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Sums, products and quotients of limits, the order rule, and the theorem that produces convergence from an inequality.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute a limit by rewriting an expression until every piece converges, name the rule that settles it and check that rule's hypothesis before using it. You will also be able to use the squeeze theorem, which concludes convergence rather than assuming it, and to say what a rule asserts when one of its hypotheses fails, which is nothing at all.
The epsilon-N definition, and the habit of producing an $N$ from an $\varepsilon$. Going back to the definition for every limit would be unbearable, so this lesson proves the rules once and then uses them — but each rule carries hypotheses, and reading those is the skill.
The algebra of limits is the collection of sum, product and quotient rules. The squeeze theorem concludes convergence from an inequality rather than assuming it. A rule's hypothesis is the condition that makes it available, and a rule whose hypothesis fails asserts nothing at all.
Let $a_n \to L$ and $b_n \to M$. Then:
The squeeze theorem. If $a_n \le c_n \le b_n$ and $a_n \to L$ and $b_n \to L$, then $c_n \to L$.
It is the exception in this list: the other rules assume the sequence in question converges and compute its limit, while the squeeze concludes convergence from an inequality. That is what makes it the tool for anything not written as an arithmetic combination — $\sin(n)/n$, $n$-th roots, factorial quotients.
The hypotheses are not decoration. Two sequences with no limits at all can have a difference that converges, and a quotient rule used on a vanishing denominator is not a weak argument but no argument. Each rule is a theorem, and a theorem is unavailable until its hypotheses have been checked.
Another way: picture
The squeeze is the only one you can draw. Two curves close in on the same height, and a third is trapped between them with nowhere else to go. The middle sequence is never described directly, and that is the point of the picture: it does not have to be.
Another way: steps
To compute a limit:
If step 1 will not come out, reach for the squeeze: bound the quantity above and below by things that do converge.
The proof of $a_n b_n \to LM$ is the splitting move from the absolute value lesson: $$|a_n b_n - LM| = |a_n b_n - a_n M + a_n M - LM| \le |a_n||b_n - M| + |M||a_n - L|.$$ The second piece is fine: $|M|$ is a constant and $|a_n - L|$ can be made small. The first has the factor $|a_n|$, which is not a constant — so the proof cannot proceed until $|a_n|$ is bounded.
It is, because $a_n$ converges, and a convergent sequence is bounded. That small theorem is not a side remark; it is the load-bearing step of the product rule, and it is why it is worth proving on its own.
The pattern recurs: a loose factor is bounded first, and only then is the fine estimate made. The same two-stage shape appeared in bounding $|x^2 - a^2|$, and it will appear again in the product rule for derivatives.
| Expression | Rewrite as | Then |
|---|---|---|
| $\dfrac{3n^2 + n}{5n^2 - 2}$ | divide by $n^2$ | quotient rule, denominator $\to 5$ |
| $\sqrt{n + 1} - \sqrt{n}$ | multiply by the conjugate over itself | $\dfrac{1}{\sqrt{n+1} + \sqrt{n}} \to 0$ |
| $\dfrac{\sin n}{n}$ | bound the numerator | squeeze between $\pm 1/n$ |
| $\dfrac{n!}{n^n}$ | bound the product of the factors | squeeze between $0$ and $1/n$ |
The first two are algebra, the last two are the squeeze. Deciding which of the two kinds of move a problem needs is most of the skill: if the expression is built from pieces with limits, rewrite; if it is not, trap it.
Applying them to divergent sequences. $\lim(a_n - b_n) = \lim a_n - \lim b_n$ is a theorem about convergent sequences. With two that diverge, the right-hand side does not name anything.
Reading a failed hypothesis as a conclusion. A denominator whose limit is zero does not make the limit infinite; it makes the rule silent. What happens then depends on the sequences, and has to be worked out.
Expecting strict inequality to survive. $a_n < b_n$ for every $n$ gives only $L \le M$. The gap can close in the limit, and usually does in the examples that matter.
