Here we will talk about how to guess the limit of a sequence. This can be applied to many sequences and uses several ingredients. First of all, one has to remember the elementary limits and the limit algebra. The second important ingredient is the understanding of interplay of powers and exponentials (and roots). We will start with little introduction.

In many sequences one finds sums of powers and similar expressions. For
instance, consider the sequence
*n*^{3} + 5*n*^{1/2} − *n*^{−2}}.*n*^{2}

Now here's a problem: What is the limit of
*n*^{3} − 5*n*^{2}}?

By the way, when we are adding and subtracting some objects and even multiplying them by real numbers like we did in the first example, we call it "linear combination". So the expression in the first 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 *n*^{−2} tends to zero and therefore does not
cause throubles 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

We will also allow for "exponentials", powers of the form
*e*^{n}, 2^{n} etc. Again, since we know
that
*q*^{n}→0*q*| < 1*q*^{n} with
*q*| > 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, factorials and even powers of the form
*n*^{n}. Now we are ready to state the problem we are
trying to address here:

**Question:**

We have an expression of the form
*α*⋅*A*(*n*) + *β*⋅*B*(*n*) +...,*A*(*n*),*B*(*n*),...*n*^{a} with
*a* > 0,*q*^{n} with
*q*| > 1,*n*)]^{a}*a* > 0,*n*! and general powers like *n*^{n}. What
happens with this expression if we let *n* go to infinity?

Note that all expressions listed above tend to infinity as
*n*→∞.

**Solution to the problem:**

If we let *n*→∞,

By a "dominant term" we mean the term that, for large *n*, 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 *n*", we really mean
that when investigating the limit of the expression
*α*⋅*A* + *β*⋅*B* for *n**B* and just worry about *A*. For more insight
into this comparison of infinities, click here.

**Example:**

We will shortly see that *n*^{2} prevails over
*n*).*n*) − *n*^{2}}.*n* grows really
large, the second term prevails, we can ignore the first one. Thus the
sequence will behave (when *n* grows to infinity) like the sequence
*n*^{2}},

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:

Now the first term converges to infinity. What about the part in parentheses? The ratio is of the type infinity over infinity, which is handled using the l'Hospital rule. By a remarkable coincidence we investigated exactly this problem in the previous section L'Hospital's rule, so we know that this ratio converges to zero. Thus we can finish the problem:

Note that the fraction that appeared after factoring out the dominant power indeed tended to a real number, exactly as stated above.

There is a way to write properly not just the correct mathematical
calculation, but also our intuitive reasoning. When we write
*A*(*n*) ∼ *B*(*n*),*A* and
*B* behave the same when
*n*→∞;*A*(*n*)*B*(*n*)*n* 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:

*n*) − *n*^{2} ∼ −*n*^{2}→−∞.

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 *n*)]^{3}"*n*)]^{3},*n*^{2}, because the type of an expression is given by the type
of its dominant term. The type tells us how the given expression
approximately behaves when *n* is really large.

This type is used when roughly comparing behaviour 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
*n*) − *n*^{2}*n*^{2}*n* gets large, but then we would have to say
that it behaves like *n*^{2}

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 *n*^{n},

(2) the factorial *n*!,

(3) the exponential *q*^{n} for
*q*| > 1,

(4) the power *n*^{ a}*a* > 0,

(5) the logarithm *n*)]^{a}*a* > 0.

For practical use people often prefer a more colloquial way of remembering this hierarchy, using phrases like "powers beat logarithms" and "exponentials beat powers" etc.

When investigating a linear combination of such terms, we always first find
the dominant expression. However, there may be more terms of this dominant
expression. Thus we also need to know mutual dominance within each category.
There the rule is simple. In categories (3), (4) and (5) there is always a
parameter (*a* or *q*) and the highest one is the dominant one.

Note that if *q* < −1,*q*^{n}}*q*|

In fact, the question of dominance is one kind of an answer to the question
"which infinity is larger". Since for
*q* < −1*q*^{n}→∞,*q* > 1.**scale of powers**:

**Example:**

What is the limit of
^{n} + *n*^{2} − 5⋅3^{n} − ln(*n*)^{1/2}}?

**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^{n} and
3^{n}. Since ^{n}. Thus we can ignore all the other terms and guess that

^{n} + *n*^{2} − 5⋅3^{n} − ln(*n*)^{1/2} ∼ −5⋅3^{n}→−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 a geometric sequence, the other two will lead to l'Hospital's rule after passing to investigation of functions:

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:
*n*)^{3} = 2^{3}⋅*n*^{3} = 8⋅*n*^{3}^{2n+1} = (3^{2})^{n}⋅3^{1} = 3⋅9^{n}.

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 individually each root. 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 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, factor it out of the dominant expression and check the limit.

Note: The factoring 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 *n*^{6}. When the root
is applied, we find that the root itself is of the type
*n*^{6/2} = *n*^{3},

We obtained our guess (on the way we saw that the root by itself behaves like
*n*^{3}*n*) 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 did 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 limit
algebra would lead to the indeterminate form

For example, if we had *n*^{6}*n*^{6}*n*^{3},*n*^{3} − 2*n*^{3} = 0

Why is it so? Becasue when we guess, each term actually represents not just
itself, but there may be other terms of lower importance hidden in it (like
the *n*^{4}"

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 *n*"*n*
gets promoted to dominance.

Rule for dominant terms:

If we are evaluating a limit 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 a sequence. There is an example in Solved Problems - Limits.

We now reached the most general type 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. If the term in the denominator prevails, we get zero as the limit.

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 (or order) 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 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 *n*. 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, 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
*n*^{3} is the dominant term in the numerator, but after we
analyzed the root, we saw that it actually behaves just like *n*.

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

Now we are supposed to confirm our guess by calculations - namely factoring. 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.

The intuitive reasoning can be applied to even more complicated expressions. A fraction as above can be put under a root and then added to some powers, exponentials etc. 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, factorials, general powers (their multiples added and/or subtracted). This basic combination can be put under some root, thus creating a new building block that can be a part of another linear combination. These linear combinations can be put together using ratios, and all these procedures can be repeated in any order.

**Warning:** Ignoring of parts can be done only in basic linear
combinations, that is, in sums with powers, exponentials, powers of
logarithms, factorials and general powers. Other expressions (like roots,
ratios) can be ignored only when they get replaced by their dominant terms
and thus become eligible parts of linear combinations. It is not permissible
to do ignoring in expressions which are not parts of roots and/or ratios,
but are mixed up with other functions. For instance, we can replace
*n* − ln(*n*) with just *n* 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. For instance, it is easy to show that
^{n − ln(n)} ∼ 2^{n}__wrong__, and without further investigation we do not know whether
*n* − ln(*n*)) ∼ ln(*n*)*n* − ln(*n*)) ∼ arctan(*n*).

For many examples we refer to Solved Problems - Limits.

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 a type of the form we 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
*n*2^{n}*n*^{2}*n*!;

Now we are comparing (after cancelling) two types,
*n*2^{n}*n*! 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
*n*! beats 2^{n}, but obviously
*n*2^{n}^{n}; could it be that it goes so much faster that it even
gets ahead of *n*!, that is, is it possible that
*n*2^{n}*n*! ?

The answer is: no. In fact, *n*2^{n}^{n} and is beaten by *n*!, so it fits
exactly between these two. The sequence above converges to zero; the proof
would be a modification of the proof that factorials beat exponentials.