1.2: Limits and Continuity

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1.2: Limits and Continuity

  1. Limits of multivariable functions
  2. Continuity
  3. Continuous functions and open sets
  4. Problems

\(\Leftarrow\)  \(\Uparrow\)  \(\Rightarrow\)

After this chapter, you will:

Limits of multivariable functions

Assume that \(S\subseteq \R^n\), and that \(\mathbf f:S\to \R^k\) is a function. The statement \[ \lim_{\bf x\to a}{\bf f}({\bf x}) ={\bf L} \] is defined to mean that \[\begin{equation}\label{lim.def} \forall \varepsilon >0, \ \ \exists \delta>0 \quad\mbox{ such that if } \mathbf x \in S \mbox{ and } 0 < |\mathbf x - \mathbf a|<\delta, \mbox{ then } |\mathbf f(\mathbf x) - {\bf L}| < \varepsilon. \end{equation}\]
In order for the definition to make sense, we need to assume that \(\mathbf a\) is a limit point of \(S\):
The point \(\mathbf a \in \R^n\) is a limit point of the set \(S\) if and only if \[\begin{equation}\label{limitpoint} \forall \delta>0, \quad \exists \mathbf x\in S \quad\mbox{ such that }\quad 0 < |\mathbf x - \mathbf a|<\delta. \end{equation}\]

That is, no matter how close you zoom in to \(\mathbf a\), you can always see a point of \(S\) nearby, and the point is distinct from \(\mathbf a\). For example, this always holds if \(\mathbf a \in S^{int}\), or even if \(\mathbf a \in \overline{(S^{int})} =\) the closure of \(S^{int}\).

Definition \(\eqref{lim.def}\) is identical to the familiar definition from single-variable calculus, except that

Despite these cosmetic differences, the mathematical structure of \(\eqref{lim.def}\) is exactly the same as in the familiar case of \(f:\R\to \R\). As a result, limits of functions \(\mathbf f:\R^n\to \R^k\) have very similar properties to limits of functions \(f:\R\to \R\), with very similar proofs. This means that you already understand most aspects of limits of multivariable functions. Congratulations!

For example,

Assume that \(S\subseteq \R^n\) and that \(\mathbf a\) is a point in \(\R^n\) satisfying \(\eqref{limitpoint}\) (for example, an interior point of \(S\)). Further assume that \(f,g:S\to \R\) are functions and \(L, M\) are numbers such that \[ \lim_{\mathbf x \to \mathbf a}f(\mathbf x) = L, \qquad \lim_{\mathbf x \to \mathbf a}g(\mathbf x) = M. \] Then \[ \lim_{\mathbf x \to \mathbf a}[f(\mathbf x)+ g(\mathbf x)] = L+M,\qquad\mbox{ and }\qquad \lim_{\mathbf x \to \mathbf a}[f(\mathbf x) g(\mathbf x)] = LM. \]
Assume that \(S\subseteq \R^n\) and that \(\mathbf a \in \R^n\) is a limit point of \(S\) (for example, an interior point of \(S\)). Further assume that \(f,g,h:S\to \R\) are functions and there exist real numbers \(\delta>0\) and \(L\) such that \[ f(\mathbf x)\le g(\mathbf x) \le h(\mathbf x)\mbox{ for all }\mathbf x\in S \mbox{ such that }0<|\mathbf x-\mathbf a|<\delta \] and \[\lim_{\mathbf x \to \mathbf a}f(\mathbf x) = \lim_{\mathbf x\to \mathbf a}h(\mathbf x) = L. \] Then \(\lim_{\mathbf x\to \mathbf a}g(\mathbf x) = L\).
A small remark We have stated the above theorems for real-valued functions rather than vector-valued functions, because for example with vector-valued functions it is not clear how to generalize the hypothesis \(f\le g\le h\), and we have several choices of how to write a product of functions: a dot product, a product of a scalar and a vector etc, a cross-product (in 3 dimensions), etc.

The proofs of these theorems are exactly like the proofs for functions of a single variable. To have Alfonso Gracia-Saz persuade you of this, consult the MAT137 videos Proof of the Limit Law for sums of functions, or Proof of the Squeeze Theorem. You will see that the proofs given there apply with no change to functions of several variables. You may also find that reviewing these proofs is a helpful reminder of basic concepts.

