If f : R → R is a differentiable function and f 2 = 6 , then l i m x → 2 ∫ 6 f x 2 t d t x…

The given limit can be written as

$l=\underset{x\to 2}{\lim }\frac{{\int }_6^{f\left(x\right)}2tdt}{\left(x-2\right)}$

Applying L' Hospital’s rule i.e. if $l=\underset{x\to a}{\lim }\frac{g\left(x\right)}{h\left(x\right)}$ and $\underset{x\to a}{\lim }g\left(x\right)\to 0 \& \underset{x\to a}{\lim }h\left(x\right)\to 0,$ then $l=\underset{x\to a}{\lim }\frac{g^{\text{'}}\left(x\right)}{h^{\text{'}}\left(x\right)}$

$l=\underset{x\to 2}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{{\displaystyle \frac{d}{dx}}{\int }_6^{f\left(x\right)}2tdt}{1}$

Now, applying Newton's Leibnitz rule i.e. $\frac{d}{dx}{\int }_a^{b}g\left(x\right)dx=g\left(b\right)·b^{\text{'}}-g\left(a\right)·a^{\text{'}},$ we get

$\Rightarrow l=\underset{x\to 2}{\lim }\left(2f\left(x\right)·f^{\text{'}}\left(x\right)-0\right)$

$\Rightarrow l=2f\left(2\right)\cdot f^{\text{'}}\left(2\right)$

Put the given value, to get

$\Rightarrow l=2\times 6\times f^{\text{'}}\left(2\right)=12f^{\text{'}}\left(2\right).$