Intuitive evaluation

Here we will try to guess the limit of a function at infinity, of course we will also show how to confirm such a guess. We will look at functions that involve all kinds of powers (polynomials, exponentials) and logarithms, we will show more precise specifications below.

Note that some of these considerations would also work for limit at negative infinity, but it may be tricky and treacherous. There's a little trick which does it safely for us: Given a limit at negative infinity, use substitution to change it into a limit at infinity and all the tricks from this section apply.

The guessing we will cover here is very useful and usually offers the shortest solution to expressions that would otherwise have to be solved in different ways, often using l'Hospital rule leading to complicated expressions. It uses two basic ingredients. The first is a good knowledge of the limit algebra, especially the basic limits, the second important ingredient is the understanding of interplay of powers and exponentials (and roots). We will start with little introduction.

In many functions one finds sums of powers and similar expressions. For instance, consider the function x³ + 5x^1/2 − x⁻², we are interested in its limit at infinity. The first two powers go to infinity (positive exponent), the third power is properly written as 1/x², which converges to 0. Using the limit algebra we easily find the limit of the whole function: ∞ + 5⋅∞ − 0 = ∞.

Now here's a problem: What is the limit at infinity of f (x) = x³ − 5x²? We know that ∞ − ∞ is an indeterminate expression. There are procedures for finding the outcome, but the appropriate methods sometimes lead to complicated calculations. In this section we will learn how to guess the correct answer and how to prove it easily (sort of).

By the way, when we are adding and subtracting some objects and even multiplying them by real numbers like we did in the above example, we call it "linear combination". So the expression in this example was a "linear combination of powers". We will use this terminology throughout this section. First we will identify what kind of problems we will address.

Note that negative powers do not cause troubles in expressions like the one above, we saw that x⁻² tends to zero at infinity and therefore does not cause troubles in the limit algebra when added to something else. Therefore we will only consider positive exponents in this section. If the exponents also happen to be all integers, we are actually talking polynomials here. But we will work in more generality and also allow for positive exponents that are not integers (for instance the exponent 1/2, that is, the square root).

We will also allow for "exponentials", that is, powers of the form a^x. Again, since we know that a^x→0 at infinity if a < 1 and 0 is not a problem in sums, we will only consider exponentials a^x with a > 1.

But that is not all. While "linear combinations" of powers and exponentials as above (powers/exponentials multiplied by numbers and then added/subtracted) appear quite often, we will include still more expressions: logarithms and even powers of the form x^x. Now we are ready to state the problem we are trying to address here:

Question:
We have an expression of the form α⋅A(x) + β⋅B(x) +..., where the functions A(x), B(x),... can be powers x^a with a > 0, exponentials a^x with a > 1, powers of logarithms [ln(x)]^a with a > 0, and general powers like x^x. What happens with this expression if we let x go to infinity?

Note that all expressions listed above tend to infinity as x→∞. When they are combined together in some linear combination, we typically get some indeterminate expression involving infinities, since we only get a result right away if all these infinities are added. What if they are subtracted? We will soon see that some infinities are "bigger" or more important then others.

Solution to the problem:
If we let x→∞, then an expression as above will behave exactly like its dominant term. To prove it mathematically, we factor the dominant term out of the expression and then apply the limit. The expression that remains after factoring should tend to a non-zero real number.

By a "dominant term" we mean the term that, for large x, prevails over all other terms in the given linear combination, and so the other terms can be ignored, they do not interfere with the tendency that the dominant power has around infinity. There is a hierarchy among all the types listed above (powers, exponentials, logarithms,...). When we say that "term A prevails over term B for large values of x", we really mean that when investigating limit of the expression α⋅A + β⋅B for x tending to infinity, we can disregard the term B and just worry about A. Actually, here we are intentionally a bit vague and imprecise, since we want to show a practical and simple approach; this topic can be done properly mathematically and is related to some other notions, if you are interested, check out the next section on order of functions.

Example:
We will shortly see that x² prevails over ln(x) at infinity. Now imagine that we need to find the limit of 23ln(x) − x² at infinity. When x grows really large, the second term prevails, we can ignore the first one. Thus this expression will behave (when x grows to infinity) like the expression -x², which we know tends to negative infinity. Consequently, also the given expression goes to negative infinity at infinity.

