# 4.2: The Limit as a Primary Tool

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- General rules about limits
- Understanding convergence

As you’ve seen from the previous sections, the formal definition of the convergence of a sequence is meant to capture rigorously our intuitive understanding of convergence. However, the definition itself is an unwieldy tool. If only there was a way to be rigorous without having to run back to the definition each time. Fortunately, there is a way. If we can use the definition to prove some general rules about limits then we could use these rules whenever they applied and be assured that everything was still rigorous. A number of these should look familiar from calculus.

Let \((c)_{n=1}^{\infty } = (c, c, c, \cdots )\) be a constant sequence. Show that \(\lim_{n \to \infty }c = c\).

In proving the familiar limit theorems, the following will prove to be a very useful tool.

**Triangle Inequality**Let \(a\) and \(b\) be real numbers. Then \[\left | a+b \right | \leq \left | a \right | + \left | b \right |\]**Reverse Triangle Inequality**Let \(a\) and \(b\) be real numbers. Then \[\left | a \right | - \left | b \right | \leq \left | a-b \right |\]

- Prove Lemma \(\PageIndex{1}\). [ Hint: For the Reverse Triangle Inequality, consider \(\left | a \right | = \left | a-b+b \right |\).]
- Show \(\left |\left | a \right | - \left | b \right | \right | \leq \left | a-b \right |\). [ Hint: You want to show \(\left | a \right | - \left | b \right | \leq \left | a-b \right |\) and \(-\left (\left | a \right | - \left | b \right | \right ) \leq \left | a-b \right |\).]

If \(\lim_{n \to \infty } a_n = a\) and \(\lim_{n \to \infty } b_n = b\), then \(\lim_{n \to \infty } \left (a_n + b_n \right ) = a + b\).

We will often informally state this theorem as “*the limit of a sum is the sum of the limits.*” However, to be absolutely precise, what it says is that if we already know that two sequences converge, then the sequence formed by summing the corresponding terms of those two sequences will converge and, in fact, converge to the sum of those individual limits. We’ll provide the scrapwork for the proof of this and leave the formal write-up as an exercise. Note the use of the triangle inequality in the proof.

**Scrapwork:**

If we let \(ε > 0\), then we want \(N\) so that if \(n > N\), then \(\left | \left (a_n + b_n \right ) - (a+b) \right | < \varepsilon\). We know that \(\lim_{n \to \infty } a_n = a\) and \(\lim_{n \to \infty } b_n = b\),so we can make \(\left | a_n - a \right |\) and \(\left | a_n - a \right |\) as small as we wish, provided we make \(n\) large enough. Let’s go back to what we want, to see if we can close the gap between what we know and what we want. We have

\[\left | \left (a_n + b_n \right ) - (a+b) \right | = \left | (a_n - a) + (b_n - b) \right | \leq \left | a_n - a \right | + \left | b_n - b \right |\]

by the triangle inequality. To make this whole thing less than \(ε\), it makes sense to make each part less than \(\frac{ε}{2}\). Fortunately, we can do that as the definitions of \(\lim_{n \to \infty } a_n = a\) and \(\lim_{n \to \infty } b_n = b\) allow us to make \(\left | a_n - a \right |\) and \(\left | a_n - a \right |\) arbitrarily small. Specifically, since \(\lim_{n \to \infty } a_n = a\), there exists an \(N_1\) such that if \(n > N_1\) then \(\left | a_n - a \right | < \frac{\varepsilon }{2}\). Also since \(\lim_{n \to \infty } b_n = b\), there exists an \(N_2\) such that if \(n > N_2\) then \(\left | b_n - b \right | < \frac{\varepsilon }{2}\). Since we want both of these to occur, it makes sense to let \(N =\max (N_1,N_2)\). This should be the \(N\) that we seek.

Prove Theorem \(\PageIndex{1}\).

If \(\lim_{n \to \infty } a_n = a\) and \(\lim_{n \to \infty } b_n = b\), then \(\lim_{n \to \infty } (a_nb_n) = ab\).

