$$$$

17  Limits

Highlights of this Chapter: we introduce the notion of a limit of a function, as well as the limit from above and the limit from below. We prove that a limit exists if and only if these right and left hand limits both exist, and are equal - a fact which will prove useful in various calculations with derivatives to come.

Sometimes we need to understand the behavior of a function near a point, without actually being able to compute the function’s value at that point (perhaps, that point is outside the functions’ domain). To do so, we use sequences once again to help us out!

First, one quick definition to make terminology easier: :::{#def-limit-point} Let $D\subset \mathbb{R}$ be a set. Then a point $p\in \mathbb{R}$ is a limit point of $D$ if there is at least one sequence of points in $D$ converging to $a$. :::

For example, $0$ is a limit point of $(0,1)$ even though it is not a point of $(0,1)$. Any point in a set is trivially a limit point of that set (just take the constant sequence equal to that point over and over).

Definition 17.1 (Limits of Functions) Let $f$ be a function defined defined on a domain $D\subset \mathbb{R}$ and let $a$ be a limit point of $D$. Then we write $$\underset{x\to a}{lim}f(x)=L$$ to mean for every sequence $\{x_n\}\subset D$ with $x_n\to a$ and $x_n\ne a$, we have $$limf(x_n)=L$$

Note, this definition looks a lot like the definition of continuity, except that we are not able to say “approaches $f(a)$” as we are not interested in what $f$ is doing at $a$ (or even if $f$ is defined at $a$), but rather only on what is happening nearby.

Let’s do some quick examples to get a feel for this definition:

Example 17.1 $$\underset{x\to 2}{lim}3x^2+4$$

Let $x_n$ be any sequence converging to $2$, for which $x_n\ne 2$ for all $n$. Then by the limit theorems we see $x_n^2\to 4$, so $3x_n^2\to 12$, and $3x_n^2+4\to 16$. Since $x_n$ was arbitrary, this holds for all such sequences, thus

$$\underset{x\to 2}{lim}3x^2+4=16$$

Example 17.2 $$\underset{x\to 2}{lim}\{\begin{array}{ll}{x}^2& x\ne 2\\ 3& x=2\end{array}$$

Let $x_n$ be any sequence converging to $2$, for which $x_n\ne 2$ for all $n$, and $f$ be the piecewise function above. Then $f(x_n)=x_n^2$ for all $n$ as we avoid the case $x=2$, and using the limit theorems $x_n^2\to 4$. Since $x_n$ was arbitrary, $$\underset{x\to 2}{lim}f(x)=4$$ Note this is true even though $f(2)=3$.

This example shows why we do not consider sequences that contain the point $a$: if we looked at sequences converging to 2 that contained infinitely many 2’s above, they either diverge, or converge to 3, whereas all other sequences converge to $4$ as we showed. The next example shows another utility of this.

Example 17.3 $$\underset{x\to 2}{lim}\frac{x^2-4}{x-2}$$

Let $x_n$ be any sequence converging to $2$, for which $x_n\ne 2$ for all $n$. Then since $x_n\ne 2$ the denominator of $(x^2-4)/(x-2)$ is never zero, and we can simplify with algebra: $$\frac{x_n^2-4}{x_n-2}=\frac{(x_n+2)(x_n-2)}{x_n-2}=x_n+2$$

Thus, for all $n$ we have $$lim\frac{x_n^2-4}{x_n-2}=lim{x}_n+2=lim(x_n)+2=4$$

Since $x_n$ was arbitrary, this holds for all sequences and

$$\underset{x\to 2}{lim}\frac{x^2-4}{x-2}=4$$

Example 17.4 $$f(x)=\{\begin{array}{ll}0& x<0\\ 17& x=0\\ x& x>0\end{array}$$ Then $\underset{x\to 0}{lim}f(x)=0$

Example 17.5 $$f(x)=\{\begin{array}{ll}0& x<0\\ 17& x=0\\ x^2+1& x>0\end{array}$$ Then $\underset{x\to 0}{lim}f(x)$ does not exist.

Here’s the familiar theorem from calculus that you can ‘plug in’ when taking limits of continuous functions.