As with functions of a single variable, the Squeeze Theorem has a useful corollary.

Assume that \(S\subseteq \R^n\) and that \(\mathbf a\) is a point in \(\R^n\) satisfying \(\eqref{limitpoint}\). Further assume that \(g,h:S\to \R\) are functions such that \[ |g(\mathbf x)|\le h(\mathbf x) \quad \mbox{ for all }\mathbf x \in S, \qquad \lim_{\mathbf x\to \mathbf a}h(\mathbf x) = 0. \] Then \(\lim_{\mathbf x\to \mathbf a}g(\mathbf x) = 0\).
Details
The hypotheses imply that \(-h(\mathbf x)\le g(\mathbf x)\le h(\mathbf x)\) for all \(\mathbf x\) and that \(\lim_{\mathbf x\to \mathbf a}-h(\mathbf x) = \lim_{\mathbf x\to \mathbf a}h(\mathbf x)= 0\), so the conclusion follows directly from the Squeeze Theorem, with \(-h(\mathbf x)\) playing the role of \(f(\mathbf x)\).

When using this corollary in proofs, it is common to write “by the Squeeze Theorem” rather than “by the Corollary of the Squeeze Theorem.”

We can use the Limit Laws and Squeeze Theorem to prove the following useful theorem. It tells us that if we have a good understanding of limits of real-valued functions, we can immediately transfer this knowledge to vector-valued functions. For this reason, and because the notation is simpler, we will often focus on real-valued functions.

Assume that \(S\subseteq \R^n\) and that \(\mathbf a\) is a limit point of \(S\). If \(\mathbf f = (f_1,\ldots, f_k)\) is a vector-valued function \(S\to \R^k\), then \[ \lim_{\mathbf x\to \mathbf a}{\bf f({\mathbf x}) = L} \qquad\mbox{ if and only if }\qquad \lim_{\mathbf x\to \mathbf a} f_j({\mathbf x}) = L_j\ \ \ \mbox{ for }j=1,\ldots, k. \] where \((L_1,\ldots, L_k)\) are the components of \({\bf L}\).

Details
First note that the definition \(\eqref{lim.def}\) of limit implies that \[\begin{equation}\label{restate} \lim_{\mathbf x\to\mathbf a}\mathbf f(\mathbf x) = {\bf L} \qquad\iff \qquad \lim_{\mathbf x\to\mathbf a}|\mathbf f(\mathbf x) -{\bf L}| = 0. \end{equation}\] (Write out the details if you are unsure.) Next, for every \(j = 1,\ldots, k\), \[ |f_j(\mathbf x) - L_j| \le \left( |f_1(\mathbf x) - L_1|^2 +\cdots + |f_k(\mathbf x) - L_k|^2\right)^{1/2} = |\mathbf f(\mathbf x)-{\bf L}|. \] Thus by \(\eqref{restate}\) and the Squeeze Theorem, \[\begin{align*} \lim_{\mathbf x\to \mathbf a} \mathbf f(\mathbf x) = {\bf L} \quad \Rightarrow \quad \lim_{\mathbf x\to \mathbf a} |\mathbf f(\mathbf x) - {\bf L}|=0 &\quad \Rightarrow \quad \lim_{\mathbf x\to \mathbf a} |f_j(\mathbf x) - L_j|=0 \\\ &\quad \Rightarrow \quad \lim_{\mathbf x\to \mathbf a} f_j(\mathbf x) = L_j. \end{align*}\] Similarly, \[ |\mathbf f(\mathbf x) - {\bf L}| \le |f_1(\mathbf x) - L_1| + \cdots + |f_k(\mathbf x) - L_k| \] by squaring both sides. Using this, the implication \[ \lim_{\mathbf x\to \mathbf a} f_j(\mathbf x) = L_j\ \ \ \mbox{ for }j=1,\ldots, k \quad\Rightarrow \quad \lim_{\mathbf x\to \mathbf a} \mathbf f(\mathbf x) = {\bf L} \] can be checked using the Limit Laws and the Squeeze Theorem.

