`lim_(x rarr 0)(int_0^x tsin(10t)dt)/x` is equal to

Options

A. 1
B. 10
C. 5
D. 0

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{\int_0^x t \sin(10t) \, dt}{x}, \]

we first note that substituting \(x = 0\) directly gives us the indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the denominator.

### Step 1: Differentiate the numerator and denominator

The numerator is

\[ \int_0^x t \sin(10t) \, dt. \]

By the Fundamental Theorem of Calculus, the derivative of this integral with respect to \(x\) is

\[ x \sin(10x). \]

The denominator is simply \(x\), and its derivative is

\[ 1. \]

### Step 2: Apply L'Hôpital's Rule

Applying L'Hôpital's Rule, we have:

\[ \lim_{x \to 0} \frac{\int_0^x t \sin(10t) \, dt}{x} = \lim_{x \to 0} \frac{x \sin(10x)}{1}. \]

### Step 3: Evaluate the limit

Now, we can substitute \(x = 0\) into the expression:

\[ \lim_{x \to 0} x \sin(10x). \]

Since \(\sin(10x) \to 0\) as \(x \to 0\), we have:

\[ \lim_{x \to 0} x \sin(10x) = 0 \cdot 0 = 0. \]

### Conclusion

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{\int_0^x t \sin(10t) \, dt}{x} = 0. \]

### Final Answer

The final answer is:

\[ \boxed{0}. \]