Definition: Open Sets

Let \(U\subset {\mathbb R}^n\) (that is, let \(U\) be a subset of \({\mathbb R}^n\)). We call \(U\) an open set when for every point \({\bf x}_0\) in \(U\) there exists some \(r>0\) such that \(D_r({\bf x}_0)\) is contained within \(U\); symbolically, we write \(D_r({\bf x}_0)\subset U\) (see Figure 2.17).

Theorem 1

For each \({\bf x}_0 \in {\mathbb R}^n\) and \(r>0,D_r({\bf x}_0)\) is an open set.

proof

Let \({\bf x} \in D_r({\bf x}_0)\); that is, let \(\| {\bf x} - {\bf x}_0 \| <r\). According to the definition of an open set, we must find an \(s > 0\) such that \(D_s({\bf x}) \subset D_r ({\bf x}_0)\). Referring to Figure 2.18, we see that \(s= r- \| {\bf x} - {\bf x}_0 \|\) is a reasonable choice; note that \(s>0\), but that \(s\) becomes smaller if \({\bf x}\) is nearer the edge of \(D_r({\bf x}_0)\).

Figure 2.18: The geometry of the proof that an open disk is an open set.

To prove that \(D_s({\bf x})\subset D_r ({\bf x}_0)\), let \({\bf y}\in D_s ({\bf x})\); that is, let \(\| {\bf y} - {\bf x} \| < s\). We want to prove that \({\bf y}\in D_r ({\bf x}_0)\) as well. Proving this, in view of the definition of an \(r\)-disk, entails showing that \(\| {\bf y} - {\bf x}_0\| < r\). This is done by using the triangle inequality for vectors in \({\mathbb R}^n\): \[ \| {\bf y} - {\bf x}_0 \| = \| ({\bf y} - {\bf x}) + ({\bf x} - {\bf x}_0) \| \leq \| {\bf y} - {\bf x} \| + \| {\bf x} - {\bf x}_0 \| < s + \| {\bf x} - {\bf x}_0 \| = r . \]

Hence, \(\| {\bf y} - {\bf x}_0 \| < r\).

example 1

Prove that \(A= \{(x,y) \in {\mathbb R}^2 \mid x>0\}\) is an open set.

solution The set is pictured in Figure 2.19.

Figure 2.19: Show that \(A\,\) is an open set.

Intuitively, this set is open, because no points on the “boundary,” \(x=0\), are contained in the set. Such an argument will often suffice after one becomes accustomed to the concept of openness. At first, however, we should give details. To prove that \(A\) is open, we show that for every point \((x,y)\in A\) there exists an \(r>0\) such that \(D_r(x,y)\subset A\). If \((x,y)\in A\), then \(x>0\). Choose \(r=x\). If \((x_1,y_1)\in D_r (x,y),\) we have \[ | x_1 - x| = \textstyle\sqrt{(x_1 - x)^2} \leq \textstyle\sqrt{(x_1 - x)^2 + (y_1 - y)^2} < r=x, \] and so \(x_1-x< x\) and \(x - x_1 < x\). The latter inequality implies \(x_1> 0\), that is, \((x_1, y_1)\in A\). Hence \(D_r(x,y)\subset A\), and therefore \(A\) is open (see Figure 2.20).

Definition: Boundary Points

Let \(A\subset {\mathbb R}^n\). A point \({\bf x}\in {\mathbb R}^n\) is called a boundary point of \(A\) if every neighborhood of \({\bf x}\) contains at least one point in \(A\) and at least one point not in \(A\).

example 2

(a) Let \(A=(a,b)\) in \({\mathbb R}\). Then the boundary points of \(A\) consist of the points \(a\) and \(b\). A consideration of Figure 2.21 and the definition will make this clear. [The reader will be asked to prove this in Exercise 28(c).]

Figure 2.21: The boundary points of the interval \(\hbox{(}a, b\hbox{)}\).

(b) Let \(A=D_r(x_0,y_0)\) be an \(r\)-disk about \((x_0,y_0)\) in the plane. The boundary consists of points \((x,y)\) with \((x-x_0)^2 + (y - y_0)^2 = r^2\) (Figure 2.22).

Figure 2.22: The boundary of \(A\,\) consists of points on the edge of \(A\,\).

