Which of the following functions is continuous at x = 0 ?

Options

A. `f(x)={("sin"(2x)/(x)",", x ne 0),(1",", x =0):}`
B. `f(x)={((1+x)^((1)/(x))",", x ne 0),(1",", x =0):}`
C. `f(x)={(e^((-1)/(x))",", x ne 0),(1",", x =0):}`
D. `f(x)={((3x+4 tan x)/(x)",", "if " x ne 0),(7",","if " x =0):}`

Explanation

To determine which of the given functions is continuous at \( x = 0 \), we need to analyze each option one by one. A function \( f(x) \) is continuous at \( x = a \) if:

1. \( f(a) \) is defined.
2. \( \lim_{x \to a} f(x) \) exists.
3. \( \lim_{x \to a} f(x) = f(a) \).

Let's evaluate each option:

### Option A: \( f(x) = \frac{\sin(2x)}{x} \)

1. **Finding the limit as \( x \to 0 \)**:
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} \frac{2\cos(2x)}{1} = 2 \]
Since \( f(0) \) is not defined (division by zero), this function is not continuous at \( x = 0 \).

### Option B: \( f(x) = 1 + x^{\frac{1}{x}} \)

1. **Finding the limit as \( x \to 0 \)**:
\[ \lim_{x \to 0} x^{\frac{1}{x}} = \lim_{x \to 0} e^{\frac{\ln(x)}{x}} = e^{-\infty} = 0 \]
Thus,
\[ \lim_{x \to 0} f(x) = 1 + 0 = 1 \]
Since \( f(0) \) is not defined, this function is not continuous at \( x = 0 \).

### Option C: \( f(x) = e^{-\frac{1}{x}} \)

1. **Finding the limit as \( x \to 0 \)**:
\[ \lim_{x \to 0} e^{-\frac{1}{x}} = e^{-\infty} = 0 \]
Since \( f(0) \) is not defined, this function is not continuous at \( x = 0 \).

### Option D: \( f(x) = \frac{3x + 4\tan(x)}{x} \)

1. **Simplifying the function**:
\[ f(x) = 3 + \frac{4\tan(x)}{x} \]

2. **Finding the limit as \( x \to 0 \)**:
\[ \lim_{x \to 0} \frac{\tan(x)}{x} = 1 \quad \text{(using the standard limit)} \]
Thus,
\[ \lim_{x \to 0} f(x) = 3 + 4 \cdot 1 = 7 \]
Since \( f(0) \) is defined as \( 7 \), this function is continuous at \( x = 0 \).

### Conclusion:
The function that is continuous at \( x = 0 \) is option D: \( f(x) = \frac{3x + 4\tan(x)}{x} \).

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