Lemma 1.9.1. The triangle inequality.
For any \(x,y \in \mathbb{R}\)
\begin{align*}
|x+y| & \leq |x| + |y|
\end{align*}
Section 1.9 (Optional) — Proving the Arithmetic of Limits
Perhaps the most useful theorem of this chapter is Theorem 1.4.3 which shows how limits interact with arithmetic. In this (optional) section we will prove both the arithmetic of limits Theorem 1.4.3 and the Squeeze Theorem 1.4.18. Before we get to the proofs it is very helpful to prove three technical lemmas that we’ll need. The first is a very general result about absolute values of numbers:
Proof.
Notice that for any real number \(x\text{,}\) we always have \(-x,x\le|x|\) and either \(|x|=x\) or \(|x|=-x\text{.}\) So now let \(x,y \in \mathbb{R}\text{.}\) Then we must have either
\begin{align*}
|x+y|&= x+y &&\le |x|+|y|\\
\end{align*}
or
\begin{align*} |x+y|&= -x-y &&\le |x|+|y| \end{align*}In both cases we end up with \(|x+y| \le |x| + |y|\text{.}\)
The second lemma is more specialised. It proves that if we have a function \(f(x) \to F\) as \(x \to a\) then there must be a small window around \(x=a\) where the function \(f(x)\) must only take values not far from \(F\text{.}\) In particular it tells us that \(|f(x)|\) cannot be bigger than \(|F|+1\) when \(x\) is very close to \(a\text{.}\)
Lemma 1.9.2.
Let \(a \in \mathbb{R}\) and let \(f\) be a function so that \(\ds \lim_{x \to a} f(x) =F\text{.}\) Then there exists a \(\delta \gt 0\) so that if \(0 \lt |x-a| \lt \delta\) then we also have \(|f(x)| \leq |F|+1\text{.}\)
The proof is mostly just manipulating the \(\epsilon\)–\(\delta\) definition of a limit with \(\epsilon=1\text{.}\)
Proof.
Let \(\epsilon = 1\text{.}\) Then since \(f(x) \to F\) as \(x \to a\text{,}\) there exists \(\delta \gt 0\) so that when \(0 \lt |x-a| \lt \delta\text{,}\) we also have \(|f(x)-F| \leq \epsilon=1\text{.}\) So now assume \(0 \lt |x-a| \lt \delta\text{.}\) Then
\begin{align*}
-\epsilon &\leq f(x) - F \leq \epsilon & \text{rearrange a little}\\
-\epsilon+F &\leq f(x) \leq \epsilon+F\\
\end{align*}
Now \(\epsilon+F \leq \epsilon+|F|\) and \(-\epsilon+F \geq -\epsilon-|F|\text{,}\) so
\begin{align*} -\epsilon-|F| &\leq f(x) \leq \epsilon+|F| \end{align*}Hence we have \(|f(x)| \leq \epsilon+|F| = |F|+1\text{.}\)
Finally our third technical lemma gives us a bound in the other direction; it tells us that when \(x\) is close to \(a\text{,}\) the value of \(|f(x)|\) cannot be much smaller than \(|F|\text{.}\)
Lemma 1.9.3.
Let \(a \in \mathbb{R}\) and \(F\ne0\) and let \(f\) be a function so that \(\ds \lim_{x \to a} f(x) = F\text{.}\) Then there exists \(\delta \gt 0\) so that when \(0 \lt |x-a| \lt \delta\text{,}\) we have \(|f(x)| \gt \frac{|F|}{2}\text{.}\)
Proof.
Set \(\epsilon = \frac{|F|}{2} \gt 0\text{.}\) Since \(f(x) \to F\text{,}\) we know there exists a \(\delta \gt 0\) so that when \(0 \lt |x-a| \lt \delta\) we have \(|f(x)-F| \lt \epsilon\text{.}\) So now assume \(0 \lt |x-a| \lt \delta\) so that \(|f(x)-F| \lt \epsilon = \frac{|F|}{2}\text{.}\) Then
\begin{align*}
|F| &= |F-f(x)+f(x)| & \text{sneaky trick}\\
& \leq |f(x) - F| + |f(x)| & \text{ but $|f(x)-F| \lt \epsilon$}\\
& \lt \epsilon + |f(x)|
\end{align*}
Hence \(|f(x)| \gt |F|-\epsilon= \frac{|F|}{2}\) as required.
Now we are in a position to prove Theorem 1.4.3. The proof has more steps than the previous \(\epsilon-\delta\) proofs we have seen. This is mostly because we do not have specific functions \(f(x)\) and \(g(x)\) and instead must play with them in the abstract — and make good use of the formal definition of limits.
