[b][size=150]Limits[/size][/b][br][br]Suppose [math]f(x,y)[/math] is a function of two variables and [math](x_0,y_0)[/math] is a point in [math]\mathbb{R}^2[/math]. We say the [b]limit[/b] of [math]f(x,y)[/math] as [math](x,y)[/math] tends to [math](x_0,y_0)[/math] is [math]L[/math], denoted by [br][br][math]\lim_{(x,y)\to(x_0,y_0)}f(x,y)=L[/math] [br][br]if [math]|f(x,y)-L|[/math] tends to [math]0[/math] as [math]|\langle x,y\rangle - \langle x_0,y_0\rangle|[/math] tends to [math]0[/math] i.e. the distance between [math](x,y)[/math] and [math](x_0,y_0)[/math] tends to [math]0[/math]. [br][br]Intuitively, it means that [math]f(x,y)[/math] can be as close to [math]L[/math] as you want if you choose [math](x,y)[/math] that is "close enough" to [math](x_0,y_0)[/math]. Actually, the rigorous definition of the limit is as follows:[br][br]For any small positive real number [math]\epsilon[/math], we can find another small positive real number [math]\delta[/math] such that whenever the distance from [math](x,y)[/math] to [math](x_0,y_0)[/math] is smaller than [math]\delta[/math], [math]|f(x,y)-L|<\epsilon[/math].[br][br][br][br][br]The limit of a function of three variables [math]g(x,y,z)[/math] can be similarly defined - we say that the limit of [math]g(x,y,z)[/math] as [math](x,y,z)[/math] tends to [math](x_0,y_0,z_0)[/math] is [math]L[/math], denoted by [br][br][math]\lim_{(x,y,z)\to (x_0,y_0,z_0)}g(x,y,z)=L[/math][br][br]if [math]|g(x,y,z)-L|[/math] tends to [math]0[/math] as [math]|\langle x,y,z\rangle - \langle x_0,y_0,z_0\rangle|[/math] tends to [math]0[/math].[br][br][br][br]In the following applet, it shows the idea of limit for a function of two variables: [br][br]First fix the point [math]P=(x_0,y_0)[/math], then set the value of [math]L[/math]. To check whether [math]\lim_{(x,y)\to (x_0,y_0)}f(x,y)=L[/math], set [math]\epsilon[/math] to be small, and then adjust [math]\delta[/math] such that the orange ball centered at point [math]P[/math] with radius [math]\delta[/math] is small enough such that every point [math](x,y)[/math] in the orange ball satisfies the inequality [math]|f(x,y)-L|<\epsilon[/math], which means the corresponding point on the graph is between the two planes [math]z=L+\epsilon[/math] and [math]z=L-\epsilon[/math].[br][br]([u]Note[/u]: The rigorous definition of limit will not be used in this course.)[br]

[u]Limit along a curve[/u][br][br]When we consider the limit of a function of one variable [math]\lim_{x\to a} f(x)[/math], we can further specify the way [math]x[/math] approaches [math]a[/math]. As we know, [math]x[/math] can approach [math]a[/math] from the left or right i.e. we have one-sided limits [math]\lim_{x\to a^-}f(x)[/math] and [math]\lim_{x\to a^+}f(x)[/math] respectively. As for the limit of a function of two variables [math]\lim_{(x,y)\to(x_0,y_0)}f(x,y)[/math], there are many different ways for [math](x,y)[/math] to approach [math](x_0,y_0)[/math]. This motivates us to consider the following definition:[br][br]Suppose a curve [math]C[/math] is parametrized by [math]\vec{r}(t)=\langle x(t),y(t)\rangle[/math] on the xy-plane such that [math]\vec{r}(t_0)=\langle x(t_0),y(t_0) \rangle=\langle x_0,y_0 \rangle[/math]. We define the limit of [math]f(x,y)[/math] when [math](x,y)[/math] tends to [math](x_0,y_0)[/math] [b]along