00:01
Okay, so for part a, we are given that f is little o of g, right, where f and g are functions, and we want to prove that f is big o of g.
00:28
Okay, well, so since f is little o of g, we know that the limit as x goes to infinity, of the ratio of f in g must be equal to zero by definition.
00:48
Since the limit approaches zero, it doesn't really matter if we take the absolute value, because the absolute value of zero is zero.
00:59
So therefore, we get that the limit as x approaches infinity of the absolute value of f over the absolute value of g is equal to the limit as x approaches infinity of the absolute value of the functions.
01:30
All right, and this has to equal zero.
01:36
Well, if the limit here is equal to zero, that implies that the absolute value of f over the absolute value of g must be less than or equal to some constant c for x greater than some k.
02:02
So when x is sufficiently large, okay, well, we can just multiply this inequality both sides by the absolute value of g, right? so it'll cancel it on the left, giving us just the absolute value of f, and then we end up multiplying it here on the right.
02:23
Okay.
02:26
And this is the definition of big o, right? so this implies then, since the absolute value of f is less than or equal to absolute value of g times some constant, then f is indeed big o, of g.
02:46
Okay, so then we need to move on to the second part here...