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## Section3.2Precise Definition of a Limit

The definition given for a limit previously is more of a working definition. In this section we pursue the actual, official definition of a limit.

###### Definition3.4. Precise Definition of Limit.

Suppose $f$ is a function. We say that $\ds \lim_{x\to a}f(x)=L$ if for every $\epsilon>0$ there is a $\delta > 0$ so that whenever $0 \lt |x-a| \lt \delta\text{,}$ $|f(x)-L|\lt \epsilon\text{.}$

The $\epsilon$ and $\delta$ here play exactly the role they did in the preceding discussion. The definition says, in a very precise way, that $f(x)$ can be made as close as desired to $L$ (that's the $|f(x)-L|\lt \epsilon$ part) by making $x$ close enough to $a$ (the $0 \lt |x-a| \lt \delta$ part). Note that we specifically make no mention of what must happen if $x=a\text{,}$ that is, if $|x-a|=0\text{.}$ This is because in the cases we are most interested in, substituting $a$ for $x$ doesn't make sense.

Make sure you are not confused by the names of important quantities. The generic definition talks about $f(x)\text{,}$ but the function and the variable might have other names. The $x$ was the variable of the original function; when we were trying to compute a slope or a velocity, $x$ was essentially a fixed quantity, telling us at what point we wanted the slope. In the velocity problem, it was literally a fixed quantity, as we focused on the time $t=2\text{.}$ The quantity $a$ of the definition in all the examples was zero: we were always interested in what happened as $\Delta x$ became very close to zero.

Armed with a precise definition, we can now prove that certain quantities behave in a particular way. The bad news is that even proofs for simple quantities can be quite tedious and complicated. The good news is that we rarely need to do such proofs, because most expressions act the way you would expect, and this can be proved once and for all.

###### Example3.5. Epsilon Delta.

Let's show carefully that $\ds \lim_{x\to 2} x+4 = 6\text{.}$

Solution

This is not something we “need” to prove, since it is “obviously” true. But if we couldn't prove it using our official definition there would be something very wrong with the definition.

As is often the case in mathematical proofs, it helps to work backwards. We want to end up showing that under certain circumstances $x+4$ is close to 6; precisely, we want to show that $|x+4-6|\lt \epsilon\text{,}$ or $|x-2|\lt \epsilon\text{.}$ Under what circumstances? We want this to be true whenever $0\lt |x-2|\lt \delta\text{.}$ So the question becomes: can we choose a value for $\delta$ that guarantees that $0\lt |x-2|\lt \delta$ implies $|x-2|\lt \epsilon\text{?}$ Of course: no matter what $\epsilon$ is, $\delta=\epsilon$ works.

So it turns out to be very easy to prove something “obvious,” which is nice. It doesn't take long before things get trickier, however.

###### Example3.6. Epsilon Delta.

It seems clear that $\ds \lim_{x\to 2} x^2=4\text{.}$ Let's try to prove it.

Solution

We will want to be able to show that $\ds |x^2-4|\lt \epsilon$ whenever $0\lt |x-2|\lt \delta\text{,}$ by choosing $\delta$ carefully. Is there any connection between $|x-2|$ and $\ds |x^2-4|\text{?}$ Yes, and it's not hard to spot, but it is not so simple as the previous example. We can write $\ds |x^2-4|=|(x+2)(x-2)|\text{.}$ Now when $|x-2|$ is small, part of $|(x+2)(x-2)|$ is small, namely $(x-2)\text{.}$ What about $(x+2)\text{?}$ If $x$ is close to 2, $(x+2)$ certainly can't be too big, but we need to somehow be precise about it. Let's recall the “game” version of what is going on here. You get to pick an $\epsilon$ and I have to pick a $\delta$ that makes things work out. Presumably it is the really tiny values of $\epsilon$ I need to worry about, but I have to be prepared for anything, even an apparently “bad” move like $\epsilon=1000\text{.}$ I expect that $\epsilon$ is going to be small, and that the corresponding $\delta$ will be small, certainly less than 1. If $\delta\le 1$ then $|x+2|\lt 5$ when $|x-2|\lt \delta$ (because if $x$ is within 1 or 2, then $x$ is between 1 and 3 and $x+2$ is between 3 and 5). So then I'd be trying to show that $|(x+2)(x-2)|\lt 5|x-2|\lt \epsilon\text{.}$ So now how can I pick $\delta$ so that $|x-2|\lt \delta$ implies $5|x-2|\lt \epsilon\text{?}$ This is easy: use $\delta=\epsilon/5\text{,}$ so $5|x-2|\lt 5(\epsilon/5) = \epsilon\text{.}$ But what if the $\epsilon$ you choose is not small? If you choose $\epsilon=1000\text{,}$ should I pick $\delta=200\text{?}$ No, to keep things “sane” I will never pick a $\delta$ bigger than 1. Here's the final “game strategy” : when you pick a value for $\epsilon\text{,}$ I will pick $\delta=\epsilon/5$ or $\delta=1\text{,}$ whichever is smaller. Now when $|x-2|\lt \delta\text{,}$ I know both that $|x+2|\lt 5$ and that $|x-2|\lt \epsilon/5\text{.}$ Thus $|(x+2)(x-2)|\lt 5(\epsilon/5) = \epsilon\text{.}$