Although the abstract theory of limits for multivariable functions is very similar to that for functions of a single variable, interesting examples show ways in which the notion of a limit is more subtle in the multivariable case.

Example 1.

Define \(f:\R^2\setminus \{(0,0)\}\to \R\) by \[ f(x,y) = \frac {xy}{x^2+y^2}. \] Does \(\lim_{(x,y)\to (0,0)} f(x,y)\) exist, and if so, what is it?

Solution For any real number \(m\) it is easy to check that \(f(x,mx) = \dfrac m{1+m^2}\) for all \(x\ne 0\). Thus \[\begin{equation}\label{line} \lim_{x\to 0} f(x,mx) = \frac m{1+m^2}. \end{equation}\] This says, informally, that if we approach the origin along any straight line, the limit depends on the slope of the line! For example, it equals \(1/2\) if \(m=1\) and \(0\) if \(m=0\).

This implies that \(\lim_{(x,y)\to (0,0)} f(x,y)\) does not exist, since \[\begin{equation}\label{fact} \mbox { if }\lim_{(x,y)\to (0,0)}f(x,y) = L, \qquad\mbox{ then }\lim_{x\to 0}f(x,mx)=L \end{equation}\] for every \(m\). (See below). This would imply that \(\frac m{1+m^2} = L\) for every \(m\), and hence that \(1/2= 0\). So the limit cannot exist.

The \((\varepsilon, \delta)\) proof of \(\eqref{fact}\) is a good exercise. In fact, questions like this, but a little harder, have been known to appear on MAT237 Term Tests! Try it before reading the details below.

Details for \(\eqref{fact}\)

Our goal is to show that \[\begin{equation}\label{facta} \forall \varepsilon >0, \exists\delta>0 \mbox{ such that } 0<|x|< \delta \Rightarrow |f(x,mx) - L|<\varepsilon \end{equation}\] Our assumption is that \(\lim_{(x,y)\to (0,0)}f(x,y) = L\). This implies that \[\begin{equation} \label{ffact} \forall \varepsilon>0, \exists \delta_1>0 \mbox{ such that } 0< |(x,y) - (0,0)|<\delta_1 \ \Rightarrow |f(x,y)-L|<\varepsilon. \end{equation}\]

Setting \(y=mx\), we observe that \[ |(x,mx) - (0,0)| = |(x, mx)| = |x|(1+m^2)^{1/2}. \] Thus, if \(|x|< \delta_1(1+m^2)^{-1/2}\), then \(|(x,mx) - (0,0)|<\delta_1\). Putting these together, it follows that for any \(\varepsilon>0\), if we choose \(\delta_1\) as in \(\eqref{ffact}\) and define \(\delta = \delta_1(1+m^2)^{-1/2}\), then \[ 0<|x|<\delta \qquad \Rightarrow \qquad |f(x,mx)-L|<\varepsilon. \] This proves \(\eqref{fact}\).

Example 2.

Define \(f:\R^2\setminus \{(0,0)\}\to \R\) by \[ f(x,y) = \frac {x^2y}{x^4+y^2}. \]Does \(\lim_{(x,y)\to (0,0)} f(x,y)\) exist, and if so, what is it?

Solution For any real number \(m\), \[ \lim_{x\to 0} f(x,mx) = \lim_{x\to 0}\frac{mx^3}{x^4+m^2 x^2} = 0 \] by first-year calculus. Despite this, the limit \(\lim_{(x,y)\to (0,0)}f(x,y)\) does not exist! Indeed, one can easily check that \[ \lim_{x\to 0} f(x,mx^2) = \frac m {1+m^2}, \] so that we see different limits as we approach the origin along different parabolas \(y=mx^2\). Then we can conclude that the limit does not exist by arguing as above. This is another good exercise to write out with \((\varepsilon, \delta)\).

Example 3.

Define \(f:\R^2\setminus \{(0,0)\}\to \R\) by \[ f(x,y) = \frac {|x|^{4/3}|y|^{8/7}}{x^2+y^2}. \] Does \(\lim_{(x,y)\to (0,0)} f(x,y)\) exist? If so, what is it?

