MCQ
$\lim _{x \rightarrow 0} \frac{48}{x^4} \int \limits_0^x \frac{t^3}{t^6+1} d t=.............$
- A$6$
- B$3$
- C$9$
- ✓$12$
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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