Question

If `f(x)` is derivable at `x = a`, then `lim_"x rarr a"(xf(a) - af(x))/(x - a)` is ??

Answer 1

`f(a)-a f'(a)`

Explanation:

`(xf(a) - af(x))/(x - a) = (a x)((f(a)/a- f(x)/x)/(x-a))` Calling now `g(x) = f(x)/x` and `x = a + h` we have

`(xf(a) - af(x))/(x - a)=(a (a+h))(g(a)-g(a+h))/h = -(a(a+h))(g(a+h)-g(a))/h`

and then

`lim_(x->a)(xf(a) - af(x))/(x - a) = -lim_(h->0)(a(a+h))(g(a+h)-g(a))/h =`

`= -a^2g'(a) = - a^2((f'(a))/a-f(a)/a^2) = f(a)-a f'(a)`