`underset(xto0)("lim")(int_(0)^(x)t tan (5t)dt)/x^3` is equal to :

Options

A. `-4/3`
B. `-5/3`
C. `4/3`
D. `5/3`

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{\int_{0}^{x} t \tan(5t) \, dt}{x^3}, \]

we will follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \), both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \). This allows us to apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
We differentiate the numerator and the denominator separately.

#### Numerator:
Let

\[ f(x) = \int_{0}^{x} t \tan(5t) \, dt. \]

Using the Fundamental Theorem of Calculus (Leibniz rule), we differentiate \( f(x) \):

\[ f'(x) = x \tan(5x). \]

#### Denominator:
The denominator is

\[ g(x) = x^3, \]

and its derivative is

\[ g'(x) = 3x^2. \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{x \tan(5x)}{3x^2}. \]

### Step 4: Simplify the limit
We can simplify this expression:

\[ \lim_{x \to 0} \frac{x \tan(5x)}{3x^2} = \lim_{x \to 0} \frac{\tan(5x)}{3x}. \]

### Step 5: Apply L'Hôpital's Rule again
This limit is still in the form \( \frac{0}{0} \). We apply L'Hôpital's Rule again:

#### Differentiate the numerator:
The derivative of \( \tan(5x) \) is

\[ 5 \sec^2(5x). \]

#### Differentiate the denominator:
The derivative of \( 3x \) is

\[ 3. \]

### Step 6: Rewrite the limit again
Now we have:

\[ \lim_{x \to 0} \frac{5 \sec^2(5x)}{3}. \]

### Step 7: Evaluate the limit
As \( x \to 0 \), \( \sec^2(5x) \) approaches \( \sec^2(0) = 1 \). Thus,

\[ \lim_{x \to 0} \frac{5 \sec^2(5x)}{3} = \frac{5 \cdot 1}{3} = \frac{5}{3}. \]

### Final Answer
Therefore, the limit is

\[ \frac{5}{3}. \]