Which of the following function is not continuous at x = 0?

Options

A. `f(x)=(1+2x)^(1//x),xne0`
`=e^(2),x=0`
B. `f(x)=sinx-cosx,xne0`
=-1,x=0
C. `f(x)=(e^(1//x)-1)/(e^(1//x)),xne0`
`=-1,x=0`
D. `f(x)=(e^(5x)-e^(2x))/(sin3x),xne0`
`=1,x=0`

Explanation

To determine which of the given functions is not continuous at \( x = 0 \), we will analyze each option step by step.

### Step 1: Analyze Option 1
Given the function:
\[ f(x) = \begin{cases} 1 + 2x^{\frac{1}{x}} & \text{if } x \neq 0 \\ e^2 & \text{if } x = 0 \end{cases} \]
We need to check if \( \lim_{x \to 0} f(x) = f(0) \).

1. **Calculate \( f(0) \)**:
\[ f(0) = e^2 \]

2. **Calculate \( \lim_{x \to 0} f(x) \)**:
\[ \lim_{x \to 0} (1 + 2x^{\frac{1}{x}}) \]
To evaluate this limit, we can rewrite \( 2x^{\frac{1}{x}} \) using logarithms:
\[ y = 2x^{\frac{1}{x}} \implies \log y = \frac{1}{x} \log(2x) = \frac{\log 2 + \log x}{x} \]
As \( x \to 0 \), \( \log x \to -\infty \), hence \( \frac{\log x}{x} \to 0 \) and \( \log y \to 0 \). Thus, \( y \to e^0 = 1 \).

Therefore,
\[ \lim_{x \to 0} f(x) = 1 + 1 = 2 \]

3. **Compare limits**:
\[ \lim_{x \to 0} f(x) = 2 \neq f(0) = e^2 \]
Thus, **Option 1 is not continuous at \( x = 0 \)**.

### Step 2: Analyze Option 2
Given the function:
\[ f(x) = \begin{cases} \sin x - \cos x & \text{if } x \neq 0 \\ -1 & \text{if } x = 0 \end{cases} \]
1. **Calculate \( f(0) \)**:
\[ f(0) = -1 \]

2. **Calculate \( \lim_{x \to 0} f(x) \)**:
\[ \lim_{x \to 0} (\sin x - \cos x) = \sin(0) - \cos(0) = 0 - 1 = -1 \]

3. **Compare limits**:
\[ \lim_{x \to 0} f(x) = -1 = f(0) \]
Thus, **Option 2 is continuous at \( x = 0 \)**.

### Step 3: Analyze Option 3
Given the function:
\[ f(x) = \begin{cases} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}}} & \text{if } x \neq 0 \\ -1 & \text{if } x = 0 \end{cases} \]
1. **Calculate \( f(0) \)**:
\[ f(0) = -1 \]

2. **Calculate \( \lim_{x \to 0} f(x) \)**:
As \( x \to 0^+ \), \( e^{\frac{1}{x}} \to \infty \):
\[ \lim_{x \to 0} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}}} = \lim_{x \to 0} \left(1 - \frac{1}{e^{\frac{1}{x}}}\right) = 1 - 0 = 1 \]

3. **Compare limits**:
\[ \lim_{x \to 0} f(x) = 1 \neq f(0) = -1 \]
Thus, **Option 3 is not continuous at \( x = 0 \)**.

### Step 4: Analyze Option 4
Given the function:
\[ f(x) = \begin{cases} e^{-2x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]
1. **Calculate \( f(0) \)**:
\[ f(0) = 1 \]

2. **Calculate \( \lim_{x \to 0} f(x) \)**:
\[ \lim_{x \to 0} e^{-2x} = e^0 = 1 \]

3. **Compare limits**:
\[ \lim_{x \to 0} f(x) = 1 = f(0) \]
Thus, **Option 4 is continuous at \( x = 0 \)**.

### Conclusion
The function that is not continuous at \( x = 0 \) is **Option 1** and **Option 3**.