Question

If $f:R\to R$ is a differentiable function and $f\left(2\right)=6$, then $\underset{x\to 2}{lim}\underset{6}{\overset{f\left(x\right)}{\int }}\frac{2tdt}{(x-2)}$ is :

• A. $0$
• B. $2f^{\prime }\left(2\right)$
• C. $12f^{\prime }\left(2\right)$
• D. $24f^{\prime }\left(2\right)$

Answer 1

The correct option is C $12f^{\prime }\left(2\right)$

$\underset{x\to 2}{lim}\underset{6}{\overset{f\left(x\right)}{\int }}\frac{2tdt}{(x-2)}$
$=\underset{x\to 2}{lim}\frac{1}{(x-2)}[t^2{]}_6^{f\left(x\right)}$
$=\underset{x\to 2}{lim}\frac{\left[f\right(x){]}^2-6^2}{(x-2)}\left(\frac{0}{0}\text{form}\right)$

Applying L'Hospital Rule, we get
$\underset{x\to 2}{lim}\frac{2f\left(x\right)f^{\prime }\left(x\right)}{1}$
$=2f\left(2\right)f^{\prime }\left(2\right)$
$=12f^{\prime }\left(2\right)$