Of course, this result is just guessing. We wrote above (in Solution) that mathematically we can do this by factoring out the dominant term. We try it:

Note that the term that remained after factoring out the dominant power indeed tended to a non-zero real number, exactly as stated above. The fact that it is non-zero becomes important for limit algebra, we can't get ∞⋅0.

There is a way to write our intuitive reasoning properly. When we write A(x) ∼ B(x), we mean that the expressions A and B behave the same when x→∞; for most practical purposes, the expressions A(x) and B(x) are the same when x is really really large. We also say that A and B are "of the same order". So the above intuitive reasoning can be written as follows:

23ln(x) − x² ∼ −x²→−∞.

Note that this handy way of writing is not universally accepted. Also, it does not constitute a proper solution - after all, we were just guessing there. Any answer that we get in this way should be confirmed by a correct mathematical calculation, for instance the factoring out procedure that we have shown above.

When we look at powers in a linear combination of powers and other similar terms, we actually do two different things. First, we look just at the powers, exponentials etc., that is, we ignore the multiplicative constants before them. In this way we determine the type of an expression. For instance, the expression "13⋅[ln(x)]³" is of the type [ln(x)]³, and the example above is of the type x², because the type of an expression is given by the type of its dominant term. The type tells us roughly how fast the given expression runs away when x is really large.

This type is used when roughly comparing behavior of different expressions, finding out which can be ignored etc. Then when we really start guessing the limit at infinity, we do have to include the multiplicative constants in our reasoning. Thus we would say that 23ln(x) − x² is of the type x² when x goes to infinity, but then we would have to say that it is of order -x² when we try to guess the limit; that is, we do have to carry the constants when using the ∼ procedure.

As you can see, the intuitive procedure can be simple, writing it properly mathematically (by factoring) may be a bit longer but should never be tricky, as long as we correctly identify and factor out the dominant term. Which brings us to the main part of this section:

Here is the list of the terms mentioned above from the most dominant to the least. That is, every listed expression prevails over all expressions listed later:

(1)   the power x^x,
(2)   the exponential a^x for a > 1,
(3)   the power x^a for a > 0,
(4)   the logarithm [ln(x)]^a for a > 0.

For practical use people often prefer a more colloquial way of remembering this hierarchy, using phrases like "powers beat logarithms at infinity" and "exponentials beat powers at infinity" etc. Proofs of this hierarchy are shown for sequences in this note, for functions it can be proved in essentially the same way.

When investigating a linear combination of such terms, we always first find the dominant expression. However, there may be more terms of this dominant category. Thus we also need to know mutual dominance within each category. There the rule is simple. In categories (2), (3) and (4) there is always a parameter (i.e. a) and the highest one is the dominant one.

In fact, the question of dominance is one kind of an answer to the question "which infinity is larger". In the following picture (not quite to scale) we will try to express symbolically the relationship between different kinds of expressions, obtaining the scale of powers:

Example:
What is the limit of 13⋅2^x + x² − 5⋅3^x − [ln(x)]^1/2 at infinity?

Solution: There are three categories there: exponentials, powers and logarithms. Exponentials are the highest on the list, so they will supply the dominant term. There are two candidates, 2^x and 3^x. Since 3 > 2, the latter exponential is the dominant term, that is, the given expression is of the type 3^x. Thus we can ignore all the other terms and guess that

13⋅2^x + x² − 5⋅3^x − [ln(x)]^1/2 ∼  − 5⋅3^x→ − 5⋅∞ = −∞.

How do we confirm this result mathematically? By factoring the dominant term out.

The three fractions in the parentheses are best handled separately. The first one is just an exponential, the other two will lead to l'Hospital's rule:

Finally we get

Note that sometimes we can apply this reasoning even for terms that do not look exactly like those above, but can be changed into them by algebra. Two most typical examples: (2x)³ = 2³⋅x³ = 8⋅x³ and 3^2x+1 = (3²)^x⋅3¹ = 3⋅9^x.

