1) 1 is not big O(1/x)
To show that 1
is not O(1/x
), we must show that for any constant c
, there is no x_0
such that 1 <= c*1/x
for all x >= x_0
. Suppose 1 is big O of 1/x. We take c = c_0 > 0
, a constant. Then we must have 1 <= c_0*1/x
for x >= x_0
. Assuming x > 0
, we can solve and get x <= c_0
. This cannot be true for all x >= x_0
(it fails for the first number that is greater than or equal to max(x_0, c_0)
), so our assumption was wrong and the first is not big O of the second.
2) e^x is not O(x^5) Big O
To show that e^x
is not O(x^5
), we must show that for any constant c
, there is no x_0
such that for all x >= x_0
, e^x <= c*x^5
. Suppose e^x
is big O of x^5
. We take c = c_0 > 0
, a constant. Then we must have e^x <= c_0*x^5
for x >= x_0
. We can rearrange this to obtain e^x / x^5 <= c_0
for all x >= x_0
. However, the limit of e^x / x^5
as x -> +inf
tends to +inf
; we can see this by iterating l'Hopital's rule:
e^x e^x e^x
lim --- = lim --- = ... = lim --- = +inf
x->+inf x^5 x->+inf 5x^4 x->+inf 120
This is a contradiction since there is no constant c_0
greater than or equal to infinity. Therefore, our assumption was wrong and the first is not big O of the second.
Note: I write lim = +inf
to mean something like "the value of the expression grows without bound".