If I is an open interval, f is differentiable on I and a is in I, then there is a sequence a_n in I\{a} such that a_n-> a and f '(a_n)->f '(a)
If I is an open interval, f is differentiable on I and a is in I, then there is a sequence a_n in I\{a} such that a_n-> a and f '(a_n)->f '(a)
we need to assume that f is a continuous, real-valued function, defined on an interval. If the derivative of f at every interior point of the interval I exists and is zero, then f is constant.
Proof: Assume the derivative of f at every interior point of the interval I exists and is zero. Let (a, b) be an arbitrary open interval in I. By the mean value theorem, there exists a point c in (a,b) such that 0=f '(c)=f(b)-f(a)/(b-a)
So f(a) = f(b). Thus, f is constant on the interior of I and thus is constant on I by continuity
If I is an open interval, f is differentiable on I and a is in I, then there is a sequence a_n in I\{a} such that a_n-> a and f '(a_n)->f '(a)
if f is differentiable on an interval then it is continuous on that interval. the mean value theorem says that there is a c in the the interval.