differentiation

1.
lim(h->0) [f(x+h)-f(x)]/h = f'(x) = g(x)
lim(x->0){lim(h->0) [f(x+h)-f(x)]/h } = lim(x->0) g(x)
lim(h->0){lim(x->0) [f(x+h)-f(x)]/h}=g(0) (since [f(x+h)-f(x)]/h is conti. at x=0)
lim(h->0) [f(0+h)-f(0)]/h = g(0)
f'(0)=g(0)

2.
It suffices to show that lim(t->0+) (d^n/dt^n) f(t) =0. (by the result of 1.)

lim(t->0-) [f(t)-f(0)]/t = lim(t->0-) (0-0)/t =0
lim(t->0+) [f(t)-f(0)]/t = lim(t->0+) [exp(-1/t)-0]/t
= lim(x->inf) exp(-x)* x (x= 1/t)
= lim(x->inf) x/exp(x) =0
so, f'(t)=
[t <=0 , 0
[t > 0 , exp(-1/t) *1/t^2

(d/dt)^(n+1) f(t) = (d/dt)^n f'(t) ( t>0)
=(d/dt)^n exp(-1/t)*(1/t^2)
(by Leibnitz's formula)
= exp(-1/t)*g(1/t) (where g(t)= poly. of t with degree 2n)
= exp(-x)* g(x) = g(x)/exp(x) ,where x= 1/t
then lim(t->0+) (d/dt)^(n+1) f(t)
=lim(x->inf) g(x)/exp(x)
(by L'Hospital's rule) =lim(x->inf) g'(x)/exp(x) = ...=0

Q.E.D.