JEE Mains · Maths · STD 11 - 12. limits
\(\lim _{x \rightarrow 0} \frac{48}{x^4} \int \limits_0^x \frac{t^3}{t^6+1} d t\) is equal to \(.......\).
- A \(6\)
- B \(3\)
- C \(9\)
- D \(12\)
Answer & Solution
Correct Answer
(D) \(12\)
Step-by-step Solution
Detailed explanation
\(48 \lim _{x \rightarrow 0} \frac{\int_0^x \frac{t^3}{t^6+1} d t}{x^4}\left(\frac{0}{0}\right)\) Applying L'Hospitals Rule \(48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}\) \(=12\)
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