(c) Let \(A=\{(x,y)\in {\mathbb R}^2 \mid x>0\}\). Then the boundary of \(A\) consists of all points on the \(y\) axis (draw a figure that depicts this).

(d) Let \(A\) be \(D_r({\bf x}_0\)) minus the point \({\bf x}_0\) (a “punctured” disk about \({\bf x}_0\)). Then \({\bf x}_0\) is a boundary point of \(A\).

Definition: Limit

Let \(f\colon\, A\subset {\mathbb R}^n \rightarrow {\mathbb R}^m,\) where \(A\) is an open set. Let \({\bf x}_0\) be in \(A\) or be a boundary point of \(A,\) and let \(N\) be a neighborhood of \({\bf b}\in {\mathbb R}^m\). We say \(f\) is eventually in \(N\) as \({\bf x}\) approaches \({\bf x}_0\) if there exists a neighborhood \(U\) of \({\bf x}_0\) such that \({\bf x}\neq {\bf x}_0,{\bf x}\in U\), and \({\bf x}\in A\) imply \(f({\bf x})\in N\). [The geometric meaning of this assertion is illustrated in Figure 2.23; note that \({\bf x}_0\) need not be in the set \(A,\) so that \(f({\bf x}_0)\) is not necessarily defined.] We say \(f({\bf x})\) approaches b as x approaches \({\bf x}_0\), or, in symbols, \[ \mathop{\hbox{limit}}_{{\bf x}\rightarrow {\bf x}_0} f({\bf x}) = {\bf b}\qquad \hbox{or}\qquad f({\bf x}) \rightarrow {\bf b}\quad \hbox{as}\quad {\bf x} \rightarrow {\bf x}_0, \] when, given any neighborhood \(N\) of b, \(f\) is eventually in \(N\) as x approaches \({\bf x}_0\) [that is, “\(f({\bf x})\) is close to \({\bf b}\) if \({\bf x}\) is close to \({\bf x}_0\)”]. It may be that as \({\bf x}\) approaches \({\bf x}_0\), the values \(f({\bf x})\) do not get close to any particular vector. In this case, we say that \({{\rm limit}_{{\bf x} \rightarrow {\bf x}_0}} f({\bf x})\) does not exist.

example 3

(a) This example illustrates a limit that does not exist. Consider the function \(f\colon\, {\mathbb R}\rightarrow {\mathbb R}\) defined by \[ f(x) = \bigg\{ \begin{array}{r@{\qquad}l} 1 & \hbox{if } x>0\\ -1 &\hbox{if } x\leq 0. \end{array} \]

The \({{\rm limit}_{x \rightarrow 0}} f(x)\) does not exist, since there are points \(x_1\) arbitrarily close to 0 with \(f(x_1)= 1\) and also points \(x_2\) arbitrarily close to 0 with \(f(x_2)= -1\); that is, there is no single number that \(f\) is close to when \(x\) is close to 0 (see Figure 2.24). If \(f\) is restricted to the domain \((0, 1)\) or \((-1, 0),\) then the limit does exist. Can you say why?

Figure 2.24: The limit of this function as \(x\to 0\) does not exist.

Figure 2.25: The limit of this function as \(x \to 0\) is zero.

(b) This example illustrates a function whose limit does exist, but whose limiting value does not equal its value at the limiting point. Define \(f\colon\, {\mathbb R}\rightarrow {\mathbb R}\) by \[ f(x) = \bigg\{ \begin{array}{l@{\qquad}l} 0 & \hbox{if }x\neq 0\\ 1 & \hbox{if }x=0. \end{array} \]

It is true that \({{\rm limit}_{x \rightarrow 0}} f(x)=0,\) since for any neighborhood \(U\) of \(0, x\in U\) and \(x \neq 0\) implies that \(f(x)=0\). We see from the graph in Figure 2.25 that \(f\) approaches \(0\) as \(x \rightarrow 0;\) we do not care that \(f\) happens to take on some other value at 0.

example 4

Use the definition to verify that the “obvious” \({{\rm limit}_{\,\,{\bf x} \rightarrow {\bf x}_0}}\, {\bf x} = {\bf x}_0\) holds, where \({\bf x}\) and \({\bf x}_0 \in {\mathbb R}^n\).