Solution One way to answer this is to write \(f(x,y) = f_1(x,y) \, f_2(x,y)\), where

\[ f_1(x,y) = \frac {|xy|}{x^2+y^2}, \text{ and } f_2(x,y) = |x|^{1/3}|y|^{1/7} \]
\(f_1(x,y)\le 1/2\) for all \((x,y)\ne (0,0)\).
Details

Note that \[ 0\le (|x| - |y|)^2 = (x^2+y^2) - 2 |xy|. \] It follows that \(2|xy|\le x^2+y^2\). If we divide both sides of this by \(2(x^2+y^2)\), which we can do if \((x,y)\ne (0,0)\), it says that \(f_1(x,y)\le 1/2\), proving our claim.

Using this lemma, we see that \[\begin{equation}\label{f1f2} 0 \le f(x,y) = f_1(x,y)\, f_2(x,y) \le \frac 12 f_2(x,y)\quad\mbox{ for all }(x,y)\ne(0,0). \end{equation}\] Also, basic properties of continuous functions imply that \(f_2\) is continuous, and we can see that \(f_2(0,0)= 0\). Thus \(\lim_{(x,y)\to (0,0)} f_2(x,y) = 0\). So the Squeeze Theorem and \(\eqref{f1f2}\) imply that \(\lim_{(x,y)\to(0,0)} f(x,y) = 0\).

Example 4.

Define \(f:\R^2\setminus \{(0,0)\}\to \R\) by \[ f(x,y) = \frac {x^{11}y}{x^{23}+y^2}. \]Does \(\lim_{(x,y)\to (0,0)} f(x,y)\) exist? If so, what is it?

Solution Here you can check that, similar to Example 2, \(\lim_{t\to 0} f(ta, tb) = 0\) for every nonzero \((a,b)\in \R^2\). However, \[ f(t^2, t^{23}) = \frac {t^{22}t^{23}}{t^{46} + t^{46}} = \frac 12 t^{-1} \] which diverges as \(t\to 0^+\). This explains why \(\lim_{(x,y)\to (0,0)}f(x,y)\) cannot exist. If we wish to give a proof, we can do so by arguing somewhat as in Example 1 above. You can fill in the details yourself if you wish.

These examples demonstrate that there are many more ways to approach a point \(\mathbf a\in\R^n\) than in the one variable case, leading to more subtle convergence behaviour. Recall that for \(f:\R\to\R\), \[\lim_{x\to a} f(x)=L \iff \lim_{x\to a^+} f(x)=L \text{ and } \lim_{x\to a^-} f(x)=L\] so we can determine the convergence behaviour of \(f\) at \(a\) by considering the limit approaching \(a\) from the left and the limit approaching \(a\) from the right.

For a function \(f:\R^n\to \R\) however, we saw in the examples that we could approach \((0,0)\) along many different paths, for example along the straight line path \((x,mx)\) as \(x\to 0\), or along the parabolic path \((x,mx^2)\) as \(x\to 0\), etc. If \(\lim_{\mathbf x\to\mathbf a} f(\mathbf x)\) exists, then it must be the case that the limit as we approach \(\bf 0\) along any two sufficiently nice paths must be equal. A precise definition of sufficiently nice will be given in the homework. Any continuous paths work, for example.

As a result, a very useful strategy for showing a limit does not exist is to find two continuous paths approaching \(\mathbf a\) along which the limit is different. Note, however, that showing a limit is equal along all straight lines, or along all parabolas, or so on is not a proof that the limit does exist, since we could always just consider another path not of this form approaching \(\mathbf a\). In the above examples, if we had found that the limit as we approach along \((x,mx)\) and \((x,mx^2)\) were all equal, this says nothing about what happens as we approach \((0,0)\) along the path \((1-e^x,\sin(x))\) as \(x\to 0\) for example. To prove a limit does exist we need to appeal to the original definition or simplify it in a way that we can apply a known result, like the Squeeze Theorem.

Continuity

If \(S\subseteq \R^n\), then a function \({\bf f}:S\to \R^k\) is continuous at \(\mathbf a\in S\) if \[\begin{equation}\label{cont.def} \forall \varepsilon >0, \ \ \exists \delta>0 \mbox{ such that if } \mathbf x \in S \mbox{ and } |\mathbf x - \mathbf a|<\delta, \mbox { then } |\mathbf f(\mathbf x) - \mathbf f(\mathbf a)| < \varepsilon. \end{equation}\]

If \(\mathbf f\) is continuous at every point in \(S\), then we say simply that \(\mathbf f\) is continuous.