Squeezing with bounds that do not meet. The two outer sequences must tend to the same limit. Bounds that tend to $0$ and $1$ trap the sequence in an interval and say nothing about convergence.
Compute $\lim \dfrac{4n^2 - n}{2n^2 + 7}$. Divide top and bottom by $n^2$: $\dfrac{4 - 1/n}{2 + 7/n^2}$.
Make every piece converge.
Numerator $\to 4$, denominator $\to 2$, and $2 \ne 0$, so the quotient rule applies.
Name the rule, check the hypothesis.
The limit is $2$. Applied before the division, the rule would have been asked about two unbounded sequences and would have said nothing.
The rewrite is what makes the rule available.
Compute $\lim \dfrac{\cos(n^2)}{n}$. Nothing here is an arithmetic combination of sequences with known limits.
The algebra of limits does not reach it.
But $-\dfrac{1}{n} \le \dfrac{\cos(n^2)}{n} \le \dfrac{1}{n}$, and both bounds tend to $0$.
Bound above and below by things that do converge.
So the limit is $0$. The behaviour of $\cos(n^2)$ was never described, only bounded — which is all the squeeze ever asks for.
Convergence produced from an inequality.
As written it is a difference of two unbounded sequences, so no rule applies; multiply by the conjugate over itself.
Rewrite until the rules can see it.
That gives $\dfrac{n}{\sqrt{n^2 + n} + n}$; dividing top and bottom by $n$ gives $\dfrac{1}{\sqrt{1 + 1/n} + 1}$.
Divide by the highest power.
Now every piece converges and the denominator's limit is $2$, so the answer is $1/2$. The naive reading — infinity minus infinity, therefore zero — gets a different and wrong answer, which is exactly why the hypotheses are checked.
Match each situation to the result that settles it.
| the sum rule | the product rule | the quotient rule, with a non-zero limit below | the squeeze theorem | |
|---|---|---|---|---|
| $a_n \to 2$ and $b_n \to 2 + 1$; what is $\lim (a_n + b_n)$? | ||||
| $a_n \to 2$ and $b_n \to 2 + 1$; what is $\lim (a_n b_n)$? | ||||
| $a_n \to 2$ and $b_n \to 2 + 1$; what is $\lim (a_n / b_n)$? | ||||
| $a_n \le c_n \le b_n$, and both $a_n$ and $b_n$ tend to $2$; what does $c_n$ do? |
What is $\displaystyle\lim_{n \to \infty} \dfrac{2n + 3}{9n - 1}$?
Answer:
Put the steps of the computation of $\displaystyle\lim \dfrac{6n + 3}{9n - 1}$ into order.
Number the steps in order (write the number in the box):
Let $a_n = 7n + 1/n$ and $b_n = 7n$. Here is an argument about $\lim (a_n - b_n)$. Mark the one step that is not permitted.
This task has no paper form; do it on a device.
$a_n \to 7$ and $b_n \to 0$, with every $b_n$ non-zero. What does the quotient rule tell you about $a_n / b_n$?
Build the proof that a convergent sequence is bounded.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each situation to the result that settles it.
| the sum rule | the product rule | the quotient rule, with a non-zero limit below | the squeeze theorem | |
|---|---|---|---|---|
| $a_n \to 8$ and $b_n \to 8 + 1$; what is $\lim (a_n + b_n)$? | ||||
| $a_n \to 8$ and $b_n \to 8 + 1$; what is $\lim (a_n b_n)$? | ||||
| $a_n \to 8$ and $b_n \to 8 + 1$; what is $\lim (a_n / b_n)$? | ||||
| $a_n \le c_n \le b_n$, and both $a_n$ and $b_n$ tend to $8$; what does $c_n$ do? |
You can compute limits with the sum, product and quotient rules, check the hypothesis each needs, and use the squeeze theorem on an expression the algebra cannot reach. Say in your own words why the product rule needs a convergent sequence to be bounded. Next: limits that can be proved to exist before anyone knows their value.
10. Your turn: $\lim \left(\sqrt{n^2 + n} - n\right)$, step 3