We will break the proof into three pieces. The minimum that is required is to prove that
\begin{align*}
\lim_{x \to a} ( f(x) + g(x) ) &= F+G\\
\lim_{x \to a} f(x) \cdot g(x) &= F\cdot G\\
\lim_{x \to a} \frac{1}{g(x)} &= \frac{1}{G}\quad\text{if } G\ne 0.
\end{align*}
From these three we can prove that
\begin{align*}
\lim_{x \to a} f(x) \cdot c &= F\cdot c\\
\lim_{x \to a} ( f(x) - g(x) ) &= F-G\\
\lim_{x \to a} \frac{f(x)}{g(x)} &= \frac{F}{G}\quad\text{if } G\ne 0.
\end{align*}
The first follows by setting \(g(x) = c\) and using \(\lim f(x) \cdot g(x)\text{.}\) The second follows by setting \(c=-1\text{,}\) putting \(h(x) = (-1)\cdot g(x)\) and then applying both \(\lim f(x) \cdot g(x)\) and \(\lim f(x)+g(x)\text{.}\) The third follows by setting \(h(x) = \frac{1}{g(x)}\) and then using \(\lim f(x) \cdot h(x)\text{.}\)
Starting with addition, in order to satisfy the definition of limit, we are going to have to show that
\begin{align*}
|( f(x) + g(x) )-(F+G) | &\text{ is small}
\end{align*}
when we know that \(|f(x)-F|, |g(x)-G|\) are small. To do this we use the triangle inequality above showing that
\begin{align*}
|( f(x) + g(x) )-(F+G) | &=
|(f(x)-F) + (g(x)-G) | \leq |f(x)-F| + |g(x)-G|
\end{align*}
This is the key technical piece of the proof. So if we want the LHS of the above to be size \(\epsilon\text{,}\) we need to make sure that each term on the RHS is of size \(\frac{\epsilon}{2}\text{.}\) The rest of the proof is setting up facts based on the definition of limits and then rearranging facts to reach the conclusion.
Proof.
Proof of Theorem 1.4.3 — limit of a sum.
Let \(a \in \mathbb{R}\) and assume that
\begin{align*}
\lim_{x \to a} f(x) &= F & \text{ and } &&
\lim_{x \to a} g(x) &= G.
\end{align*}
We wish to show that
\begin{align*}
\lim_{x \to a} f(x)+g(x) &= F+G.
\end{align*}
Let \(\epsilon \gt 0\) — we have to find a \(\delta \gt 0\) so that when \(|x-a| \lt \delta\) we have \(|(f(x)+g(x))-(F+G)| \lt \epsilon\text{.}\)
Let \(\epsilon \gt 0\) and set \(\epsilon_1 = \epsilon_2 = \frac{\epsilon}{2}\text{.}\) By the definition of limits, because \(f(x) \to F\) there exists some \(\delta_1 \gt 0\) so that whenever \(|x-a| \lt \delta_1\text{,}\) we also have \(|f(x)-F| \lt \epsilon_1\text{.}\) Similarly there exists \(\delta_2 \gt 0\) so that if \(|x-a| \lt \delta_2\text{,}\) then we must have \(|g(x)-G| \lt \epsilon_2\text{.}\) So now choose \(\delta = \min\{ \delta_1, \delta_2 \}\) and assume \(|x-a| \lt \delta\text{.}\) Then we must have that \(|x-a| \lt \delta_1, \delta_2\) and so we also have
\begin{align*}
|f(x)-F|& \lt \epsilon_1 & |g(x)-G|& \lt \epsilon_2
\end{align*}
Now consider \(|(f(x)+g(x))-(F+G)|\) and rearrange the terms:
\begin{align*}
|(f(x)+g(x))-(F+G)| &= |(f(x)-F)+(g(x)-G)|\\
\end{align*}
now apply triangle inequality
\begin{align*} &\leq |f(x)-F| + |g(x)-G| & \text{ use facts from above}\\ & \lt \epsilon_1 + \epsilon_2\\ &= \epsilon. \end{align*}Hence we have shown that for any \(\epsilon \gt 0\) there exists some \(\delta \gt 0\) so that when \(|x-a| \lt \delta\) we also have \(|(f(x)+g(x))-(F+G)| \lt \epsilon\text{.}\) Which is exactly the formal definition of the limit we needed to prove.