The intuitive procedure also works if there are some roots mixed in the expression; that is, if some parts of the expression are closed under roots. We then follow the following procedure. First we handle each root individually. For every root, we find the dominant term of the expression inside this root, this determines the type of the expression under the root. When we apply the root to this expression, we obtain the type of the root as a whole. We can confirm this by factoring it out, the resulting root then should have a proper non-zero limit at infinity.

After handling the roots (if any), we put all the types together (those that were by themselves, and types of roots) and determine the dominant term of the whole expression. This procedure may be repeated several times, in case there is a root under which there is an expression with another root and so on. Finally, having determined the dominant term of the whole given expression, we can handle it as above; that is, we factor out the dominant expression and check the limit.

Note: The factoring out part is usually easier when one repeats the guessing part; that is, first factor dominant terms from the roots and then work your way out.

Example: Find (if it exists)

First we check on the root. There are just powers under it, so they are in the same category and the one with higher exponent wins. In our case, the expression under the root is of the type x⁶. When the root is applied to this, we find that the root itself is of the type x^6/2 = x³, just like the second part of the given expression. Thus we get two terms of the same type, two dominant terms, and therefore none can be ignored. We try to do the guessing now, first ignoring the term under the root that we know can be ignored.

We obtained our guess (on the way we saw that the root by itself behaves like 3x³ for large x) and now we prove it. We start by pulling out the dominant factor from the root, as recommended.

In fact, the guessing part above might have been wrong and we would not know it, because we did not cover an important topic yet. We got lucky there, but now it is a time for

Warning: What happens if we get several dominant terms? If they are added, we can add them safely. If they are subtracted (that is, if the limit algebra would lead to the indeterminate form ∞ − ∞), then we have to be very careful. If applying the usual algebra to dominant terms in the guessing phase would preserve this dominant term, then we can do it as usual. If algebra would make this term disappear, we cannot use guessing!

For example, if we had 4x⁶ instead of 9x⁶ in this last example, the root would behave like 2x³, which would cancel with the other term. Then the step 2x³ − 2x³ = 0 cannot be done.

Why is it so? Because when we guess, each term actually represents not just itself, but there may be other terms of lower importance hidden in it (like "9x⁶" represents also the "+ x⁴" part in the above example). When we subtract dominant terms and some is left, then these parts that we ignored before can still be ignored (the dominant term that overshadowed them before is still there) and the guessing calculation is correct. However, if algebra would cause the dominant term to disappear, then one of the terms we previously ignored would suddenly get promoted to dominance, therefore this new dominant term would now determine the outcome of the limit! In such cases we have to use a more precise, more careful method of evaluation, a method that would not ignore terms that might be temporarily unimportant.

Compare the following two examples; they might look silly, but they illustrate the point well. In each of them we do the guessing first (even if it might be wrong) and then do the proper calculation.

In the first example, the guessing is correct; note that in the proper calculation, after putting the terms together, we can still ignore the "+ x" part, so it really plays no role in the final outcome. In the second calculation this x gets promoted to dominance.

Rule for dominant terms: If we are evaluating a limit at infinity by guessing and there are more dominant terms in an expression, then we can put them together by algebra only if they do not cancel as a result.

If they do cancel, we have to give up on intuitive calculations and try some other method. However, even then this guessing part helps as a preparation, since it is often useful to know the types of terms in an expression.

Remark: We talked about roots, but these are just special powers (the square root is just the 1/2-th power etc.). Of course, the procedure described above works for all such powers, so we should not be taken aback if we get an expression involving something like (x² − x + 1)¹³. This term would then belong among those that we start with, we would find the dominant term x² of the inside, the whole power is then of the type x²⁶.

We now reached the most typical expression that can be handled using the intuitive calculations: a fraction whose numerator and denominator are of the type we described above. How do we handle such fractions? First we separately investigate the numerator and denominator: We determine the dominant term of each of them, then factor them out. Then we have one dominant term in the numerator, one in the denominator, so we cancel them if we can and finally we find the limit of the resulting ratio. The scale of powers can again help here. If the term in the numerator prevails, we get infinity as the limit at infinity. If the term in the denominator prevails, we get zero as the limit at infinity.