This has been a long discussion, but most of it was explanation and scratch work. If this were written down as a proof, it would be quite short, like this:

Proof that $\ds \lim_{x\to 2}x^2=4\text{.}$ Given any $\epsilon\text{,}$ pick $\delta=\epsilon/5$ or $\delta=1\text{,}$ whichever is smaller. Then when $|x-2|\lt \delta\text{,}$ $|x+2|\lt 5$ and $|x-2|\lt \epsilon/5\text{.}$ Hence $\ds |x^2-4|=|(x+2)(x-2)|\lt 5(\epsilon/5) = \epsilon\text{.}$

It probably seems obvious that $\ds \lim_{x\to2}x^2=4\text{,}$ and it is worth examining more closely why it seems obvious. If we write $\ds x^2=x\cdot x\text{,}$ and ask what happens when $x$ approaches 2, we might say something like, “Well, the first $x$ approaches 2, and the second $x$ approaches 2, so the product must approach $2\cdot2\text{.}$” In fact this is pretty much right on the money, except for that word “must.” Is it really true that if $x$ approaches $a$ and $y$ approaches $b$ then $xy$ approaches $ab\text{?}$ It is, but it is not really obvious, since $x$ and $y$ might be quite complicated. The good news is that we can see that this is true once and for all, and then we don't have to worry about it ever again. When we say that $x$ might be “complicated” we really mean that in practice it might be a function. Here is then what we want to know:

We must use the Precise Definition of a Limit to prove the Produce Law for Limits. So given any $\epsilon$ we need to find a $\delta$ so that $0\lt |x-a|\lt \delta$ implies $|f(x)g(x)-LM|\lt \epsilon\text{.}$ What do we have to work with? We know that we can make $f(x)$ close to $L$ and $g(x)$ close to $M\text{,}$ and we have to somehow connect these facts to make $f(x)g(x)$ close to $LM\text{.}$

We use, as is often the case, a little algebraic trick:

\begin{align*} |f(x)g(x)-LM|\amp =|f(x)g(x)-f(x)M+f(x)M-LM|\\ \amp =|f(x)(g(x)-M)+(f(x)-L)M|\\ \amp \leq |f(x)(g(x)-M)|+|(f(x)-L)M|\\ \amp =|f(x)||g(x)-M|+|f(x)-L||M|\text{.} \end{align*}

This is all straightforward except perhaps for the “$\leq$”. That is an example of the triangle inequality, which says that if $a$ and $b$ are any real numbers then $|a+b|\leq|a|+|b|\text{.}$ If you look at a few examples, using positive and negative numbers in various combinations for $a$ and $b\text{,}$ you should quickly understand why this is true. We will not prove it formally.