solution Let \(f\) be the function defined by \(f({\bf x}) = {\bf x}\), and let \(N\) be any neighborhood of \({\bf x}_0\). We must show that \(f({\bf x})\) is eventually in \(N\) as \({\bf x} \rightarrow {\bf x}_0\). According to the definition, we must find a neighborhood \(U\) of \({\bf x}_0\) with the property that if \({\bf x} \neq {\bf x}_0\) and \({\bf x} \in U\), then \(f({\bf x}) \in N.\) Pick \(U=N\). If \({\bf x} \in U\), then \({\bf x}\in N\); because \({\bf x} = f({\bf x})\), it follows that \(f({\bf x}) \in N\). Thus, we have shown that \({\rm limit}_{{\bf x} \to {\bf x}_0}\) \({\bf x}= {\bf x}_0\). In a similar way, we have \[ {\displaystyle \mathop {{\rm limit}\ }_{(x,y) \rightarrow (x_0, y_0)}} x = x_0, \hbox{etc.} \]

example 5

(This example demonstrates another case in which the limit cannot simply be “read off” from the function.) Find \({\rm limit}_{x\rightarrow 1} g(x)\) where \[ g\colon \,x \mapsto \frac{x-1}{\sqrt{x}-1}. \]

solution This function is graphed in Figure 2.26.

Figure 2.26: These graphs are the same except that in part (a), \(g\,\) is undefined at \(x = 1\), whereas in part (b), \(g^{\ *}\) is defined for all \(x \ge 0\).

We see that \(g(1)\) is not defined, because division by zero is not defined. However, if we multiply the numerator and denominator of \(g(x)\) by \(\sqrt{x} +1\), we find that for all \(x\) in the domain of \(g\) we have \[ g(x) = \frac{x-1}{\sqrt{x}-1} = \sqrt{x} + 1, \qquad x \neq 1. \]

The expression \(g^*(x) = \sqrt{x} +1\) is defined and takes the value 2 at \(x=1;\) from one-variable calculus, \(g^*(x) \rightarrow 2\) as \(x \rightarrow 1\). But because \(g^*(x) = g(x)\) for all \(x \geq 0, x \neq 1\), we must have as well that \(g(x) \rightarrow 2\) as \(x \rightarrow 1\).

Theorem 2 Uniqueness of Limits

\[ \hbox{If}\qquad\displaystyle \mathop{\rm limit}_{{\bf x} \rightarrow {\bf x}_0} f({\bf x}) = {\bf b}_1\qquad\hbox{and}\qquad\displaystyle \mathop{\rm limit}_{{\bf x} \rightarrow {\bf x}_0} f({\bf x}) = {\bf b}_2,\qquad\hbox{then}\qquad{\bf b}_1 \,{=}\,\, {\bf b}_2. \]

Theorem 3 Properties of Limits

Let \(f\colon\, A\subset {\mathbb R}^n \rightarrow {\mathbb R}^m\), \(g\colon\, A\) \(\subset {\mathbb R}^n \rightarrow {\mathbb R}^m, {\bf x}_0\) be in \(A\) or be a boundary point of \(A, {\bf b} \in {\mathbb R}^m\), and \(c \in {\mathbb R}\); then

• (i) If \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0} f({\bf x}) = {\bf b}\), then \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0} cf({\bf x})= c{\bf b}\), where \(cf\colon\, A \rightarrow {\mathbb R}^m\) is defined by \({\bf x} \mapsto c(f({\bf x}))\).
• (ii) If \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0}f({\bf x}) = {\bf b}_1\) and \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0} g({\bf x}) = {\bf b}_2\), then \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0}\,(f+g)({\bf x}) = {\bf b}_1 + {\bf b}_2\), where \((f+g)\colon\, A \rightarrow {\mathbb R}^m\) is defined by \({\bf x}\mapsto f({\bf x}) + g({\bf x})\).
• (iii) If \(m=1, {\rm limit}_{{\bf x} \rightarrow {\bf x}_0} f({\bf x}) = b_1\), and \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0} g({\bf x}) = b_2\), then \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0}\, (fg)({\bf x}) = b_1b_2\), where \((fg)\colon\, A \rightarrow {\mathbb R}\) is defined by \({\bf x}\mapsto f({\bf x}) g({\bf x})\).
• (iv) If \(m=1\), \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0} f({\bf x}) = b \neq 0\), and \(f({\bf x})\neq 0\) for all \({\bf x}\in A\), then \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0} 1/f({\bf x}) = 1/b\), where \(1/f\colon\, A \rightarrow {\mathbb R}\) is defined by \({\bf x}\mapsto 1/f({\bf x})\).
• (v) If \(f({\bf x}) = (f_1({\bf x}), \ldots, f_m({\bf x}))\), where \(f_i\colon\, A \rightarrow {\mathbb R}\), \(i=1, \ldots, m,\) are the component functions of \(f\), then \({\rm limit}_{{\bf x} \rightarrow {\bf x}_0} f({\bf x}) = {\bf b} = (b_1, \ldots, b_m)\) if and only if \(\,{\rm limit}_{{\bf x} \rightarrow {\bf x}_0} f_i({\bf x}) = b_i\) for each \(i = 1, \ldots, m\).

