`lim_(x rarr 0)(48)/(x^(4))int_(0)^(x)(t^(3))/(t^(6)+1)dt` is equal to

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6 + 1} dt, \]

we can use L'Hôpital's rule since both the numerator and the denominator approach 0 as \(x\) approaches 0.

### Step 1: Identify the form of the limit
As \(x \to 0\), the integral \(\int_0^x \frac{t^3}{t^6 + 1} dt\) approaches 0, and \(x^4\) also approaches 0. Thus, we have a \( \frac{0}{0} \) form.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's rule, we differentiate the numerator and the denominator with respect to \(x\):

- **Numerator**: \( \int_0^x \frac{t^3}{t^6 + 1} dt \) differentiated with respect to \(x\) gives us \( \frac{x^3}{x^6 + 1} \) (by the Fundamental Theorem of Calculus).
- **Denominator**: \( x^4 \) differentiated with respect to \(x\) gives us \( 4x^3 \).

Now we rewrite the limit:

\[ \lim_{x \to 0} \frac{\frac{x^3}{x^6 + 1}}{4x^3}. \]

### Step 3: Simplify the expression
We can simplify the expression:

\[ \lim_{x \to 0} \frac{x^3}{4x^3(x^6 + 1)} = \lim_{x \to 0} \frac{1}{4(x^6 + 1)}. \]

### Step 4: Evaluate the limit
As \(x\) approaches 0, \(x^6\) approaches 0, hence:

\[ \lim_{x \to 0} \frac{1}{4(x^6 + 1)} = \frac{1}{4(0 + 1)} = \frac{1}{4}. \]

### Step 5: Multiply by the constant in front
Now, we multiply this result by the constant 48 from the original limit:

\[ 48 \cdot \frac{1}{4} = 12. \]

Thus, the final result is:

\[ \lim_{x \to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6 + 1} dt = 12. \]

### Final Answer
The limit is equal to \(12\).

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