Let us do similarly for the limit of a product. Some of the details of the proof are very similar, but there is a little technical trick in the middle to make it work. In particular we need to show that
\begin{align*}
|f(x) \cdot g(x) - F\cdot G| & \text{ is small}
\end{align*}
when we know that \(|f(x)-F|\) and \(|g(x)-G|\) are both small. Notice that
\begin{align*}
f(x) \cdot g(x) - F\cdot G
&= f(x) \cdot g(x) - F\cdot G + \underbrace{ f(x)\cdot G - f(x) \cdot G
}_{=0}\\
&= f(x)\cdot g(x)- f(x) \cdot G + f(x) \cdot G - F\cdot G\\
&= f(x)\cdot( g(x)-G) + (f(x)-F)\cdot G
\end{align*}
So if we know \(|f(x)-F|\) is small and \(|g(x)-G|\) is small then we are done — except that we also need to know that \(f(x)\) doesn’t become really large near \(a\) — this is exactly why we needed to prove Lemma 1.9.2.
As was the case in the previous proof, we want the LHS to be of size at most \(\epsilon\text{,}\) so we want, for example, the two terms on the RHS to be of size at most \(\frac{\epsilon}{2}\text{.}\) This means
- we need \(|G|\cdot|f(x)-F|\) to be of size at most \(\frac{\epsilon}{2}\text{,}\) and
- we need \(|g(x)-G|\) to be of size at most \(\frac{\epsilon}{2(|F|+1)}\) since we know that \(|f(x)| \leq |F|+1\) when \(x\) is close to \(a\text{.}\)
Armed with these tricks we turn to the proofs.
Proof.
Proof of Theorem 1.4.3 — limit of a product.
Let \(a \in \mathbb{R}\) and assume that
\begin{align*}
\lim_{x \to a} f(x) &= F & \text{ and } &&
\lim_{x \to a} g(x) &= G.
\end{align*}
We wish to show that
\begin{align*}
\lim_{x \to a} f(x)\cdot g(x) &= F\cdot G.
\end{align*}
Let \(\epsilon \gt 0\text{.}\) Set \(\epsilon_1 = \frac{\epsilon}{2(|G|+1)}\) (the extra \(+1\) in the denominator is just there to make sure that \(\epsilon_1\) is well–defined even if \(G=0\)), and \(\epsilon_2 = \frac{\epsilon}{2(|F|+1)}\text{.}\) From this we establish the existence of \(\delta_1, \delta_2, \delta_3\) which we need below.
- By assumption \(f(x) \to F\) so there exists \(\delta_1 \gt 0\) so that whenever \(|x-a| \lt \delta_1\text{,}\) we also have \(|f(x)-F| \lt \epsilon_1\text{.}\)
- Similarly because \(g(x) \to G\text{,}\) there exists \(\delta_2 \gt 0\) so that whenever \(|x-a| \lt \delta_2\text{,}\) we also have \(|g(x)-G| \lt \epsilon_2\text{.}\)
- By Lemma 1.9.2 there exists \(\delta_3 \gt 0\) so that whenever \(|x-a| \lt \delta_3\text{,}\) we also have \(|f(x)| \leq |F|+1\text{.}\)
Let \(\delta = \min\{\delta_1, \delta_2, \delta_3 \}\text{,}\) assume \(|x-a| \lt \delta\) and consider \(|f(x) \cdot g(x) - F\cdot G|\text{.}\) Rearrange the terms as we did above:
\begin{align*}
| f(x) \cdot g(x) - F\cdot G |
&= |f(x)\cdot( g(x)-G) + (f(x)-F)\cdot G |\\
& \leq |f(x)| \cdot |g(x)-G| + |G| \cdot |f(x)-F|
\end{align*}
By our three dot-points above we know that \(|f(x)-F| \lt \epsilon_1\) and \(|g(x)-G| \lt \epsilon_2\) and \(|f(x)| \leq |F|+1\text{,}\) so we have
\begin{align*}
| f(x) \cdot g(x) - F\cdot G | & \lt |f(x)| \cdot \epsilon_2 + |G| \cdot \epsilon_1\\
\end{align*}
sub in \(\epsilon_1,\epsilon_2\) and bound on \(f(x)\)
\begin{align*} & \lt (|F|+1) \cdot \frac{\epsilon}{2(|F|+1)} + |G|\cdot\frac{\epsilon}{2(|G|+1)}\\ &\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*}Thus we have shown that for any \(\epsilon \gt 0\) there exists \(\delta \gt 0\) so that when \(|x-a| \lt \delta\) we also have \(|f(x)\cdot g(x)-F\cdot G| \lt \epsilon\text{.}\) Hence \(f(x)\cdot g(x) \to F\cdot G\text{.}\)
Finally we can prove the limit of a reciprocal. Notice that
\begin{align*}
\frac{1}{g(x)} - \frac{1}{G} &= \frac{G-g(x)}{g(x) \cdot G}
\end{align*}
We need to show the LHS is of size at most \(\epsilon\) when \(x\) is close enough to \(a\text{,}\) so if \(G-g(x)\) is small we are done — except if \(g(x)\) or \(G\) are close to zero. By assumption (go back and read Theorem 1.4.3) we have \(G \neq 0\text{,}\) and we know from Lemma 1.9.3 that \(|g(x)|\) cannot be smaller than \(\frac{|G|}{2}\text{.}\) Together these imply that the denominator on the RHS cannot be zero and indeed must be of magnitude at least \(\frac{|G|^2}{2}\text{.}\) Thus we need \(|G-g(x)|\) to be of size at most \(\epsilon \cdot \frac{|G|^2}{2}\text{.}\)
Proof.
