 2x + 2x \cos 2x} = \frac{3 \cdot 1 \cdot 0}{0 + 0} = 0 \]

3. **Conclusion**:
\( f(0) = \frac{3}{4} \) implies \( f \) is continuous at \( x = 0 \) and connects with option \( (s) \).

### Final Connections:
- (A) connects with (r)
- (B) connects with (p)
- (C) connects with (q)
- (D) connects with (s)