**Scrapwork:**

Given \(ε > 0\), we want \(N\) so that if \(n > N\), then \(\left | a_nb_n - ab \right | < \varepsilon\). One of the standard tricks in analysis is to “*uncancel*.” In this case we will subtract and add a convenient term. Normally these would “*cancel out*,” which is why we say that we will uncancel to put them back in. You already saw an example of this in proving the Reverse Triangle Inequality. In the present case, consider

\[\begin{align*} \left | a_nb_n - ab \right | &= \left | a_nb_n - a_nb + a_nb -ab \right |\\ &\leq \left |a_nb_n - a_nb \right | + \left |a_nb -ab \right | \\ &= \left | a_n \right |\left | b_n - b \right | + \left | b \right |\left | a_n - a \right | \end{align*}\]

We can make this whole thing less than \(ε\), provided we make each term in the sum less than \(\frac{ε}{2}\). We can make \(\left | b \right |\left | a_n - a \right | < \frac{\varepsilon}{2}\) if we make \(\left | a_n - a \right | < \frac{\varepsilon}{2\left | b \right |}\). But wait! What if \(b = 0\)? We could handle this as a separate case or we can do the following “*slick trick*.” Notice that we can add one more line to the above string of inequalities: \(\left | a_n \right |\left | b_n - b \right | + \left | b \right |\left | a_n - a \right | < \left | a_n \right |\left | b_n - b \right | + \left (\left | b \right | + 1 \right ) \left | a_n - a \right |\). Now we can make \(\left | a_n - a \right | < \frac{\varepsilon}{2\left (\left | b \right | +1 \right )}\) and not worry about dividing by zero.

Making an \(\left | a_n \right |\left | b_n - b \right | < \frac{\varepsilon}{2}\) requires a bit more finesse. At first glance, one would be tempted to try and make \(\left | b_n - b \right | < \frac{\varepsilon}{2\left | a_n \right |}\). Even if we ignore the fact that we could be dividing by zero (which we could handle), we have a bigger problem. According to the definition of \(\lim_{n \to \infty } b_n = b\), we can make \(\left |b_n - b \right |\) smaller than any given fixed positive number, as long as we make n large enough (larger than some \(N\) which goes with a given epsilon). Unfortunately, \[(frac{\varepsilon}{2\left | a_n \right |}\) is not fixed as it has the variable \(n\) in it; there is no reason to believe that a single \(N\) will work with all of these simultaneously. To handle this impasse, we need the following

If \(\lim_{n \to \infty } a_n = a\), then there exists \(B > 0\) such that \(|a_n|≤ B\) for all \(n\).

**End of Scrapwork.**

Prove Lemma \(\PageIndex{2}\). [ *Hint*: We know that there exists \(N\) such that if \(n > N\), then \(|a_n - a| < 1\). Let \(B = \max \left (|a_1|,|a_2|,..., |a_{\left \lceil N \right \rceil}|,|a|+ 1 \right )\), where \(\left \lceil N \right \rceil\) represents the smallest integer greater than or equal to \(N\). Also, notice that this is not a convergence proof so it is not safe to think of \(N\) as a large number.]

Armed with this bound \(B\), we can add on one more inequality to the above scrapwork to get

\[\begin{align*} \left | a_n\cdot b_n - a\cdot b \right | &= \left | a_n\cdot b_n - a_n\cdot b + a_n\cdot b -a\cdot b \right |\\ &\leq \left |a_n\cdot b_n - a_n\cdot b \right | + \left |a_n\cdot b -a\cdot b \right | \\ &= \left | a_n \right |\left | b_n - b \right | + \left | b \right |\left | a_n - a \right | \\ &< B\left | b_n - b \right | + \left (\left | b \right | + 1 \right )\left | a_n - a \right | \end{align*}\]

At this point, we should be able to make the last line of this less than \(ε\).

End of **Scrapwork**.

Prove Theorem \(\PageIndex{2}\).

If \(\lim_{n \to \infty } a_n = a\) and \(c \in \mathbb{R}\) then \(\lim_{n \to \infty } c\cdot a_n = c\cdot a\)

Prove the above corollary to Theorem \(\PageIndex{2}\)

Just as Theorem \(\PageIndex{2}\) says that the limit of a product is the product of the limits, we can prove the analogue for quotients.

Suppose \(\lim_{n \to \infty } a_n = a\) and \(\lim_{n \to \infty } b_n = b\). Also suppose \(b \neq 0\) and \(b_n \neq 0, \forall n\). Then \(\lim_{n \to \infty }\left ( \frac{a_n}{b_n} \right ) = \frac{a}{b}\).

**Scrapwork:**

To prove this, let’s look at the special case of trying to prove \(\lim_{n \to \infty }\left ( \frac{1}{b_n} \right ) = \frac{1}{b}\). The general case will follow from this and Theorem \(\PageIndex{2}\). Consider \(\left | \frac{1}{b_n} - \frac{1}{b} \right | = \frac{\left | b - b_n \right |}{\left | b_n \right |\left | b \right |}\). We are faced with the same dilemma as before; we need to get \(\left | \frac{1}{b_n} \right |\) bounded above. This means we need to get \(|b_n|\) bounded away from zero (at least for large enough \(n\)).

