06 Limits of the form $\ds\lim_{x\to a}f(x)$

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Sections 2.2 and 2.3 cover the idea of limit of a function at a point, using the so-called "epsilon-delta" definition of limit, which is named after the Greek letters that are used by convention in the definition. (But it's important to remember that when the definition is applied, the corresponding quantities are not always named epsilon and delta.) You might have already seen this definition. In this course, we are less interested in using the definition to prove that a given limit has a certain value and more interested in giving rigorous proofs of theorems about limits. Here, as is usually true in our textbook, "function" means "real-valued function of a single real variable."

(My definition here leaves out the condition in the book's definition that $f$ be defined on some open interval containing $a$ except possibly at $a$ itself. Instead, it says that $f$ is defined for all $x$ satisfying $0<|x-a|<\delta$, which is the same thing.)

Let's apply this definition to prove that $\ds\lim_{x\to 3}x^2=9.$ I want to explore in some detail how one might think about this problem. Here, $f(x)=x^2$, $a=3$, and $L=9.$ The function $f(x)=x^2$ is defined everywhere, so the condition that the function is defined is satisfied everywhere. (Proofs will sometimes not explicitly mention checking this condition, if it seems obvious, but it is an important part of the definition.) We must begin with an arbitrary positive number $\eps.$ We must find some positive number $\delta$ that satisfies: for any $x$, if $0<|x-3|<\delta$, then $|x^2-9|<\eps.$ Note that our proof has to work for any $\eps.$ We have picked some specific but arbitrary $\eps$ to work with. We don't know anything about it except that it is greater than zero. We have to produce a specific $\delta$ that will work for that specific $\eps$, but because $\eps$ is arbitrary, our value for $\delta$ will depend on the specific $\eps$ that we are given. This is a roundabout way of saying that we actually want to define $\delta$ as a function of $\eps.$

So, our goal is to force the quantity $|x^2-9|$ to be small, and we have to do that by making $|x-3|$ small. We need to figure out what $|x-3|$ might have to do with $|x^2-9|$. Hopefully, we notice that $$|x^2-9|=|(x-3)(x+3)|=|x-3|\cdot|x+3|$$ We can make $|x-3|$ as small as we like, but what to do about the $|x+3|$? Notice that we don't have to make $|x+3|$ small, since we can make $|x-3|$ as small as we like. We just have to make sure that $|x+3|$ is not too big. That is, we want to bound $|x+3|$. In fact, since we are going to make $x$ close to 3, it should follow that $x+3$ is close to $6$. Specifically, if we make sure that $|x-3|\le 1$, then $2\le x\le 4$ and $5\le x+3\le 7$. In particular, as long as $|x-3|\le 1$, we will have $|x+3|\le7$. Then to make the product $|x-3|\cdot|x+3|$ be less than $\eps,$ we only have to make $|x-3|<\frac{\eps}{7}$. That is, as long as $|x-3|$ is less than 1 and also less than $\frac{\eps}{7}$, we will have $|x^2-9|=|x-3|\cdot|x+3|<\frac{\eps}{7}\cdot 7=\eps$. So, if we choose our $\delta$ to be the function of $\eps$ given by $\delta=\min\big(1,\frac\eps7\big)$, then $|x-3|<\delta$ will imply $|x^2-9|<\eps,$ which is just what we needed to show. Let's put this into a formal proof—which should not include any of the exploratory thinking that we have just worked through:

Proof that $\ds\lim_{x\to3}x^2=9\colon$  Let $\eps>0$. We want to find $\delta>0$ such that for any $x$, if $|x-3|<\delta$, then $|x^2-9|<\eps$. Define $\delta=\min\big(1,\frac\eps7\big).$ Suppose $x$ satisfies $|x-3|<\delta$. Since $|x-3|\le 1$, we have $2\le x\le 4$ and $5\le x+3\le 7$, and it follows that $|x+3|\le 7.$ Therefore, $$\begin{align*} |x^2-9| & = |(x-3)(x+3)|\\ & = |x-3|\cdot|x+3|\\ & < \textstyle\frac\eps 7\cdot 7\\ & = \eps \end{align*}$$ as we wanted to show. $\qed$

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Let's apply similar reasoning to the proof of the sum and product rules for limits: If $\ds\lim_{x\to a}f(x)=L$ and $\ds\lim_{x\to a}g(x)=M$, then $\ds\lim_{x\to a}\big(f(x)+g(x)\big)=L+M,$ and $\ds\lim_{x\to a}\big(f(x)g(x)\big)=LM.$ The two limits that we are given in the hypothesis mean that we can force $|f(x)-L|$ and $|g(x)-M|$ to be as small as we like by making $|x-a|$ small enough (but non-zero, since knowing the two limits does not tell us anything about what happens to $f(x)$ and $g(x)$ when $x$ is exactly equal to $a$.)

