If a, b and c are real numbers, then the value of `lim_(trarr0) ((1)/(t)int_(0)^(t)(1+asinbx)^(c//x)dx)` equals

Options

A. abc
B. `(ab)/(c)`
C. `(bc)/(a)`
D. `(ca)/(b)`

Explanation

To solve the limit problem given in the question, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Limit Expression**:
We need to evaluate:
\[ \lim_{t \to 0} \frac{1}{t} \int_{0}^{t} (1 + a \sin(bx))^{\frac{c}{x}} \, dx \]

2. **Rewrite the Expression**:
The limit can be rewritten using the logarithm:
\[ \lim_{t \to 0} \ln\left(\frac{1}{t} \int_{0}^{t} (1 + a \sin(bx))^{\frac{c}{x}} \, dx\right) \]

3. **Apply L'Hôpital's Rule**:
Since substituting \( t = 0 \) directly leads to the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:
\[ \lim_{t \to 0} \frac{\frac{d}{dt} \left( \int_{0}^{t} (1 + a \sin(bx))^{\frac{c}{x}} \, dx \right)}{\frac{d}{dt}(t)} \]

4. **Differentiate the Integral**:
By the Fundamental Theorem of Calculus:
\[ \frac{d}{dt} \left( \int_{0}^{t} (1 + a \sin(bx))^{\frac{c}{x}} \, dx \right) = (1 + a \sin(bt))^{\frac{c}{t}} \]

5. **Evaluate the Limit**:
Now we have:
\[ \lim_{t \to 0} (1 + a \sin(bt))^{\frac{c}{t}} \]
This is of the form \( 1^{\infty} \). We can rewrite it using the exponential function:
\[ \lim_{t \to 0} e^{\frac{c}{t} \ln(1 + a \sin(bt))} \]

6. **Simplify the Logarithm**:
As \( t \to 0 \), \( \sin(bt) \to bt \), hence:
\[ \ln(1 + a \sin(bt)) \approx a \sin(bt) \text{ for small } t \]
Therefore:
\[ \frac{c}{t} \ln(1 + a \sin(bt)) \approx \frac{c}{t} (a \sin(bt)) \approx \frac{c}{t} (a bt) = abc \]

7. **Final Limit Evaluation**:
Thus, we can conclude:
\[ \lim_{t \to 0} e^{abc} = e^{abc} \]

8. **Final Answer**:
The value of the limit is:
\[ abc \]