Proof of Theorem 1.4.3 — limit of a reciprocal.
Let \(\epsilon \gt 0\) and set \(\epsilon_1 = \epsilon|G|^2 \cdot \frac{1}{2}\text{.}\) We now use this and Lemma 1.9.3 to establish the existence of \(\delta_1, \delta_2\text{.}\)
- Since \(g(x) \to G\) we know that there exists \(\delta_1 \gt 0\) so that when \(|x-a| \lt \delta_1\) we also have \(|g(x)-G| \lt \epsilon_1\text{.}\)
- By Lemma 1.9.3 there exists \(\delta_2\) so that when \(|x-a| \lt \delta_2\) we also have \(|g(x)| \gt \frac{|G|}{2}\text{.}\) Equivalently, when \(|x-a| \lt \delta_2\) we also have \(\left|\frac{G}{2g(x)}\right| \lt 1\text{.}\)
Set \(\delta = \min\{\delta_1,\delta_2\}\) and assume \(|x-a| \lt \delta\text{.}\) Then
\begin{align*}
\left| \frac{1}{g(x)} - \frac{1}{G} \right| &= \left| \frac{G - g(x)}{g(x) \cdot G} \right|\\
&= |g(x) - G| \cdot \frac{1}{|G|\cdot |g(x)|} & \text{ by assumption}\\
& \lt \frac{\epsilon_1}{|G| \cdot |g(x)|} & \text{ sub in $\epsilon_1$}\\
& = \epsilon \cdot \frac{|G|}{2 |g(x)|} & \text{ since $\left|\frac{G}{2g(x)}\right| \lt 1$}\\
& \lt \epsilon
\end{align*}
Thus we have shown that for any \(\epsilon \gt 0\) there exists \(\delta \gt 0\) so that when \(|x-a| \lt \delta\) we also have \(|\frac{1}{g(x)} - \frac{1}{G}| \lt \epsilon\text{.}\) Hence \(\frac{1}{g(x)} \to \frac{1}{G}\text{.}\)
We can also now prove the Squeeze / sandwich / pinch theorem.
Proof.
Proof of Theorem 1.4.18 — Squeeze / sandwich / pinch.
In the squeeze theorem, we are given three functions \(f(x)\text{,}\) \(g(x)\) and \(h(x)\) and are told that
\begin{align*}
f(x) \leq g(x) \leq h(x) \quad\text{and}\quad \lim_{x \to a} f(x) &= \lim_{x \to a} h(x) = L
\end{align*}
and we must conclude from this that \(\lim\limits_{x \to a} g(x) = L\) too. That is, we are given some fixed, but unspecified, \(\epsilon \gt 0\) and it is up to us to find a \(\delta \gt 0\) with the property that \(\left| g(x) - L \right| \lt \epsilon\) whenever \(|x-a| \lt \delta\text{.}\) Now because we have been told that \(f\) and \(h\) both converge to \(L\text{,}\) there exist \(\delta_1 \gt 0\) and \(\delta_2 \gt 0\) such that
- \(\left| f(x) - L \right| \lt \epsilon\text{,}\) i.e. \(L-\epsilon \lt f(x) \lt L+\epsilon\text{,}\) whenever \(|x-a| \lt \delta_1\text{,}\) and
- \(\left| h(x) - L \right| \lt \epsilon\text{,}\) i.e. \(L-\epsilon \lt h(x) \lt L+\epsilon\text{,}\) whenever \(|x-a| \lt \delta_2\)
So set \(\delta = \min\{\delta_1,\delta_2\}\) and assume \(|x-a| \lt \delta\text{.}\) Then both \(L-\epsilon \lt f(x) \lt L+\epsilon\) and \(L-\epsilon \lt h(x) \lt L+\epsilon\) so that
\begin{align*}
L-\epsilon \lt f(x) &\le g(x) \le h(x) \lt L+\epsilon & \text{which implies that}\\
L-\epsilon \lt & g(x) \lt L+\epsilon & \text{which in turn gives us}\\
&\left| g(x) - L \right| \lt \epsilon
\end{align*}
as desired.