For the limit of the sum $f(x)+g(x),$ we need to force $|(f(x)+g(x))-(L+M)|<\eps$, using our control over $|f(x)-L|$ and $|g(x)-M|$. Here, we should notice that $$|(f(x)+g(x))-(L+M)|=|(f(x)-L)+(g(x)-M)|$$ and the triangle inequality allows us to relate this back to $|f(x)-L|$ and $|g(x)-M|\colon$ $$|(f(x)-L)+(g(x)-M)|\le |f(x)-L| +|g(x)-M|$$ Since we can force both $|f(x)-L|$ and $|g(x)-M|$ to be small by choosing $|x-a|$ small enough, we are good! But we need to do it formally. Given any $\eps>0$, we can force $|f(x)-L|<\frac\eps2$ by making $0<|x-a|<\delta_1$ for some $\delta_1>0$. And we can force $|g(x)-M|<\frac\eps2$ by making $0<|x-a|<\delta_2$ for some $\delta_2>0$. We can force both conditions by making $0<|x-a|<\min(\delta_1,\delta_2).$ So, we choose $\delta=\min(\delta_1,\delta_2).$ Then for any $x$ satisfying $0<|x-a|<\delta$, we have $$\begin{align*} |(f(x)+g(x))-(L+M)| & = |(f(x)-L)+(g(x)-M)|\\ &\le |f(x)-L| +|g(x)-M|\\ &<\textstyle \frac\eps2+\frac\eps2\\ &=\eps \end{align*}$$

For the product $f(x)g(x)$,we need to force $|f(x)g(x)-LM|$ to be small. The proof is probably not something that you would come up with quickly on your own, but it uses the same general ideas that we have been discussing. Again, we want to relate the quantity $|f(x)g(x)-LM|$ back to $|f(x)-L|$ and $|g(x)-M|.$ Note that if we had $f(x)g(x)-Lg(x),$ we could factor out the $g(x)$ to get $(f(x)-L)M$, and we see the $f(x)-L$ that we are looking for. There is no $Lg(x)$ in $|f(x)g(x)-LM|$, but we can use the trick of adding and subtracting the same quantity to introduce $Lg(x)$ into the formula, and things work out nicely using the triangle inequality: $$\begin{align*} |f(x)g(x)-LM| & = |(f(x)g(x)-Lg(x)) + (Lg(x)-LM)| \\ & \le |f(x)g(x)-Lg(x)| + |Lg(x)-LM| \\ & = |(f(x)-L)g(x)| + |L(g(x)-M)| \\ & = |f(x)-L|\cdot|g(x)| + |L|\cdot|g(x)-M| \end{align*}$$ We can force $|f(x)-L|$ and $|g(x)-M|$ to be as small as we like. As long as we can also put some limit on $|g(x)|$, the quantity as a whole will be small. But $g(x)\to M$ as $x\to a$, which means that if $x$ is close to $a$, then $|g(x)|$ will not be too much bigger than $|M|$.

So, given $\eps>0$, using the fact that $\ds\lim_{x\to a}g(x)=M$, choose $\delta_1>0$, such that $0<|x-a|<\delta_1$ implies $|g(x)-M|<1$, which in turn implies $|g(x)|<|M|+1.$ (The last inequality follows from the last theorem in the fourth reading guide.) Choose $\delta_2>0$ such that $0<|x-a|<\delta_2$ implies $|f(x)-L|<\frac{\eps}{2(|M|+1)}.$ And choose $\delta_3>0$ such that $0<|x-a|<\delta_3$ implies $|g(x)-M|<\frac{\eps}{2(|L|+1)}$. (The denominator uses $|L|+1$ rather than $|L|$ to avoid the problem of division by zero in the case where $L$ is zero.) Finally, let $\delta=\min(\delta_1,\delta_2,\delta_3)$. Then, for any $x$ that satisfies $0<|x-a|<\delta$, we have

$$\begin{align*} |f(x)g(x)-LM| & \le |f(x)-L|\cdot|g(x)| + |L|\cdot|g(x)-M|\\ & < \textstyle \frac{\eps}{2(|M|+1)}\cdot(|M|+1) + |L|\cdot\frac{\eps}{2(|L|+1)}\\ & = \textstyle \frac\eps2 + \frac{|L|}{|L|+1}\cdot\frac\eps2\\ & < \textstyle \frac\eps2 + \frac\eps2\\ & = \eps \end{align*}$$

Note that the strange denominators in $\frac{\eps}{2(|M|+1)}$ and $\frac{\eps}{2(|L|+1)}$ were chosen so that we could end up with $\eps$ at the end of the preceding calculation. Of course, I had to look ahead to see what was needed to make that happen.

