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Let $f: R \rightarrow R$ be a differentiable function having $f(2)=6, f^{\prime}(2)=\left(\frac{1}{48}\right)$. Then $\lim _{x \rightarrow 2} \int_6^{f(x)} \frac{4 t^3}{x-2} d t$ equals
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Verified Answer
The correct answer is:
18
18
$$
\lim _{x \rightarrow 2} \int_0^{f(x)} \frac{4 t^3}{x-2} d t
$$
Applying $L$ Hospital rule
$$
\begin{aligned}
& \lim _{x \rightarrow 2}\left[4 f(x)^2 f^{\prime}(x)\right]=4 f(2)^3 f^{\prime}(2) \\
& =4 \times 6^3 \times \frac{1}{48}=18 .
\end{aligned}
$$
\lim _{x \rightarrow 2} \int_0^{f(x)} \frac{4 t^3}{x-2} d t
$$
Applying $L$ Hospital rule
$$
\begin{aligned}
& \lim _{x \rightarrow 2}\left[4 f(x)^2 f^{\prime}(x)\right]=4 f(2)^3 f^{\prime}(2) \\
& =4 \times 6^3 \times \frac{1}{48}=18 .
\end{aligned}
$$
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