My professor gave the following: f(x) = f(x) - f(a) + f(a) f(x) =[f(x) - f(a) /
ID: 2847939 • Letter: M
Question
My professor gave the following:
f(x) = f(x) - f(a) + f(a)
f(x) =[f(x) - f(a) / (x-a) ] + [f(a) / (x-a)]
f(x) = lim(x->a) f(x) - f(a) / (x-a) + lim(x->a) f(a)/ (x-a) -----------------(1)
f'(x) = lim(x->a) f(x) - f(a) / (x-a) ----------------------------------------(2)
Then, asked the class to prove that equation (1) = (2). I have no idea how to proof this, and I have never seen this in calculus I. This is worth 10 additional points for an exams. If you know, please help. I would really be grateful. I'm not even sure if the question make sense.
Explanation / Answer
f is differentiable at point " a " then ( f(x) - f(a) ) / (x-a)
has a limit A when x converges towards " a " called f ' (a)
lim (x --> a) ( f(x) - f(a) ) / (x-a) = A = f ' (a)
Suppose that f is differentiable at the point x = a. Then we know that
lim
h?0 f(a+h) - f(a)/ h
exists, and equals f'(a).
Thus,
limit
when h?0 of f(a+h) - f(a) = lim when h?0 of f(a+h) - f(a)/h . h = f'(a). 0 = 0.
because
Limit of product = product of limits
This gives
lim h?0 f(a+h) = lim h?0 [f(a+h) - f(a)] + f(a) = 0 + f(a) = f(a).
Limit of sum = sum of limits
If we take x to be a+h, then h = x-a, and the above result can be written as
lim when x-a?0 of f(x) = f(a).
In other words,
lim when x?a of f(x) = f(a),
which means that f is continuous at x = a
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.