Does the limit below always equal f'(a)?

Question

Suppose f:R-->R is differentiable at a. Let u and v be real valued functions defined on a neighborhood I of a such that u - v doesn't vanish on I - {a} and such that u(x) --> 0 and v(x) --> 0 as x --> a. Then, is it true that

lim ( x -->a) [f(a + u(x)) - f(a + v(x))]/(u(x) - v(x)) = f'(a)?

Answer 1

Yes.

a) u(x) => 0 as x => a, and v(x) => 0 as x => a
That means that I can make both u(x) and v(x) arbitrarily close to 0, by restricting x to be close enough to a. Therefore, for every positive value of epsilon, I can find a positive value of delta, such that u(x) and v(x) are both less (in magnitude) than epsilon whenever (x-a) is less than delta (in magnitude). Then (u(x)-v(x)) must (in magnitude) be less than 2*epsilon; so this is easily "controlled".

b) Since f is differentiable at x = a,
f(x) - f(a) = f'(a)*(x-a) + o(x-a) , where o(x-a) means that:
[(f(x)-f(a)) - f'(a)*(x-a)] < epsilon' *(x-a) (in magnitude), where epsilon' can be made as small as desired be restricting (x-a) to be small enough (in magnitude); say, less than delta'.

c) So, since you can make the ratio
(f(a+u(x))-f(a+v(x))/(u(x)-v(x)) as close as desired to f'(a) as long as (u(x)-v(x)) is small enough; and you can make (u(x)-v(x)) as small as desired as long as (x-a) is small enough; then it is clear that you cannot escape f'(a).