Let f(x) be a differentiable function at x=a with f'(a)=2 and f(a)=4.

Question

Let $f(x)$ be a differentiable function at $x=a$ with $f'(a)=2$ and $f(a)=4$. Then $\displaystyle\lim_{x	o a}\frac{xf(a)-af(x)}{x-a}$ equals

Accepted Answer

$$L = \mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$$ [$${0 \over 0}$$ form]

Using L' Hospital rule we get

$$L = \mathop {\lim }\limits_{x \to a} {{f(a) - af'(x)} \over 1}$$

$f(a) - af'(a) = 4 - 2a$

Solutions · 1