1.If f(2)=4f(2)=4f(2)=4 and f′(2)=1f'(2)=1f′(2)=1, then limx→2xf(2)−2f(x)x−2\displaystyle\lim_{x\to2}\frac{xf(2)-2f(x)}{x-2}x→2limx−2xf(2)−2f(x) is equal toa.−2-2−2b.111c.222d.333Login to continueOnly logged in users canattempt or see the solution.