Remarks

  1. Note that, as with limits, the mathematical structure of this definition is exactly the same as in the case of functions of one variable. This means that you already understand most aspects of continuous functions of several variables, as we will see in more detail below.

  2. If \(\mathbf a\) is a point where it makes sense to talk about \(\displaystyle\lim_{\mathbf x\to \mathbf a}\mathbf f(\mathbf x)\) (that is, a limit point) then by comparing the definitions, we see that \(\mathbf f:S\to \R^k\) is continuous at \(\mathbf a\in S\) if and only if for \(\mathbf x\in S\), \(\displaystyle\lim_{\mathbf x\to \mathbf a} \bf f(x) = f(a)\).

  3. Students should commit the definition of continuous to memory. It is essentially the same as the definition of continuity in MAT137.

Basic properties of continuity

First we summarize some basic facts about continuity. Many are familiar from single-variable calculus.

Assume that \(S\subseteq \R^n\) and that \(\mathbf a\in S\).

  1. If \({\bf f}:S\to \R^k\) is a vector-valued function with components \((f_1,\ldots, f_k)\), then \(\bf f\) is continuous at \(\mathbf a\) if and only if every component function \(f_j\) is continuous at \(\mathbf a\).
  2. If \(\mathbf f, {\bf g}:S\to \R^k\) are continuous at \(\mathbf a\), then the sum \(\mathbf f + {\bf g}\) is continuous at \(\mathbf a\).
  3. If \(f,g :S\to \R\) are continuous at \(\mathbf a\), then the product \(fg\) is continuous at \(\mathbf a\).
    If in addition \(g(\mathbf a)\ne 0\), then the quotient \(f/g\) is continuous at \(\mathbf a\).
  4. A composition of continuous functions is continuous. That is, assume that \(S\subseteq \R^n\) and \(T\subseteq \R^k\), and that \(\mathbf f:S\to \R^k\) and \({\bf g}:T\to \R^{\ell}\) are functions such that \({\bf g}\circ \mathbf f\) is well-defined (i.e., the image of \(\mathbf f\) is a subset of \(T\).) If \(\mathbf f\) is continuous at \(\mathbf a\) and \({\bf g}\) is continous at \(\mathbf f(\mathbf a)\), then \({\bf g}\circ \mathbf f\) is continuous at \(\mathbf a\).
  5. The elementary functions of a single variable (trigonometric functions and their inverses, polynomials, exponential and log) are continuous on their domains.

Details
  1. This is a direct consequence of Theorem 3.
  2. This is a direct consequence of the Limit Law for sums in Theorem 1.
  3. The continuity of the product follows directly from the Limit Law for products in Theorem 1. For the quotient, for a nonzero real number \(x\), let \(h(x)=1/x\). Then \(f(x)/g(x) = f(x) h(g(x))\). We know (see point (5) below) that \(h\) is continuous on its domain, so it follows from point (4) below that \(h\circ g\) is continuous wherever \(g\ne 0\). Now the conclusion follows from the case of products.
  4. Try to prove this yourself.
    Hint The argument is exactly like the proof for functions of a single variable, familiar from MAT137.
  5. We assume that this is known from MAT137.

How to recognize (many) continuous functions

From Theorem 4, we see that any multivariable function is continuous, as long as

Using this, we are able to instantly recognize many functions as continuous. For example,

Remark. Suppose that you are writing a proof, and you want to use reasoning like that given above. How should you write it? For this class, we recommend writing something like “By basic properties of continuity, we can see that the function \(f(x,y) = \log(x-3y+2)\) is continuous on the set \(\{(x,y)\in \R^2 : x-3y+2>0\}\).” Statements like this will be accepted as long as the domain is identified correctly.

Continuous functions and open sets

There is an intimate connection between continuity and open/closed sets, summarized in the following theorem.