Suppose $\epsilon>0\text{.}$ Since $\ds\lim_{x\to a} f(x)=L\text{,}$ there is a value $\delta_1$ such that $0\lt |x-a|\lt \delta_1$ implies $\ds|f(x)-L|\lt \frac{\epsilon}{2(1+|M|)}\text{.}$ This means that $0\lt |x-a|\lt \delta_1$ implies $|f(x)-L||M|\lt |f(x)-L|(1+|M|)\lt \epsilon/2\text{.}$

Now we focus our attention on the other term in the inequality, $|f(x)||g(x)-M|\text{.}$ We can make $|g(x)-M|$ smaller than any fixed number by making $x$ close enough to $a\text{;}$ unfortunately, $\epsilon/(2f(x))$ is not a fixed number, since $x$ is a variable. Here we need another little trick, just like the one we used in analyzing $x^2\text{.}$ We can find a $\delta_2$ so that $|x-a|\lt \delta_2$ implies that $|f(x)-L|\lt 1\text{,}$ meaning that $L-1 \lt f(x) \lt L+1\text{.}$ This means that $|f(x)|\lt N\text{,}$ where $N$ is either $|L-1|$ or $|L+1|\text{,}$ depending on whether $L$ is negative or positive. The important point is that $N$ doesn't depend on $x\text{.}$ Finally, we know that there is a $\delta_3$ so that $0\lt |x-a|\lt \delta_3$ implies $|g(x)-M|\lt \epsilon/(2N)\text{.}$ Let $\delta$ be the smallest of $\delta_1\text{,}$ $\delta_2\text{,}$ and $\delta_3\text{.}$ Then $|x-a|\lt \delta$ implies that $|f(x)-L|\lt \epsilon/(2(1+|M|))\text{,}$ $|f(x)|\lt N\text{,}$ and $|g(x)-M|\lt \epsilon/(2N)\text{.}$ Then

\begin{align*} |f(x)g(x)-LM|\amp \leq |f(x)||g(x)-M|+|f(x)-L||M|\\ \amp \lt N\frac{\epsilon}{2N}+\frac{\epsilon}{2(1+|M|)}(1+|M|)\\ \amp =\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\text{.} \end{align*}

This is just what we needed, so by the official definition, $\ds \lim_{x\to a}f(x)g(x)=LM\text{.}$

The concept of a one-sided limit can also be made precise.

###### Definition3.8. One-sided Limit.

Suppose that $f(x)$ is a function. We say that $\ds \lim_{x\to a^-}f(x)=L$ if for every $\epsilon>0$ there is a $\delta > 0$ so that whenever $0 \lt a-x \lt \delta\text{,}$ $|f(x)-L|\lt \epsilon\text{.}$ We say that $\ds\lim_{x\to a^+}f(x)=L$ if for every $\epsilon>0$ there is a $\delta > 0$ so that whenever $0 \lt x-a \lt \delta\text{,}$ $|f(x)-L|\lt \epsilon\text{.}$

##### Exercises for Section 3.2.

Give an $\epsilon$–$\delta$ proof of the fact that $\ds \lim_{x\to 4} (2x-5) = 3\text{.}$

Solution

Suppose $\epsilon > 0\text{,}$ and let $\delta = \epsilon/2\text{.}$ Then $\delta > 0$ and for all $0 \lt |x-4| \lt \delta\text{,}$ we see that

\begin{align*} \amp \amp |x-4| \lt \epsilon/2\\ \amp \Rightarrow \amp |2x-8| \lt \epsilon\\ \amp \Rightarrow \amp |(2x-5)-3| \lt \epsilon \end{align*}

Therefore, $\lim\limits_{x \to 4}(2x-5) = 3\text{.}$

Let $\epsilon$ be a small positive real number. How close to 2 must we hold $x$ in order to be sure that $3x+1$ lies within $\epsilon$ units of 7?

Solution

Let $\epsilon > 0$ and suppose

\begin{equation*} |3x+1 -7| \lt \epsilon \end{equation*}

That is, $|3x-6| \lt \epsilon\text{.}$

\begin{align*} \amp \amp |3x-6| \lt \epsilon\\ \amp \Rightarrow \amp 3|x-2| \lt \epsilon\\ \amp \Rightarrow \amp |x-2| \lt \epsilon/3 \end{align*}

Therefore, we require $x$ to be within $\epsilon/3$ units of $2\text{.}$