Download Order of Magnitude Analysis for Infinite Limits and more Study notes Data Structures and Algorithms in PDF only on Docsity!
0.1 Finding Infinite Limits: Order of Magnitude Analysis
In order to work with infinite limits it is helpful to introduce the set of affinely extended real numbers, which contains all of the real numbers as well as ±∞. Using set notation, we write
R∗^ = R ∪ {∞, −∞}.
One way in which we encounter limits where
xlim→x 0
f (x) = ∞
is when x 0 ∈ R, and f contains a division by 0 at the point x 0. In such a situation f (x 0 ) is often undefined, but because the denominator of f (x) approaches 0 as x → x 0 , the output of f (x) grows without bound as x → x 0.
Example 1. Find lim x→− 4
(x + 4)^4
Solution In analyzing this function, the first thing to notice is that we have a fourth power in the denominator, (x + 4)^4. Since (x + 4) is raised to an even power, the denominator is always non-negative (it is 0 when x = −4 and positive otherwise). Since the numerator is always positive as well, it means that this function is always non-negative. As x ↗ −4 and x ↘ 4 the denominator approaches 0, so correspondingly the values of the function grow without bound (so the limit does not exist). We write
lim x→− 4
(x + 4)^4
to signify the function values grow without bound as x → −4 from both sides.
Example 2. Find lim x→ 3
3 − x (x − 3)^4
Solution The first thing for us to do is simplify this fraction as much as possible. If we factor −1 from the numerator, we find that
lim x→ 3
3 − x (x − 3)^4 = lim x→ 3
−(x − 3) (x − 3)^4 = lim x→ 3
(x − 3)^3
Note that above we were able to cancel the factor of (x − 3) because in looking at the limit as x → 3, we know that x 6 = 3, so the factor will never be 0 (we, of course, cannot divide by 0). In this case, unlike the previous one, we have an odd power in the denominator; that is, (x − 3) is raised to the third power. Because of this, we will have
(x − 3)^3 < 0 when x < 3 and (x − 3)^3 > 0 when x > 3.
Since the sign of the denominator is different depending on what side x → 3 from, we should look at the one-sided limits rather than trying to calculate a two-sided limit directly (because they will have different signs, so unless they are both 0, the two-sided limit will not exist). As x ↗ 3 the denominator is negative, so the entire fraction is positive (because there is a −1 in the numerator). As x ↘ 3 the denominator is always positive, so the entire function is negative. In both cases the denominator approaches 0, so we find that
lim x↗ 3
3 − x (x − 3)^4 = lim x↗ 3
(x − 3)^3 = ∞ and lim x↘ 3
3 − x (x − 3)^4 = lim x↘ 3
(x − 3)^3
Since our function has no consistent behavior as x → 3, all we can say is that the limit does not exist.
Example 3. Find lim x→ 3
4 x^2 − 9 x − 9 x − 3
Solution As x → 3, the denominator approaches 0, but so does the numerator. For any polynomial to have a zero at 3 means it has a factor of (x − 3). In fact,
4 x^2 − 9 x − 9 = (x − 3)(4x + 3).
Now we can cancel a factor of (x − 3) from numerator and denominator, to find that
lim x→ 3
4 x^2 − 9 x − 9 x − 2
= lim x→ 3 (4x + 3) = 15.
Suppose that we have two functions, f and g, with
xlim→x 0 |f (x)| = lim x→x 0 |g(x)| = ∞,
where x 0 ∈ R∗. What happens when we look at
xlim→x 0
f (x) g(x)
In this situation we cannot apply the quotient rule for limits, because
xlim→x 0 f (x) and (^) xlim→x 0 g(x)
do not actually exist (recall that writing a limit is ∞ means that the limit doesn’t exist precisely because the function values grow without bound). Nevertheless, just as when we were faced with a limit of a quotient where both the limits of the numerator and denominator were 0, we suspect that it is possible for the limit of this quotient to be finite. For instance, if we let
f (x) = g(x) = x,
then we have two functions both approaching ∞ as x → ∞, yet the limit of their ratio is 1. Before delving into this problem in more generality, we will note that there is a simple trick for evaluating this type of limit when we are dealing with rational functions (a ratio of polynomials). If we divide by the highest power of x in the denominator we can evaluate the limit.