Assume that \(\mathbf f\) is a function \(\R^n\to \R^k\). Then the following are equivalent:

  1. \(\mathbf f\) is continuous.
  2. For every open set \(U\subseteq \R^k\), the set \(\mathbf f^{-1}(U)=\{ \mathbf x\in \R^n : \mathbf f(\mathbf x)\in U\}\) is open.
  3. For every closed set \(K\subseteq \R^k\), the set \(\mathbf f^{-1}(K)=\{ \mathbf x\in \R^n : \mathbf f(\mathbf x)\in K\}\) is closed.
Remark about \(\mathbf f^{-1}(U), \mathbf f^{-1}(K)\) Recall that for a function \(\mathbf f:\R^n\to\R^k\), and a subset \(S\subseteq \R^k\), the notation \(\mathbf f^{-1}(S)\) simply means \(\{ \mathbf x \in \R^n : \mathbf f(\mathbf x)\in S\}\), the preimage of \(S\). The function \(\mathbf f\) does not need to be invertible for this to make sense!
Details
  • \((1) \Rightarrow (2)\): (sketch). Assume that \(\mathbf f:\R^n\to \R^k\) is continuous, and that \(U\) is an open subset of \(\R^k\). Let \(V =\{ \mathbf x\in \R^n : \mathbf f(\mathbf x)\in U\}\). Let \(\mathbf a\) be an arbitrary point in \(V\); thus \(\mathbf b = \mathbf f(\mathbf a)\in U\).

According to Theorem 3 in Section 1.1.3, it suffices to verify that there exists \(\varepsilon>0\) such that \(B(\varepsilon, \mathbf a)\subseteq V\). You should be able to do this, by using * the fact that \(\mathbf b\in U\), * the fact that \(U\) is open (and properties of open sets), and * the assumption that \(\mathbf f\) is continuous.

  • \((2) \Rightarrow (1)\): (sketch). Assume that \(\mathbf f:\R^n\to \R^k\) has the property that
    \[\begin{equation}\label{cc0} \mbox{ for every open } U\subseteq \R^k, \mbox{ the set }\{ \mathbf x\in \R^n : \mathbf f(\mathbf x)\in U\} \mbox{ is open}. \end{equation}\] We must show that \(\mathbf f\) is continuous at \(\mathbf a\), for any \(\mathbf a\in \R^n\). So, consider some \(\mathbf a\in \R^n\). Given \(\varepsilon>0\), we must find \(\delta>0\) such that \[\begin{equation}\label{cc1} |\mathbf x - \mathbf a| < \delta \qquad\Rightarrow \qquad |\mathbf f(\mathbf x) - \mathbf f(\mathbf a)|<\varepsilon. \end{equation}\] Let’s write \(\mathbf b = \mathbf f(\mathbf a)\). Then we can rewrite \(\eqref{cc1}\) as \[\begin{equation}\label{cc2} \mathbf x\in B(\delta, \mathbf a) \qquad\Rightarrow \qquad \mathbf f(\mathbf x)\in B(\varepsilon, \mathbf b). \end{equation}\] We can in turn rewrite \(\eqref{cc2}\) as \[\begin{equation}\label{cc3} B(\delta, \mathbf a)\subset\{ \mathbf x\in \R^n : \mathbf f(\mathbf x) \in B(\varepsilon, \mathbf b)\}. \end{equation}\] So our goal is now to find \(\delta\) such that \(\eqref{cc3}\) holds. You should be able to do this, using
    • assumption \(\eqref{cc0}\), using a particular open set \(U\).
    • the fact that \(\mathbf a \in \{ \mathbf x\in \R^n : \mathbf f(\mathbf x) \in B(\varepsilon, \mathbf b)\}\), which is true since \(\mathbf b = \mathbf f(\mathbf a)\).
    • properties of open sets.
  • \((2) \Rightarrow (3)\): (sketch) Assume that (2) holds, and let \(K\subseteq \R^k\) be closed. Then \(K^c\) is open, so

\[ \{ \mathbf x\in \R^n : \mathbf f(\mathbf x) \in K^c\} \quad\mbox{ is open}. \] Thus

\[ \{ \mathbf x\in \R^n : \mathbf f(\mathbf x) \in K^c\}^c \quad\mbox{ is closed}. \] So it suffices to verify that

\[ \{ \mathbf x\in \R^n : \mathbf f(\mathbf x) \in K^c\}^c = \{ \mathbf x\in \R^n : \mathbf f(\mathbf x) \in K\} \] which you can do.