After all that, you should probably carefully reread the proofs from the textbook and make sure that you understand how the proofs of the limit theorems work.

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Occasionally, you might need to prove that a limit does not exist. That is, for some function $f$ and some $a\in\R$, $\ds \lim_{x\to a}f(x)$ is not $L$ for any number $L$. The definition of $\ds\lim_{x\to a}f(x)=L$ can be expressed symbolically as $$\forall \eps>0, \exists \delta>0, \forall x\in\R, \big(0<|x-a|<\delta \Rightarrow |f(x)-L|<\eps\big)$$ The negation of this statement is $\ds\lim_{x\to a}f(x)\ne L,$ which is expressed as $$\exists \eps>0, \forall \delta>0, \exists x\in\R, \big(0<|x-a|<\delta \mbox{ and } |f(x)-L|\ge\eps\big)$$ To say that $\ds \lim_{x\to a}f(x)$ does not exist is to say $\forall L\in\R$, $\ds\lim_{x\to a}f(x)\ne L.$ This adds another quantifier to the previous statement. So, $\ds\lim_{x\to a}f(x)$ does not exist if and only if $$\forall L\in\R, \exists \eps>0, \forall \delta>0, \exists x\in\R, \big(0<|x-a|<\delta \mbox{ and } |f(x)-L|\ge\eps\big)$$ This statement is quite complicated logically, and it is not surprising that it can be difficult to keep things straight when proving that a limit does not exist.

However, the basic idea is that $\ds\lim_{x\to a}f(x)\ne L$ if there are points, $x,$ arbitrarily close to $a$ for which the value of $f(x)$ is bounded away from $L$ by some positive distance, $\eps.$ For a simple example, $\ds\lim_{x\to 1}x^2 \ne 3$ because there are values of $x$ arbitrarily close to 1 such that the distance of $x^2$ from 3 is greater than 1. (In fact, of course, in this case, for all $x$ close to 1, the distance $|x^2-3|$ is greater than 1 — which is a stronger condition that we need.)

So, let's show that ${\ds\lim_{x\to0}}\sin\big(\frac1x\big)\ne 0.$ Now, for $x=\frac{1}{\pi},\frac{1}{2\pi},\frac{1}{3\pi},\frac{1}{4\pi},\dots,$ $\sin\big(\frac1x\big)=0$, so there are points $x$ arbitrarily close to 0 for which $\sin\big(\frac1x\big)$ is very close (in fact equal to) 0. However, if we look at the points $x=\frac{2}{\pi},\frac{2}{3\pi},\frac{2}{5\pi},\frac{2}{7\pi},\dots,$ we have $\sin\big(\frac1x\big)=1.$ So we also get points $x$ arbitrarily close to 0 for which $\sin\big(\frac1x\big)$ is bounded away from 0. In particular, letting $\eps=\frac12$, we see that for any $\delta>0$, we can find a point $x$ such that $0<|x-0|<\delta$ and $\big|\sin\big(\frac1x\big)-0\big|>\eps$ (namely, choose an odd natural number $n$ so that $\frac{2}{n\pi}<\delta$, and let $x=\frac{2}{n\pi}$). Convince yourself that this is exactly what we need to satisfy the definition given above for $\ds\lim_{x\to a}f(x)\ne L.$

Of course, to show that ${\ds\lim_{x\to0}}\sin\big(\frac1x\big)$ does not exist, we need to show that for any number $L$, ${\ds\lim_{x\to0}}\sin\big(\frac1x\big)\ne L.$ Here is a more formal proof of that fact: Let $L$ be any number. Let $\eps=\frac12.$ We want to show that for any $\delta>0$, there is a number $x$ such that $0<|x-0|<\delta$ but $\big|\sin\big(\frac1x\big)-0\big|>\eps.$ Let $\delta$ be any positive number. In the case $L<0$, let $n$ be an odd natural number such that $\frac{2}{n\pi}<\delta$, which exists by the Archimedian property of $\R,$ and let $x=\frac{2}{n\pi}$. Then $0<|x-0|<\delta$ and $\sin\big(\frac1x\big)=1$. Since $L<0$ and $\sin\big(\frac1x\big)=1$, we see that $\big|\sin\big(\frac1x\big)-0\big|>1>\frac12=\eps,$ as we wanted to show. In the case $L\ge 0$, let $n$ be an odd natural number such that $\frac{2}{n\pi}<\delta$, and let $x=-\frac{2}{n\pi}$. Then $0<|x-0|<\delta$ and $\sin\big(\frac1x\big)=-1$. Since $L\ge0$ and $\sin\big(\frac1x\big)=-1$, we see that $\big|\sin\big(\frac1x\big)-0\big|\ge1>\frac12=\eps,$ so that we are done in this case as well. $\qed$

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