This can be done as follows. Since \(b \neq 0\), then \(\frac{\left | b \right |}{2} > 0\). Thus, by the definition of \(\lim_{n \to \infty } b_n = b\), there exists \(N_1\) such that if \(n > N_1\), then \(\left | b \right | - \left | b_n \right | \leq \left | b - b_n \right | < \frac{\left | b \right |}{2}\). Thus when \(n > N_1\), then \(\frac{\left | b \right |}{2} < \left | b_n \right |\) and so \(\frac{1}{\left | b_n \right |} < \frac{2}{b}\). This says that for \(n > N_1\), \(\frac{\left | b - b_n \right |}{\left | b_n \right |\left | b \right |} < \frac{2}{\left | b \right |^2}\left | b - b_n \right |\). We should be able to make this smaller than a given \(ε > 0\), provided we make \(n\) large enough.

End of **Scrapwork**.

Prove Theorem \(\PageIndex{3}\)

These theorems allow us to compute limits of complicated sequences and rigorously verify that these are, in fact, the correct limits without resorting to the definition of a limit.

Identify all of the theorems implicitly used to show that

\[\lim_{n \to \infty } \frac{3n^3 - 100n + 1 }{5n^3 + 4n^2 - 7} = \lim_{n \to \infty } \frac{n^3\left ( 3 - \frac{100}{n^2} + \frac{1}{n^3} \right )}{n^3\left ( 5 + \frac{4}{n} - \frac{7}{n^3} \right )} = \frac{3}{5}\]

Notice that this presumes that all of the individual limits exist. This will become evident as the limit is decomposed.

There is one more tool that will prove to be valuable.

Let \((r_n)\), \((s_n)\), and \((t_n)\) be sequences of real numbers with \(r_n ≤ s_n ≤ t_n,∀\) positive integers \(n\). Suppose \(\lim_{n \to \infty }r_n = s = \lim_{n \to \infty } t_n\). Then \((s_n)\) must converge and \(\lim_{n \to \infty }s_n = s\).

Prove Theorem \(\PageIndex{4}\). [*Hint*: This is probably a place where you would want to use \(s - ε < s_n < s + ε\) instead of \(|s_n - s| < ε\).]

The Squeeze Theorem holds even if \(r_n ≤ s_n ≤ t_n\) holds for only sufficiently large \(n\); i.e., for \(n\) larger than some fixed \(N_0\). This is true because when you find an \(N_1\) that works in the original proof, this can be modified by choosing \(N =\max (N_0,N_1)\). Also note that this theorem really says two things: \((s_n)\) converges and it converges to \(s\). This subtle point affects how one should properly use the Squeeze Theorem.

Prove \(\lim_{n \to \infty }\frac{n + 1}{n^2} = 0\).

**Proof:**

Notice that \(0 \leq \frac{n + 1}{n^2} \leq \frac{n + n}{n^2} = \frac{2}{n}\).

Since \(\lim_{n \to \infty }0 = 0 = \lim_{n \to \infty }\frac{2}{n}\), then by the Squeeze Theorem, \(\lim_{n \to \infty }\frac{n + 1}{n^2} = 0\).

Notice that this proof is completely rigorous. Also notice that this is the proper way to use the Squeeze Theorem. Here is an example of an improper use of the Squeeze Theorem.

How **not **to prove Example \(\PageIndex{1}\). Notice that

\[0 \leq \frac{n + 1}{n^2} \leq \frac{n + n}{n^2} = \frac{2}{n}\]

so

\[0 = \lim_{n \to \infty }0 \leq \lim_{n \to \infty } \frac{n + 1}{n^2} \leq \lim_{n \to \infty } \frac{2}{n} = 0\]

and

\[\lim_{n \to \infty } \frac{n + 1}{n^2} = 0\]

This is incorrect in form because it presumes that \(\lim_{n \to \infty } \frac{n + 1}{n^2}\) exists, which we don’t yet know. If we knew that the limit existed to begin with, then this would be fine. The Squeeze Theorem proves that the limit does in fact exist, but it must be so stated.

These general theorems will allow us to rigorously explore convergence of power series in the next chapter without having to appeal directly to the definition of convergence. However, you should remember that we used the definition to prove these results and there will be times when we will need to apply the definition directly. However, before we go into that, let’s examine divergence a bit more closely.

## Contributor

Eugene Boman (Pennsylvania State University) and Robert Rogers (SUNY Fredonia)