  • \((3) \Rightarrow (2)\): (sketch) This is very much like the proof that \((2) \Rightarrow (3)\). You can work out the details.

Remark. It is very important in the above theorem that the domain of \(\mathbf f\) is all of \(\R^n\) and not a proper subset \(S\subsetneq \R^n\). To understand what can go wrong with a proper subset of \(\R^n\), consider \(f:[0,\infty)\to \R\) defined by \(f(x)=\sqrt{x}\). This is continuous. However, the preimage of the open set \((-1,1)\) is \(f^{-1}((-1,1)) = [0,1)\) which is not open in \(\R\).

In order to find a true statement of this type for functions \(\mathbf f:S\to \R^k\), we could

With a new definition of open in \(S\) coming from defining an open ball in \(S\) to be a set of the form \(S\cap B(r,\mathbf a)\), it is true that \(\mathbf f^{-1}(U)\) is open in \(S\) for \(U\) open and similarly \(\mathbf f^{-1}(K)\) is closed in S for \(K\) closed.

How to recognize many closed and open sets

For us, the most important consequence of the theorem is “\((1)\Rightarrow (2),(3)\)”, since this allows us to instantly recognize many sets as open or closed.

Examples. Consider the sets

These sets all have the form

\[ S_j = \{ \mathbf x \in \R^3 : f(\mathbf x)\in T_j\},\quad\mbox{ for }f(x,y,z) = \arctan(z^3 e^{x \sin y})\ \ \ \]

where

Since we recognize \(f\) as continuous, we conclude that

\[ S_1 \mbox{ is open}, \ \ S_2, S_3 \mbox{ are closed}. \]

The general principle is:

You are free to use both of these at any time to show that a set is open or closed.

Problems

Basic

Determine whether the following limits exist. Explain your answer.

What do we mean by “explain”? In this class, “Explain your answer” means that you are not necessarily being asked for a full mathematical proof, but you are being asked to communicate how you know your answer is correct. For example, for a limit that does not exist, a typical explanation might be something like “This limit does not exist, because \(\displaystyle\lim_{x\to 0} f(x, 2x) = c\), and \(\displaystyle\lim_{x\to 0} f(x, 3x) = d\).” For a limit that does exist, an explanation might involve using the Squeeze Theorem.

Of course, when you are asked to explain something, you are welcome to give a full \((\varepsilon, \delta)\) proof, and sometimes that is the best explanation.

  1. \(\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{x^2y^3}{x^2+y^4}\).

  2. \(\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{|x|^{\pi}|y|^{e}}{x^6+y^6}\).

  3. \(\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{xy }{x^4+y^4}\).

  4. \(\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{x^3y^3 }{x^4+y^4}\).

  5. \(\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{xy + x^3y^3 }{x^4+y^4}\).

  6. \(\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{ x^2+7xy^2 - 12 x^3y +4x^{11} - \frac 37 xy^{27} +y^2}{x^2+y^2 }\).

  7. \(\displaystyle\lim_{(x,y)\to (0,0)} \dfrac {\sin^2( x\sqrt{|y|} )}{x^2+y^2}\).

  8. \(\displaystyle\lim_{\mathbf x \to {\bf 0}} \left(\dfrac{x_1}{|\mathbf x|}\right) \quad\) in \(\R^n\), \(n>1\).

Recognizing continuous functions and open/closed sets

Use basic properties of continuity to answer questions like the ones below. These may be one step in the solution of a more complicated problem, so it is important that you are comfortable with them.

  1. Is the function \(f(x,y,z) = x^2 e^{y z \cos\left(\frac{xy}{1+x^2}\right)}\) continuous?
    Answer Yes, by basic properties of continuous functions, this is continuous. You can use an acknowledgement like this when you want to indicate that such a function is continuous. It lets the reader know that you know that you need continuity, and that you know how to derive it.
  2. Is the function \(f(x_1, x_2, x_3, x_4) = x_1 x_3^2 \cos( x_2 \log (3 + \sin(x_1 x_4^3)))-4x_3\) continuous?
    Answer Yes, by basic properties of continuous functions, this is continuous.
  3. Is the set \(\{(x,y,z)\in \R^3: -3< x^2 e^{y z \cos\left(\frac{xy}{1+x^2}\right) } < 2\}\) open?
    Answer Yes, since the function is continuous, and the set is defined by strict inequalities.
  4. Is the set \(\{(x_1,x_2,x_3,x_4)\in \R^4: x_1 x_3^2 \cos( x_2 \log (3 + \sin(x_1 x_4^3))) \le 4 x_3 \}\) closed?
    Answer Yes, since the function is continuous, and the set is defined by non-strict inequalities.

