If `f(x) ={((5^(x)-e^(x))/(sin 2x),",",x != 0),(1/2(log 5+1),",",x = 0):}` , then

Options

A. f(x) is continuous at `x = 0`
B. f(x) is discontinuous at `x = 0`
C. `lim_(x to 0)f(x)` does not exist
D. none of these

Answer

To determine if the function \( f(x) \) is continuous at \( x = 0 \), we need to check if the following condition holds:

\[ \lim_{x \to 0} f(x) = f(0) \]

### Step 1: Identify the function values
The function is defined as:
\[ f(x) = \begin{cases} \frac{5^x - e^x}{\sin(2x)} & \text{if } x \neq 0 \\ \frac{1}{2}(\log 5 + 1) & \text{if } x = 0 \end{cases} \]

### Step 2: Calculate \( f(0) \)
From the definition, we have:
\[ f(0) = \frac{1}{2}(\log 5 + 1) \]

### Step 3: Calculate the limit as \( x \) approaches 0
We need to find:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{5^x - e^x}{\sin(2x)} \]

### Step 4: Substitute \( x = 0 \) into the limit
Substituting \( x = 0 \) directly gives:
\[ \frac{5^0 - e^0}{\sin(2 \cdot 0)} = \frac{1 - 1}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 5: Apply L'Hôpital's Rule
Differentiate the numerator and denominator:
- The derivative of the numerator \( 5^x - e^x \) is \( 5^x \ln(5) - e^x \).
- The derivative of the denominator \( \sin(2x) \) is \( 2 \cos(2x) \).

Thus, we have:
\[ \lim_{x \to 0} \frac{5^x \ln(5) - e^x}{2 \cos(2x)} \]

### Step 6: Substitute \( x = 0 \) again
Now substituting \( x = 0 \):
\[ \frac{5^0 \ln(5) - e^0}{2 \cos(0)} = \frac{1 \cdot \ln(5) - 1}{2 \cdot 1} = \frac{\ln(5) - 1}{2} \]

### Step 7: Compare the limit with \( f(0) \)
Now we compare:
\[ \lim_{x \to 0} f(x) = \frac{\ln(5) - 1}{2} \]
with
\[ f(0) = \frac{1}{2}(\log 5 + 1) \]

### Step 8: Check for equality
We need to check if:
\[ \frac{\ln(5) - 1}{2} = \frac{1}{2}(\log 5 + 1) \]

Multiplying both sides by 2 gives:
\[ \ln(5) - 1 = \log 5 + 1 \]
This simplifies to:
\[ \ln(5) - \log(5) = 2 \]
Since \( \ln(5) \) and \( \log(5) \) are equal, this statement is false.

### Conclusion
Since the limit does not equal \( f(0) \), the function \( f(x) \) is discontinuous at \( x = 0 \).