JEE Advanced · Mathematics · 16. Limits
The value of \(\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \log (1+t)}{t^4+4} d t\) is
- A
0
- B
\(\frac{1}{12}\)
- C
\(\frac{1}{24}\)
- D
\(\frac{1}{64}\)
Answer & Solution
Correct Answer
(B)
\(\frac{1}{12}\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^{x t} \frac{\log (1+t)}{4+t^4} d t\)
Using L' Hospital's rule,
\[
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\frac{x \log (1+x)}{4+x^4}}{3 x^2} \\
= & \lim _{x \rightarrow 0} \frac{\log (1+x)}{3 x} \cdot \frac{1}{4+x^4} \\
= & \frac{1}{3} \cdot \frac{1}{4}=\frac{1}{12}\left[\text { using, } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]
\end{aligned}
\]
Using L' Hospital's rule,
\[
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\frac{x \log (1+x)}{4+x^4}}{3 x^2} \\
= & \lim _{x \rightarrow 0} \frac{\log (1+x)}{3 x} \cdot \frac{1}{4+x^4} \\
= & \frac{1}{3} \cdot \frac{1}{4}=\frac{1}{12}\left[\text { using, } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]
\end{aligned}
\]
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