If function #f# is differentiable at #c#, 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#