example 6

Let \(f\colon\, {\mathbb R}^2 \rightarrow {\mathbb R}, (x,y) \mapsto x^2 + y^2 +2\). Compute the limit \[ \displaystyle\mathop{\rm limit}_{(x,y) \,\rightarrow\, (0,1)} f(x,y). \]

solution Here \(f\) is the sum of the three functions \((x,y) \mapsto x^2, (x,y) \mapsto y^2\), and \((x,y)\mapsto 2\). The limit of a sum is the sum of the limits, and the limit of a product is the product of the limits (Theorem 3). Hence, using the fact that \({\rm limit}_{(x,y) \rightarrow (x_0,y_0)} x = x_0\) (Example 4), we obtain \[ {\displaystyle \mathop {{\rm limit}\ }_{(x,y) \rightarrow (x_0,y_0)}} x^2 = \left( {\displaystyle \mathop {{{\rm limit}\ } x}_{(x,y) \rightarrow (x_0,y_0)}}\right) \left( {\displaystyle \mathop {{{\rm limit}\ }x}_{(x,y) \rightarrow (x_0,y_0)}}\right) = x^2_0 \] and, using the same reasoning, \({\rm limit}_{(x,y) \rightarrow (x_0,y_0)} y^2=y^2_0\). Consequently, \[ {\displaystyle \mathop {{\rm limit}\ }_{(x,y) \rightarrow (0,1)}} f(x,y) = 0^2 + 1^2 + 2 = 3. \]

Definition: Continuity

Let \(f\colon\, A \subset {\mathbb R}^n \rightarrow {\mathbb R}^m\) be a given function with domain \(A\). Let \({\bf x}_0 \in A\). We say \(f\) is continuous at \({\bf x}_0\) if and only if \[ {\displaystyle \mathop {{\rm limit}\ }_{{\bf x} \rightarrow {\bf x}_0}} f({\bf x}) = f({\bf x}_0). \]

If we just say that \(f\) is continuous, we shall mean that \(f\) is continuous at each point \({\bf x}_0\) of \(A\). If \(f\) is not continuous at \({\bf x}_0\), we say \(f\) is discontinuous at \({\bf x}_0\). If \(f\) is discontinuous at some point in its domain, we say \(f\) is discontinuous.

example 7

Any polynomial \(p(x) = a_0 + a_1x + \cdots + a_nx^n\) is continuous from \({\mathbb R}\) to \({\mathbb R}\). Indeed, from Theorem 3 and Example 4, \begin{eqnarray*} {\displaystyle \mathop {{\rm limit}\ }_{x \rightarrow x_0}}\big(a_0 + a_1 x + \cdots + a_nx^n\big) &=& {\displaystyle \mathop {{\rm limit}\ }_{x \rightarrow x_0}} a_0 + {\displaystyle \mathop {{\rm limit}\ }_{x \rightarrow x_0}} a_1 x + \cdots + {\displaystyle \mathop {{\rm limit}\ }_{x \rightarrow x_0}} a_nx^n \\ &=& a_0 + a_1 x_0 + \cdots + a_n x^n_0, \end{eqnarray*} because the limit of a product is the product of the limits, which gives \[ {\displaystyle \mathop {{\rm limit}\ }_{x \rightarrow x_0}} x^n = \Big( {\displaystyle \mathop {{\rm limit}\ }_{x \rightarrow x_0}}x\Big)^{n} = x_0^n. \]