Advanced

  1. Let \(f(x,y) = \frac {y^3 - x^8y}{x^6+y^2}\) for \((x,y)\ne (0,0)\). Does \(\lim_{(x,y)\to (0,0)} \frac{ f(x,y) - y}{ \sqrt{x^2+y^2}}\) exist? Explain your answer. Note that this involves the same kind of reasoning as in some of the questions above, but it is considerably more complicated than most of them.

  2. Suppose that \(f,g\) are functions \(\R^n\to \R\). If \(f\) is continuous and \(g\) is discontinuous at a point \(\mathbf a\in \R^n\), which of the following statements is true? Justify your answer with a proof if you believe it is always continuous/discontinuous, or examples of each if you believe the third option is true.

    • The function \(f+g\) must be discontinuous at \(\mathbf a\).
    • The function \(f+g\) must be continuous at \(\mathbf a\).
    • We cannot determine if \(f+g\) is continuous at \(\mathbf a\).

  3. Same question, but assume instead that both \(f\) and \(g\) are discontinuous at \(\mathbf a\).

  4. Same questions as 14 and 15, but consider \(fg\) instead of \(f+g\).

  5. Prove that if \(f:\R^n\to \R\) is continuous, then \(\{\mathbf x\in \R^n : f(\mathbf x)>0\}\) is open. This is a special case of part of Theorem 6, but it is good practice to try to prove it “from scratch”, without using the theorem or consulting its proof.

  6. Given \(f:S\to \R\), in order for the definition of limit of \(f\) at \(\mathbf a\) to make sense, we needed to assume that \(\mathbf a\) satisfies \[ \forall \delta>0, \quad \exists \mathbf x\in S \quad\mbox{ such that }\quad 0 < |\mathbf x - \mathbf a|<\delta. \] For the following sets, find all points where this condition holds, and all points where it does not. That is, find the limit points of the following sets:

    • \(S = \{ (x,y)\in \R^2 : x^2 + y^2 < 1 \} \cup \{(0,2)\}\)
    • \(S = \{ (x,y)\in \R^2 : x^2 + y^2 < 1 \} \cup \{(0,1)\}\)
    • \(S = \{ (0, 2^{-n})\in \R^2 : n\in \mathbb N \}\)
    • \(S = \{ (0, 2^{-n})\in \R^2 : n\in \mathbb N \} \cup \{(0,0)\}\)

Note that we do not need to limit our attention to \(\mathbf a\in S\).

  1. Define \(f:\R\to \R\) by \[ f(x) = \begin{cases}x&\mbox{ if }x\in {\mathbb Q} \\\ 0&\mbox{ if not}. \end{cases} \] At which points, if any, is \(f\) continuous? Prove that your answer is correct.

  2. Construct a function \(f:\R\to \R\) that is continuous at \(x=-1\) and \(x=1\) and discontinuous everywhere else.

  3. Construct a function \(f:\R\to \R\) that is discontinuous at \(x=-1\) and \(x=1\) and continuous everywhere else.

  4. Assume that \(\mathbf f:\R^n\to \R^k\) and \({\bf g}:\R^k\to \R^{\ell}\) are functions, and that \(\mathbf a\in \R^n\) is a point such that \(\mathbf f\) is continuous at \(\mathbf a\) and \({\bf g}\) is continuous at \(\mathbf f(\mathbf a)\). Prove that \({\bf g}\circ \mathbf f\) is continuous at \(\mathbf a\). Note: This is part 4 of Theorem 4, and the proof was left as an exercise. This is the exercise.

\(\Leftarrow\)  \(\Uparrow\)  \(\Rightarrow\)

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