If f(2) = 4 and f'(2) = 1, then $\lim\limits_{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}=$
Answer & explanation
Correct answer: option 1
We have,
$\lim\limits_{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$
$= \lim\limits_{x \rightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2}$
$= \lim\limits_{x \rightarrow 2} \frac{(x-2) f(2)-2(f(x)-f(2))}{x-2}$
$=\lim\limits_{x \rightarrow 2} \frac{(x-2) f(2)}{x-2}-2 \lim\limits_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}$
= f(2) - 2f'(2) ${\left[∵ f'(2)=\lim\limits_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}\right]}$
= 4 - 2 × 1 = 2 [∵ f(2) = 4 and f'(2) = 1]