- #1
redyelloworange
- 20
- 0
Homework Statement
Two functions f,g:R->R are equal up to nth order at if
lim h-->0 [f(a+h) - g(a+h) / hn ]= 0
Show that f is differentiable at a iff there is a function g(x)=b+m(x-a) such that f and g are equal up to first order at a.
Homework Equations
f is differentiable at a if the following limit exists:
lim h-->0 [f(a+h) - f(a) / h]
The Attempt at a Solution
Direction 1:
=>
Assume f is differentiable. Show there exists a function g(x)=b+m(x-a) such that f and g are equal up to first order at a.
So, lim h-->0 [f(a+h) - f(a) / h] exists and equals some L.
=>lim h-->0 [f(a+h) - f(a) - g(a+h) + g(a+h)/ h]
=>lim h-->0 [(f(a+h)- g(a+h)) - f(a) + g(a+h)/ h]
this implies that we want to show
lim h-->0 [g(a+h) - f(a)/h = L]
I'm not sure where to go from here.
Direction 2:
<=
Assume that lim h-->0 [f(a+h) -g(a+h) / h = 0]
show lim h-->0 [f(a+h) - f(a) / h] is equal to some constant.
Take lim h-->0 [f(a+h) -g(a+h) / h = 0]:
=> lim h-->0 [f(a+h) -g(a+h) -f(a) + f(a) - g(a) + g(a) / h = 0]
=> lim h-->0 [(f(a+h)-f(a)) -(g(a+h) - g(a)) + (f(a) - g(a))/ h = 0]
but lim h-->0 [f(a+h)-f(a)/h] is just f'
and lim h-->0 [g(a+h) - g(a)/h] is -m
so, we just need to show that
f(a) = g(a) and then lim h-->0 [f(a) - g(a)/h] = 0 and f' = m
Not sure how to do that.
That's all.
Thanks for your help.