Example 4. Find (^) xlim→∞
2 x^2 − 1 5 x^2 − x
Solution As before, we want to multiply by a convenient choice of 1 in order to rewrite the expression into something we can manipulate. Since the highest power of x in the denominator is 2 (ie, x^2 ) we multiply by 1 /x^2 1 /x^2
in order to remove all terms in the denominator approaching ±∞, reducing the limit into something we can analyze (because considering something of the form ∞/∞ is meaningless). Doing so we find
xlim→∞
2 x^2 − 1 5 x^2 − x = lim x→∞
2 x^2 − 1 5 x^2 − x
1 /x^2 1 /x^2 = lim x→∞
2 − 1 /x^2 5 − 1 /x
Theorem 0.1.1 (Relative Rates of Growth). As x → ∞, the magnitude of exponential, power, and logarithmic functions approach ∞. Moreover, the order of magnitude with which they approach ∞ is given in the following order, with exponential functions approaching the fastest, and logarithmic functions approaching the slowest.
- Exponentials of the form ax, where a > 0 , a 6 = 1. If a > b, then ax^ approaches ∞ faster than bx^ as x → ∞.
- Power functions, xn, n ∈ N, If n > m, then xn^ approaches ∞ faster than xm.
- Logarithmic Functions, loga(x), a > 0 , a 6 = 1. All logarithmic functions approach ∞ at the same rate, because loga(x)/ logb(x) = ln(b)/ ln(a).
If any of the above functions is multiplied by a nonzero constant, it does not change the relative order of magnitude at which the magnitude of the function approaches ∞.
Using this information, we can immediately evaluate limits such as
lim x→∞
ex ln(x)
= ∞ and lim x→∞
ln(x) ex^
There is one additional fact we need in order to really utilize this tool. The key thing is that when we look at sums and differences of these functions, it is only the term that grows that fastest that matters - this is called the dominant term.
Definition 0.1.2 (Dominant Term). Suppose f can be written as a linear combination of functions {f 1 , f 2 ,... , fn}, and
xlim→x 0
|f (x)| = ∞,
where x 0 ∈ R∗. The dominant term of f is the function fi which approaches ∞ on the highest order of magnitude, denoted by fˆ.
Recall that a linear combination of functions is simply a sum of constant multiples of the functions. For instance, a polynomial is simply a linear combination of power functions.
Theorem 0.1.2 (Dominant Terms). Consider f (x) and g(x), with
lim x→x 0 |f (x)| = lim x→x 0 |g(x)| = ∞.
If fˆ and ˆg are the dominant terms of f and g respectively, then
x^ lim→x 0
f (x) g(x) = lim x→x 0
f^ ˆ (x) gˆ(x)
Now for any given limit we simply need to look at the dominant terms. For instance,
xlim→∞
ex^ − 12 x^2 + x x^4 + 1 = lim x→∞
ex x^4
We can also apply this same idea to limits in which both the numerator and denominator approach 0.
Definition 0.1.3 (Relative Rates of Decay). Let f, g : R → R, x 0 ∈ R∗, and
lim x→x 0 f (x) = lim x→x 0 g(x) = 0.
We say f approaches 0 on a higher order of magnitude than g if
lim x→x 0
f (x) g(x)
We say f approaches 0 on a lower order of magnitude than g if
xlim→x 0
∣∣ f^ (x) g(x)
We say that f and g approach 0 on the same order of magnitude if
lim x→x 0
f (x) g(x)
= a, a ∈ R, a 6 = 0.
The most simple case is to look at the reciprocals of exponential and power functions. The relative rates of decay of such functions follow immediately from the relative rates at which they grow. For instance, consider the functions e−x^ and x−^2. As x → ∞, ex^ grows faster than x^2 , so it follows that its reciprocal decays faster, because in viewing e−x^ as 1/ex^ and x−^2 as 1/x^2 , we see that the magnitude of the denominator of the first fraction is growing the fastest. Thus, the magnitude of the overall fraction is decaying the fastest. Another way to evaluate limits involving these reciprocal functions is simply to rewrite them in terms of the growing functions.
Example 7. Evaluate lim x→∞
e−x^ + 1 x−^2
Solution In order evaluate this limit we simply rewrite the constituent functions in terms of their reciprocals. e−x^ + 1 x−^2
e−x x−^2
x−^2
x^2 ex^
Now we simply evaluate the limit, using our knowledge of the relative rates of growth of the given functions, and see that
lim x→∞
x^2 ex^
In some more complicated cases we can also evaluate infinite limits by using substitution.
Example 8. Find lim x→∞ sin(
x
Solution Here we will use the fact that as x → ∞ we have 1/x → 0, and introduce the variable t = 1/x. Thus,
xlim→∞ sin(
x ) = lim t→ 0 +^
sin(t) = 0
The reason we need to write this as a one-sided limit is because x → ∞ from only one side.