example 8

Let \(f\colon\, {\mathbb R}^2 \rightarrow{\mathbb R},f(x,y)=xy\). Then \(f\) is continuous, because, by the limit theorems and Example 4, \[ \mathop {{\rm limit}\ }_{(x,y)\rightarrow(x_0,y_0)} xy=\left( \mathop {{\rm limit}\ }_{(x,y)\rightarrow(x_0,y_0)} x\right)\! \left(\mathop {\rm limit}_{(x,y)\rightarrow(x_0,y_0)} y\right)=x_0y_0. \]

example 9

The function \(f\colon\, {\mathbb R}^2 \rightarrow {\mathbb R}\) defined by \[ f(x,y)=\bigg\{ \begin{array}{c@{\qquad}l} 1 & \hbox{if } x\leq 0 \hbox{ or } y\leq 0\\[1.2pt] 0 & \hbox{otherwise} \end{array} \] is not continuous at (0, 0) or at any point on the positive \(x\) axis or positive \(y\) axis. Indeed, if \((x_0,y_0)={\bf u}\) is such a point (i.e., \(x_0=0\) and \(y_0\ge 0\), or \(y_0=0\) and \(x_0\ge 0\)) and \(\delta > 0\), there are points \((x,y)\in D_\delta({\bf u})\), a neighborhood of \({\bf u}\), with \(f(x,y)=1\) and other points \((x,y)\in D_\delta({\bf u})\) with \(f(x,y)=0\). Thus, it is not true that \(f(x,y)\rightarrow f(x_0,y_0)=1\) as \((x,y)\rightarrow(x_0,y_0)\).

Theorem 4 Properties of Continuous Functions

Suppose that \(f\colon\, A\subset{\mathbb R}^n\rightarrow {\mathbb R}^m ,g\colon\, A\subset {\mathbb R}^n\rightarrow {\mathbb R}^m\), and let \(c\) be a real number.

• (i) If \(f\) is continuous at \({\bf x}_0\), so is \(cf\), where \((cf)({\bf x})=c[f({\bf x})]\).
• (ii) If \(f\) and \(g\) are continuous at \({\bf x}_0,\) so is \(f+g\), where the sum of \(f\) and \(g\) is defined by \((f+g)({\bf x}) = f({\bf x})+g({\bf x})\).
• (iii) If \(f\) and \(g\) are continuous at \({\bf x}_0\) and \(m=1\), then the product function \(f\! g\) defined by \((f\! g)({\bf x})=f({\bf x})g({\bf x})\) is continuous at \({\bf x}_0\).
• (iv) If \(f\colon\, A\subset{\mathbb R}^n\rightarrow {\mathbb R}\) is continuous at \({\bf x}_0\) and nowhere zero on \(A\), then the quotient \(1/f\) is continuous at \({\bf x}_0\), where \((1/f)({\bf x})=1/f({\bf x})\).
• (v) If \(f\colon\, A\subset{\mathbb R}^n\rightarrow {\mathbb R}^m\) and \(f({\bf x})=(f_1({\bf x}),\ldots, f_m({\bf x}))\), then \(f\) is continuous at \({\bf x}_0\) if and only if each of the real-valued functions \(f_1, \ldots , f_m\) is continuous at \({\bf x}_0\).

example 10

Let \(f\colon\, {\mathbb R}^2 \rightarrow {\mathbb R}^2, (x,y)\mapsto (x^2y,(y+x^3)/(1+x^2))\). Show that \(f\) is continuous.

solution To see this, it is sufficient, by property (v) of Theorem 4, to show that each component is continuous. As we have mentioned, any polynomial in two variables is continuous; thus, the map \((x, y) \mapsto x^2y\) is continuous. Because \(1+x^2\) is continuous and nonzero, by property (iv), we know that \(1/(1+x^2)\) is continuous; hence, \((y+x^3)/(1+x^2)\) is a product of continuous functions, and by (iii) is continuous.

