00:01
Okay, so here of part a, we have that a sub n is equal to n times f of 1 over n.
00:06
So then we take the limit as n goes to infinity of a sub n.
00:09
That's equal to the limit as n goes to infinity of n times f of 1 over n, which is equal to the limit as delta x approaches zero from the right of f of delta x over delta x, which is then going to be equal to by the definition here we have the limit as delta x approaches zero from the right of f of 0 plus delta x and then minus f of 0, all divided by delta x, which is just equal to f, first derivative here, f1 of 0, f prime of 0, where we have that delta x is equal to 1 over n.
00:57
Okay, now, for part b, we have that a sub n is equal to n times the inverse tangent of 1 over n.
01:12
So we take the limit as n goes to infinity of a sub n.
01:16
That's equal to the limit as n goes to infinity of n times the inverse tangent of 1 over n, which is just equal to f prime of 0, so which is going to be equal to, to 1 over 1 plus 0 squared, which is equal to 1.
01:37
So that then implies that f of x is going to be equal to the inverse tangent of x.
01:48
All right...