Algebra of Little o Terms
The properties of little o terms give rise to what is known as the algebra of little o, governed by the following rules: $$ o(x^n) + o(x^n) = o(x^n) $$ $$ o(x^n) - o(x^n) = o(x^n) $$ $$ c \cdot o(x^n) = o(c \cdot x^n) = o(x^n), \:\: \text{if} \: c \ne 0 $$ $$ x^m \cdot o(x^n) = o(x^{m+n}) $$ $$ o(x^m) \cdot o(x^n) = o(x^{m+n}) $$ $$ o(o(x^n)) = o(x^n) $$ $$ o(x^n + o(x^n)) = o(x^n) $$
Sum of Little o Terms
If two functions are infinitesimal of order higher than xn as x tends to x0, then their sum satisfies: $$ o(x^n) + o(x^n) = o(x^n) $$
Proof
Let f(x) and g(x) be two infinitesimal functions:
$$ f(x) = o(x^n) $$ $$ g(x) = o(x^n) $$
Both are infinitesimal of order higher than xn as x tends to zero:
$$ \lim_{x \rightarrow 0} \frac{f(x)}{x^n} = 0 $$
$$ \lim_{x \rightarrow 0} \frac{g(x)}{x^n} = 0 $$
By the linearity of limits, the limit of their sum is also zero:
$$ \lim_{x \rightarrow 0} \left( \frac{f(x)}{x^n} + \frac{g(x)}{x^n} \right) = 0 $$
$$ \lim_{x \rightarrow 0} \frac{f(x) + g(x)}{x^n} = 0 $$
Thus, the sum f(x) + g(x) is also infinitesimal of order higher than xn as x → 0:
$$ o(f(x) + g(x)) $$
This establishes the rule for the sum of little o terms:
$$ o(x^n) + o(x^n) = o(x^n) $$
Difference of Little o Terms
If two functions are infinitesimal of order higher than xn as x tends to x0, then their difference also satisfies: $$ o(x^n) - o(x^n) = o(x^n) $$
Proof
The argument is identical to that for the sum - one simply replaces addition with subtraction.
Product of a Little o Term by a Scalar
If f(x) is infinitesimal of order higher than xn as x tends to x0, then multiplying it by a nonzero scalar c yields: $$ c \cdot o(x^n) = o(c \cdot x^n) = o(x^n), \:\: \text{if} \: c \ne 0 $$
Proof
This result follows directly from elementary properties of limits.
Since f(x) is infinitesimal of order higher than xn as x → x0:
$$ f(x) = o(x^n) $$
which means:
$$ \lim_{x \rightarrow 0} \frac{f(x)}{x^n} = 0 $$
Multiplying both sides by any constant c ≠ 0 gives:
$$ c \cdot \lim_{x \rightarrow 0} \frac{f(x)}{x^n} = 0 $$
which proves the identity:
$$ c \cdot o(x^n) = o(c \cdot x^n) = o(x^n), \:\: \text{if} \: c \ne 0 $$
Product of a Little o Term by a Function
If f(x) is infinitesimal of order higher than xn as x tends to x0, then multiplying it by xm yields: $$ x^m \cdot o(x^n) = o(x^{m+n}) $$
Proof
This can be shown in the same way as the product with a scalar:
$$ x^m \cdot o(x^n) $$
$$ = o(x^m \cdot x^n) $$
$$ = o(x^{m+n}) $$
Product of Little o Terms
If two functions are infinitesimal of order higher than xn as x approaches x0, their product satisfies: $$ o(x^m) \cdot o(x^n) = o(x^{m+n}) $$
Proof
The argument is analogous to the case of a scalar multiple:
$$ o(x^m) \cdot o(x^n) $$
$$ = o(x^m \cdot x^n) $$
$$ = o(x^{m+n}) $$
Little o of Little o
If f(x) is an infinitesimal of order higher than a little o term o(xn) as x approaches x0, then: $$ o(o(x^n)) = o(x^n) $$
Proof
Here, the infinitesimal function as x → x0 is: $$ f(x) = o(o(x^n)) $$
We aim to show that:
$$ \lim_{x \rightarrow 0} \frac{f(x)}{x^n} = 0 $$
That is: $$ \lim_{x \rightarrow 0} \frac{o(o(x^n))}{x^n} = 0 $$
Multiplying numerator and denominator by o(xn) does not affect the limit:
$$ \lim_{x \rightarrow 0} \frac{o(o(x^n))}{x^n} \cdot \frac{o(x^n)}{o(x^n)} = 0 $$
Rewriting this expression:
$$ \lim_{x \rightarrow 0} \frac{o(o(x^n))}{o(x^n)} \cdot \frac{o(x^n)}{x^n} = 0 $$
Since o(xn)/xn tends to zero, it follows that: $$ o(o(x^n)) = o(x^n) $$
Note: The limit is finite and equals zero by assumption. Therefore, the ratio o(o(xn)) / o(xn) cannot diverge to infinity - otherwise, the expression would become an indeterminate form of the type "infinity times zero". The term o(o(xn)) must therefore be an infinitesimal of xn, of order equal to or higher than xn.
Hence, as x → 0:
$$ f(x) = o(o(x^n)) = o(x^n) $$
This completes the proof of the "little o of little o" rule:
$$ o(o(x^n)) = o(x^n) $$
Little o of xn + o(xn)
If f(x) is an infinitesimal of order higher than xn + o(xn) as x approaches x0, then: $$ o(x^n + o(x^n)) = o(x^n) $$
Proof
Here, the infinitesimal function as x → x0 is: $$ f(x) = o(x^n + o(x^n)) $$
We aim to show that:
$$ \lim_{x \rightarrow 0} \frac{f(x)}{x^n} = 0 $$
That is: $$ \lim_{x \rightarrow 0} \frac{o(x^n + o(x^n))}{x^n} = 0 $$
Multiplying numerator and denominator by o(xn):
$$ \lim_{x \rightarrow 0} \frac{o(x^n + o(x^n))}{x^n} \cdot \frac{o(x^n)}{o(x^n)} = 0 $$
Rearranging terms:
$$ \lim_{x \rightarrow 0} \frac{o(x^n + o(x^n))}{o(x^n)} \cdot \frac{o(x^n)}{x^n} = 0 $$
Since o(xn)/xn tends to zero, it follows that: $$ o(x^n + o(x^n)) = o(x^n) $$
Note: The limit is finite and equal to zero by assumption. Thus, the ratio o(xn + o(xn)) / o(xn) cannot diverge to infinity - otherwise, the expression would yield an indeterminate form of the type "infinity times zero". The term o(xn + o(xn)) must therefore be an infinitesimal of xn, of order equal to or higher than xn.
Thus, as x → 0:
$$ f(x) = o(x^n + o(x^n)) = o(x^n) $$
This concludes the proof:
$$ o(x^n + o(x^n)) = o(x^n) $$
And so on.