Theorem 5 Continuity of Compositions

Let \(g\colon\, A \subset {\mathbb R}^n\) \(\rightarrow {\mathbb R}^m\) and let \(f\colon\, B \subset {\mathbb R}^m \rightarrow {\mathbb R}^p\). Suppose \(g(A)\subset B,\) so that \(f \circ g\) is defined on \(A\). If \(g\) is continuous at \({\bf x}_0\in A\) and \(f\) is continuous at \({\bf y}_0=g({\bf x}_0),\) then \(f\circ g\) is continuous at \({\bf x}_0\).

example 11

Let \(f(x,y,z)=(x^2+y^2+z^2)^{30}+\sin\, z^3\). Show that \(f\) is continuous.

solution Here we can write \(f\) as a sum of the two functions \((x^2+y^2+z^2)^{30}\) and \(\sin\, z^3\), so it suffices to show that each is continuous. The first is the composite of \((x,y,z)\mapsto (x^2+y^2+z^2)\) with \(u\mapsto u^{30}\), and the second is the composite of \((x,y,z)\mapsto z^3\) with \(u\mapsto \sin\, u\), and so we have continuity by Theorem 5.

Theorem 6

Let \(f\colon\, A\subset {\mathbb R}^n \rightarrow {\mathbb R}^m\) and let \({\bf x}_0\) be in \(A\) or be a boundary point of \(A\). Then \({\rm limit}_{{\bf x}\to {\bf x}_0} f({\bf x})={\bf b}\) if and only if for every number \(\varepsilon > 0\) there is a \(\delta >0\) such that for any \({\bf x}\in A\) satisfying \(0 < \| {\bf x}-{\bf x}_0 \| < \delta\), we have \(\| f({\bf x})-{\bf b} \| < \varepsilon\) (see Figure 2.31).

example 12

Show that \({\rm limit}_{(x,y)\rightarrow (0,0)} x=0\) using the \(\varepsilon\)-\(\delta\) method.

solution Note that if \(\delta > 0, \| (x,y)-(0,0) \| = \sqrt{x^2+ y^2}< \delta\) implies \(|x-0|=|x|=\sqrt{x^2}\leq \sqrt{x^2+y^2} <\delta\). Thus, if \(\| (x,y)-(0,0) \| < \delta\), then \(|x-0|\) is also less than \(\delta\). Given \(\varepsilon >0\), we are required to find a \(\delta >0\) (generally depending on \(\varepsilon\)) with the property that \(0< \| (x,y)-(0,0) \| <\delta\) implies \(|x-0|<\varepsilon\). What are we to pick as our \(\delta\)? From the preceding calculation, we see that if we choose \(\delta =\varepsilon\), then \(\| (x,y)\,{-}\,(0,0) \| <\delta\) implies \(|x\,{-}\,0|\,{<}\,\varepsilon\). This shows that \({\rm limit}_{(x,y)\rightarrow (0,0)} x\,{=}\,0\). Given \(\varepsilon > 0\), we could have also chosen \(\delta=\varepsilon/2\) or \(\varepsilon/3\), but it suffices to find just one \(\delta\) satisfying the requirements of the definition of a limit.

example 13

Consider the function \[ f(x,y)=\frac{\sin\,(x^2+y^2)}{x^2+y^2}. \]

Even though \(f\) is not defined at (0, 0), determine whether \(f(x,y)\) approaches some number as \((x,y)\) approaches (0, 0).

solution From one-variable calculus or L’Hôpital’s rule we know that \[ \mathop {{\rm limit}\ }_{\alpha\rightarrow 0}\frac{\sin\, \alpha}{\alpha}=1. \]

Thus, it is reasonable to guess that \[ \mathop {{\rm limit}\ }_{{\bf v} \to (0,0)}f({\bf v})=\mathop {{\rm limit}\ }_{{\bf v}\rightarrow (0,0)} \frac{\sin\, \| {\bf v} \| ^2}{ \| {\bf v} \| ^2}=1. \]

Indeed, because \({\rm limit}_{\alpha\rightarrow 0}\,(\sin\, \alpha)/\alpha =1\), given \(\varepsilon >0\) we are able to find a \(\delta >0\), with \(0\,{<}\,\delta\,{<}\,1\), such that \(0<|\alpha|<\delta\) implies that \({|(}\sin\, \alpha)/ \alpha-1|<\varepsilon\). If \(0< \| {\bf v} \| <\delta\), then \(0< \|{\bf v} \| ^2<\delta^2<\delta\), and therefore \[ |f({\bf v})-1|=\bigg|\frac{\sin\, \| {\bf v} \|^2}{ \| {\bf v}\|^2}-1\bigg|<\varepsilon. \]

Figure 2.32: Graph of the function \(f\hbox{(}x,y\hbox{)}= [\sin\, \hbox{(}x^2+y^{\,2}\hbox{)}]/\hbox{(}x^2+y^{\,2}\hbox{)}\).

