`lim_(x rarr 0)(int_0^x tsin(10t)dt)/x` is equal to (A) 1 (B) 10 (C) 5 (D) 0

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 start by observing that both the numerator and denominator approach 0 as \( x \) approaches 0. This gives us the indeterminate form \( \frac{0}{0} \), which allows us to apply L'Hôpital's Rule.

### Step 1: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

**Numerator:**
We need to differentiate the integral \( \int_0^x t \sin(10t) \, dt \). By the Fundamental Theorem of Calculus, the derivative of this integral with respect to \( x \) is:

\[ \frac{d}{dx} \left( \int_0^x t \sin(10t) \, dt \right) = x \sin(10x). \]

**Denominator:**
The derivative of \( x \) with respect to \( x \) is simply:

\[ \frac{d}{dx}(x) = 1. \]

### Step 2: Rewrite the Limit
Now we can rewrite the limit using the derivatives we found:

\[ \lim_{x \to 0} \frac{x \sin(10x)}{1}. \]

### Step 3: Evaluate the Limit
Next, we evaluate the limit as \( x \) approaches 0:

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

As \( x \) approaches 0, \( \sin(10x) \) approaches \( 10x \) (using the small angle approximation \( \sin(t) \approx t \) when \( t \) is small). Thus, we have:

\[ \lim_{x \to 0} x \sin(10x) = \lim_{x \to 0} x \cdot 10x = \lim_{x \to 0} 10x^2 = 0. \]

### Conclusion
Therefore, the limit evaluates to:

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

The answer is (D) 0.