Question

\( \lim_{x \to 0} \frac{48 x^4}{x} \int_0^x \frac{t^3}{t^6 + 1} \, dt \) is equal to:

Answer 1

Correct Answer: 12

We are tasked to evaluate the limit:

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

Step 1: Check for Indeterminate Form

Substituting \(x = 0\):

\[ \int_0^x \frac{t^3}{t^6 + 1} dt = 0 \quad \text{and} \quad x^4 = 0. \]

This results in the indeterminate form \(\frac{0}{0}\).

Step 2: Apply L’Hôpital’s Rule

Using L’Hôpital’s Rule, differentiate the numerator and denominator with respect to \(x\):

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

Step 3: Simplify the Expression

Simplify the limit:

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

Step 4: Evaluate the Limit

As \(x \to 0\), \(x^6 \to 0\), so:

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

Step 5: Multiply by 48

Finally, multiply the result by 48:

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

Conclusion

The value of the limit is:

\[ \boxed{12}. \]