Question

If f(2) = 5 and f '(2) = 2, then the value of $\underset{x\to 2}{\lim }\frac{\mathrm{xf}\left(2\right)-2f\left(x\right)}{x-2}$ is

Answer 1

According to the definition of Differentiability, we have
$f\text{'}\left(2\right)=\underset{x\to 2}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}\Rightarrow 1=\underset{x\to 2}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}\mathrm{Now},\underset{x\to 2}{\lim }\frac{\mathrm{xf}\left(2\right)-2f\left(x\right)}{x-2}=\underset{x\to 2}{\lim }\frac{\mathrm{xf}\left(2\right)-2f\left(2\right)+2f\left(2\right)-2f\left(x\right)}{x-2}=\underset{x\to 2}{\lim }\frac{\left(x-2\right)f\left(2\right)}{x-2}-2\underset{x\to 2}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}=f\left(2\right)-2f\text{'}\left(2\right)=5-2\times 2=1$