Theorem 17.1 (Limit of Continuous Functions) If $f$ is continuous at $a$ then $$\underset{x\to a}{lim}f(x)=f(a)$

Proof. Let $f$ be continuous at $a$ and $x_n\to a$ be an arbitrary sequence with $x_n\ne a$ for all $n$. By the assumption of continuity, we know that for all sequences converging to $a$, $$limf(x_n)=f(lim{x}_n)=f(a)$$ Since our $x_n$ is such a sequence (just with the extra condition that $x_n\ne a$) the same holds. And, as $x_n$ was an arbitrary such sequence $$\underset{x\to a}{lim}f(x)=f(a)$$

17.0.1 Epsilons and Deltas

Just like for continuity - whatever we can do with arbitrary sequences we can also do with $ϵ$s and $\delta$s. Here’s an alternative conception of functional limits:

Definition 17.2 (Function Limits and $ϵ-\delta$) Let $f$ be a function with domain $D$ and $a$ a limit point of $D$. Then we write $\underset{x\to a}{lim}f(x)=L$ if for all $ϵ>0$ there is a $\delta >0$ such that for all $x\in D,x\ne a$ with $|x-a|<\delta$, we have $|f(x)-f(a)|<ϵ$.

Exercise 17.1 Reprove the above theorem using the $ϵ-\delta$ definition of limit, and the $ϵ-\delta$ definition of continuity.

Exercise 17.2 (Equivalence of Limit Definitions) Prove that the $ϵ-\delta$ definition of functional limits is equivalent to the sequence definition (using the same ideas we used to prove the analogous definitions of continuity are equivalent).

17.1 Limits from Above and Below

In some cases, we want to consider a more restricted notion of limit: not one that considers all sequences but rather one that only notices sequences larger than, or smaller than the target value. It is easy to modify the sequence definition for

Definition 17.3 (Limit From Above) Let $a$ be a limit point of the domain of a function $f$. Then we write $$\underset{x\to a^{+}}{lim}f(x)=L$$ and say “The limit from above is $L$” if for all sequences $x_n$ in the domain with $x_n\to a$ and $x_n>a$, we have $limf(x_n)=L$.

Definition 17.4 (Limit From Below) Let $a$ be a limit point of the domain of a function $f$. Then we write $$\underset{x\to a^{-}}{lim}f(x)=L$$ and say “The limit from below is $L$” if for all sequences $x_n$ in the domain with $x_n\to a$ and $x_n<a$, we have $limf(x_n)=L$.

Exercise 17.3 Give an $ϵ-\delta$ definition of limits from above, and limits from below.

One sided limits are very useful as they can be easier to compute, but their values completely determine the value of the overall limit. In particular:

Exercise 17.4 Let $f$ be a function and $a$ a limit point of its domain. Then if $$\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)$$ The overall limit $\underset{x\to a}{lim}f(x)$ exists, and has the same value as the two one sided limits.

*Hint: take an arbitrary sequence $x_n\to a$ with $x_n\ne a$, and decompose it into a union of two subsequences, the “right subsequence” $r_n$ of terms $>a$ and the “left subsequence” ${\ell }_n$ of terms $<a$. Use what you know about limits of sequences and subsequences!

The converse of this is immediate, as the sequences ranged over in the definitions of $\underset{x\to a^{+}}{lim}$ and $\underset{x\to a^{-}}{lim}$ are just particular cases of the sequences ranged over in the definition of $\underset{x\to a}{lim}$. Together, these yield the following theorem:

Theorem 17.2 Let $f$ be a function and $a$ a point of its domain. Then $\underset{x\to a}{lim}f(x)$ exists if and only if both one sided limits exist and are equal. And, in this case $$\underset{x\to a}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)=\underset{x\to a^{-}}{lim}f(x)$$

We can generalize this to formulate a condition on when gluing two continuous functions together remains continuous. While elementary, this result proves extremely useful in analysis and topology, and is called the pasting lemma as it allows you to paste continuous functions together.

Exercise 17.5 (The Pasting Lemma) Let $f,g$ be two continuous functions and $a\in \mathbb{R}$ is a point such that $f(a)=g(a)$. Prove that the piecewise function below is continuous at $a$. $$h(x)=\{\begin{array}{ll}f(x)& x\le a\\ g(x)& x>a\end{array}$$

When do one-sided limits exist at all? One useful assurance of existence is monotonicity:

Exercise 17.6 Let $f$ be a bounded monotone function on the interval $(a,b)$. Then both of the one sided limits exist $$\underset{x\to a^{+}}{lim}f(x)\underset{x\to b^{-}}{lim}f(x)$$ Hint: show they are the inf and sup of $\{f(x)\mid x\in (a,b)\}$

This proves useful in many cases where we know only that our function is monotone, but cannot compute its values. For us, the most important application is Proposition 23.2 where we show exponential functions are differentiable, when we have only assumed they are continuous.

One result that will be useful later on is describing continuity with function limits. These definitions are very similar, except that $\underset{x\to a}{lim}f(x)$ only considers sequences with $x_n\ne a$, whereas the definition of continuity at $a$ requires we consider all sequences that converge to $a$. This does not pose any serious issue, as the only difference is terms literally equal to $a$!

Theorem 17.3 $f$ is continuous at $a$ if and only if $\underset{x\to a}{lim}f(x)=f(a)$.