If function ff is differentiable at cc, simplify lim_(h->0)((f(c+h^2)-f(c))/h)?

1 Answer
Jun 25, 2018

lim_(h->0) ((f(c+h^2)-f(c))/h) = 0

Explanation:

Multiply and divide by h:

lim_(h->0) ((f(c+h^2)-f(c))/h) = lim_(h->0) h ((f(c+h^2)-f(c))/h^2)

Substitute eta = h^2, clearly when h->0 also eta ->0 and:

h = sqrteta for h >0
h= -sqrteta for h < 0

Then:

lim_(h->0^+) ((f(c+h^2)-f(c))/h) = lim_(eta->0) sqrt eta ((f(c+eta)-f(c))/eta)

As f is differentiable in c:

lim_(eta->0) ((f(c+eta)-f(c))/eta) = f'(c)

so:

lim_(h->0^+) ((f(c+h^2)-f(c))/h) = lim_(eta->0) sqrt eta lim_(eta->0)((f(c+eta)-f(c))/eta) = f'(c) lim_(eta->0) sqrt eta = 0

and similarly:

lim_(h->0^-) ((f(c+h^2)-f(c))/h) = -lim_(eta->0) sqrt eta lim_(eta->0)((f(c+eta)-f(c))/eta) = -f'(c) lim_(eta->0) sqrt eta = 0

Then:

lim_(h->0) ((f(c+h^2)-f(c))/h) = 0