Thus, \({\rm limit}_{{\bf v}\to (0,0)} f({\bf v})=1\). If we plot \([\sin\, (x^2+y^2)]/(x^2+y^2)\) on a computer, we get a graph that is indeed well behaved near (0, 0) (Figure 2.32).

example 14

Show that \[ \mathop {{\rm limit}\ }_{(x,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}}=0. \]

solution We must show that \(x^2/\sqrt{x^2+y^2}\) is small when \((x,y)\) is close to the origin. To do this, we use the following inequality: \begin{eqnarray*} 0\leq \frac{x^2}{\sqrt{x^2+y^2}} &\leq& \frac{x^2+y^2}{\sqrt{x^2+y^2}}\qquad \hbox{(because \(y^2\geq 0\))}\\ &=& \textstyle\sqrt{x^2+y^2}.\\[-22pt] \end{eqnarray*}

Given \(\varepsilon >0\), choose \(\delta =\varepsilon\). Then \( \| (x,y)-(0,0) \| = \| (x,y) \| =\sqrt{x^2+y^2}\), and so \(\| (x,y)-(0,0) \| <\delta\) implies that \[ \Bigg|\frac{x^2}{\sqrt{x^2+y^2}}-0\Bigg|=\frac{x^2}{\sqrt{x^2+y^2}}\,\leq \,\textstyle\sqrt{x^2+y^2}= \| (x,y)-(0,0) \| <\delta=\varepsilon. \]

Thus, the conditions of Theorem 6 have been fulfilled and the limit is verified.

example 15

(a) Does \[ \displaystyle \mathop{\rm limit}_{(x,y)\to (0,0)} x^2/(x^2+y^2) \] exist? [See Figure 2.33.]

(b) Prove that [see Figure 2.33] \[ \displaystyle{\mathop{\rm limit}_{(x,y)\,\rightarrow\, (0,0)}}\, \,\frac{2x^2y}{x^2+y^2}=0. \]

Figure 2.33: (a) The function \(z=x^2/\hbox{(}x^2+y^{\,2}\hbox{)}\) has no limit at (0, 0). (b) The function \(z=\hbox{(}2x^2y\hbox{)}/\hbox{(}x^2+y^{\,2}\hbox{)}\) has limit 0 at (0, 0).

solution (a) If the limit exists, \(x^2/(x^2+y^2)\) should approach a definite value, say \(a\), as \((x,y)\) gets near (0, 0). In particular, if \((x,y)\) approaches zero along any given path, then \(x^2/(x^2+y^2)\) should approach the limiting value \(a\). If \((x,y)\) approaches (0, 0) along the line \(y=0\), the limiting value is clearly 1 (just set \(y=0\) in the preceding expression to get \(x^2/x^2=1\)). If \((x,y)\) approaches (0, 0) along the line \(x=0\), the limiting value is \[ \mathop {\rm lim}_{y\,\rightarrow\, 0}\,\,\frac{0^2}{0^2+y^2}=0 \not=1. \]

Hence, \({\rm limit}_{(x,y)\to (0,0)} x^2/(x^2+y^2)\) does not exist.

(b) Note that \[ \bigg| \frac{2x^2y}{x^2+y^2}\bigg|\leq \bigg|\frac{2x^2y}{x^2}\bigg|=2 |y|. \]

Thus, given \(\varepsilon > 0\), choose \(\delta=\varepsilon/2\); then \(0< \| (x,y)-(0,0) \| =\sqrt{x^2+y^2}<\delta\) implies \(|y|<\delta\), and thus \[ \bigg|\frac{2x^2y}{x^2+y^2}-0\bigg|<2\delta = \varepsilon. \]

Theorem 7

Let \(f\colon\, A\subset {\mathbb R}^n\rightarrow {\mathbb R}^m\) be given. Then \(f\) is continuous at \({\bf x}_0\in A\) if and only if for every number \(\varepsilon > 0\) there is a number \(\delta>0\) such that \[ {\bf x}\in A\qquad \hbox{and}\qquad \| {\bf x}-{\bf x}_0 \| <\delta \qquad \hbox{implies} \qquad \| f({\bf x})-f({\bf x}_0) \| <\varepsilon. \]