JEE MAIN - Mathematics (2023 - 30th January Morning Shift - No. 11)
$$\lim_\limits{x \rightarrow 0} \frac{48}{x^{4}} \int_\limits{0}^{x} \frac{t^{3}}{t^{6}+1} \mathrm{~d} t$$ is equal to ___________.
Answer
12
Explanation
$$
48 \lim\limits_{x \rightarrow 0} \frac{\int_0^x \frac{t^3}{t^6+1} d t}{x^4}\left(\frac{0}{0}\right)
$$
Applying L' Hospitals Rule
$$ \begin{aligned} 48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}\\\\ \end{aligned} $$
Applying L' Hospitals Rule
$$ \begin{aligned} 48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}\\\\ \end{aligned} $$
= $${{48} \over 4}$$$$\mathop {\lim }\limits_{x \to 0} {{1} \over {{x^6} + 1}} = 12$$
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