This is quite natural. We usually get an infinity over infinity situation, and prevailing means that one infinity is larger than the other infinity, so it wins. For instance, when the infinity in the denominator prevails over the infinity in the numerator, it means that the denominator is eventually much much bigger than the numerator and the resulting ratio is thus very tiny, which suggests that it goes to zero.

We mentioned that sometimes the dominant terms of the denominator and numerator can be cancelled when they are factored from the ratio; we then obtain the type of the fraction as a whole.

Sometimes when facing a ratio, people prefer cancelling to factoring out and comparing. It works - but only sometimes. We offer a thorough discussion of how to handle ratios of polynomials at infinity in this note.

Example: Find (if it exists)

First we will guess the answer intuitively, then do it by proper calculations. We should start by handling the roots.

In the cubic root, the dominant term is the cube, therefore we can ignore the other terms for large values of x. In the square root in the denominator, the exponential prevails over the square and so we can ignore the power.

Now we know the types of the roots at infinity, so we can compare them with the remaining terms and find the dominants, separately for the numerator and denominator. Then we use the hierarchy of powers to guess the outcome:

What was the reasoning? In the numerator, powers beat logarithms, and the highest power is the square. By the way, this shows why it is important to always handle roots first. At the first glance one would guess that the x³ is the dominant term in the numerator, but after we analyzed the root, we saw that it actually behaves just like x.

In the denominator, the dominant term was the exponential 2^x, so we ignored the rest. Finally, since exponentials beat powers at infinity, we concluded that the ratio converges to zero.

Now we are supposed to confirm our guess by calculations - namely factoring out. Although an experienced student would do it on a few lines, we prefer to show more detail and also add a remark concerning the two roots; therefore we offer the calculations here.

Although it does not happen often, sometimes expressions of the above type are multiplied together. In such a case one applies the intuitive procedure to each term of the product, finding the type of each. The type of the whole product is then the product of individual types. However, usually one does not get one of the types studied above (powers, exponentials etc.), rather a product of such types. Such expressions are not listed in our scale of powers, therefore it is rare to get a ready answer this way. However, often one can use experience with types to find out something useful anyway.

Example:

Note that we found out that the numerator is of the type x2^x and the denominator is of the type x²2^x; fortunately, we could cancel the two types smartly and still got the answer using the scale of powers. However, now we will make a small change and it will not work any more:

Now we are comparing (after cancelling) two types, x2^x in the numerator and x^x in the denominator. Since the former is not a part of the usual scale of powers, we do not know the outcome when the two types get divided. We know that x^x beats 2^x, but obviously x2^x goes to infinity much faster than just 2^x; could it be that it goes so much faster that it even gets ahead of x^x, that is, is it possible that x2^x beats x^x?

The answer is: no. In fact, x2^x is a special type that beats 2^x and is beaten by x^x, so it fits exactly between these two. The function above converges to zero at infinity; the proof would be a modification of the proof that factorials beat exponentials, see Appendix in this note.

The intuitive reasoning can be applied to even more complicated expressions. It would be difficult to express precisely where we can apply this kind of reasoning, but less precisely it goes like this: The basic building block is a linear combination of powers, exponentials, powers of logarithms, and general powers (their multiples added and/or subtracted). This basic combination can be put under some root/into a power, thus creating a new building block that can be a part of another linear combination. These linear combinations can be put together using ratios/product and so create new members of further linear combinations. All these procedures can be repeated in any order.

Warning: Above we freely ignored parts of linear combinations and later even roots when they became unimportant compared to some dominant terms. However, there are two things one should be careful about. First, we can ignore only in linear combinations; roots/powers and ratios/products can be ignored only when they get replaced by their dominant terms and thus become eligible parts of linear combinations. Second and more importantly, this ignoring business can be only done in expressions as outlined above (linear combinations, roots/powers and ratios/products mixed up). It is not permissible to do ignoring in expressions which are mixed up with other functions. For instance, we can replace x − ln(x) with just x if it is under a root or a part of a fraction, but we cannot do it if it is an argument of, say, exponential. Thus 2^x-ln(x) ∼ 2^x is wrong, sinh(x − ln(x)) ∼ sinh(x) is wrong, etc.

For many examples we refer to Solved Problems - Limits.

Order of functions, asymptotes
Back to Theory - Limits