If `f(x)` is continuous at `x=0`, where `f(x)=(e^(5x)-e^(2x))/(sin 3x)`, for `x!=0` then `f(0)=`

Options

A. 1
B. -1
C. 3
D. None of these

Answer

To find \( f(0) \) such that the function \( f(x) = \frac{e^{5x} - e^{2x}}{\sin(3x)} \) is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-step Solution:

1. **Identify the limit**:
Since \( f(x) \) is defined for \( x \neq 0 \), we need to find:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{5x} - e^{2x}}{\sin(3x)} \]

2. **Apply L'Hôpital's Rule**:
As \( x \to 0 \), both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{e^{5x} - e^{2x}}{\sin(3x)} = \lim_{x \to 0} \frac{(5e^{5x} - 2e^{2x})}{3\cos(3x)} \]

3. **Evaluate the limit**:
Now substituting \( x = 0 \) into the new limit:
\[ = \frac{5e^{0} - 2e^{0}}{3\cos(0)} = \frac{5 - 2}{3 \cdot 1} = \frac{3}{3} = 1 \]

4. **Conclusion**:
Since we have found the limit, we can conclude that for \( f(x) \) to be continuous at \( x = 0 \), we define:
\[ f(0) = 1 \]

### Final Answer:
Thus, \( f(0) = 1 \).

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