$$\lim_\limits{x \rightarrow 0} \frac{48}{x^{4}} \int_\limits{0}^{x} \frac{t^{3}}{t^{6}+1} \mathrm{~d} t$$ is equal to ___________.
Answer (integer)
12
Solution
$$
48 \lim\limits_{x \rightarrow 0} \frac{\int_0^x \frac{t^3}{t^6+1} d t}{x^4}\left(\frac{0}{0}\right)
$$
<br/><br/>Applying L' Hospitals Rule
<br/><br/>$$
\begin{aligned}
48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}\\\\
\end{aligned}
$$
<p>= ${{48} \over 4}$$\mathop {\lim }\limits_{x \to 0} {{1} \over {{x^6} + 1}} = 12$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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