the curve C[/b] to be[br][br][math]\lim_{(x,y)\to(x_0,y_0) \atop \text{along } C}f(x,y)=\lim_{t\to t_0}f(x(t),y(t))[/math][br][br][br]The limit of a function of three variables [math]g(x,y,z)[/math] along a curve can be similarly defined - we say that the limit of [math]g(x,y,z)[/math] as [math](x,y,z)[/math] tends to [math](x_0,y_0,z_0)[/math] along a curve [math]C[/math] in [math]\mathbb{R}^3[/math] is[br][br][math]\lim_{(x,y,z)\to (x_0,y_0,z_0) \atop \text{along } C}g(x,y,z)=\lim_{t\to t_0} g(x(t),y(t),z(t))[/math][br][br]where [math]C[/math] is parametrized by [math]\vec{r}(t)=\langle x(t),y(t),z(t)\rangle[/math] with [math]\vec{r}(t_0)=\langle x_0,y_0,z_0\rangle[/math].[br][br][br][u]Example[/u]:[br][br]Let [math]f(x,y)=\frac{xy}{x^2+y^2}[/math] for [math](x,y)\ne (0,0)[/math] and [math](x_0,y_0)=(0,0)[/math]. Fine the limit of [math]f(x,y)[/math] as [math](x,y)[/math] tends to [math](0,0)[/math] along each of the following curves with parametrization [math]\langle x(t),y(t)\rangle[/math] such that [math]\langle x(0),y(0)\rangle=\langle 0,0 \rangle[/math]:[br][br][list][*] [math]C_1[/math] is the x-axis i.e. [math]\langle x(t),y(t)\rangle=\langle t, 0\rangle[/math][/*][br][*] [math]C_2[/math] is the y-axis i.e. [math]\langle x(t),y(t)\rangle=\langle 0, t\rangle[/math][/*][br][*] [math]C_3[/math] is the straight line [math]y=x[/math] i.e. [math]\langle x(t),y(t)\rangle=\langle t, t\rangle[/math][/*][br][*] [math]C_4[/math] is the curve [math]y=x^2[/math] i.e. [math]\langle x(t),y(t)\rangle=\langle t, t^2\rangle[/math][/*][/list][br][br]Find [math]\lim_{(x,y)\to (0,0) \atop \text{along } C_i}f(x,y)[/math] for [math]i=1,2,3,4[/math].[br][br][u]Answer[/u]:[br][br][math]\lim_{(x,y)\to (0,0) \atop \text{along } C_1}f(x,y)=\lim_{t\to 0}f(t,0)=\lim_{t\to 0}\frac {t\cdot 0}{0^2+t^2}=0[/math][br][br][math]\lim_{(x,y)\to (0,0) \atop \text{along } C_2}f(x,y)=\lim_{t\to 0}f(0,t)=\lim_{t\to 0}\frac {0\cdot t}{t^2+0^2}=0[/math][br][br][math]\lim_{(x,y)\to (0,0) \atop \text{along } C_3}f(x,y)=\lim_{t\to 0}f(t,t)=\lim_{t\to 0}\frac {t\cdot t}{t^2+t^2}=\frac 12[/math][br][br][math]\lim_{(x,y)\to (0,0) \atop \text{along } C_4}f(x,y)=\lim_{t\to 0}f(t,t^2)=\lim_{t\to 0}\frac {t\cdot t^2}{t^2+t^4}=\lim_{t\to 0}\frac{t}{1+t^2}=0[/math][br] [br]The applet below visualizes the limit of the function along various curves in the example.

As we know, the limit of a function of one variable exists if and only if its left and right hand limits exist and are equal. We have a similar result for functions of two variables as follows:[br][br][u]Theorem[/u]: [math]\lim_{(x,y)\to(x_0,y_0)}f(x,y)=L[/math] if and only if [math]\lim_{(x,y)\to(x_0,y_0) \atop \text{along } C}f(x,y)=L[/math] for any curve [math]C[/math] [br][br][u]Proof[/u]: Omitted.[br][br]The following is an important consequence of the above theorem: Suppose there exist two curves [math]C_1[/math] and [math]C_2[/math] such that[br][br][math]\lim_{(x,y)\to(x_0,y_0) \atop \text{along } C_1}f(x,y) \ne \lim_{(x,y)\to(x_0,y_0) \atop \text{along } C_2}f(x,y)[/math][br][br]Then [math]\lim_{(x,y)\to(x_0,y_0)}f(x,y) [/math] does not exist.[br][br][br]In the above example, [math]\lim_{(x,y)\to (0,0)}f(x,y) [/math] does not exist because the limits along [math]C_1[/math] and [math]C_3[/math] are not equal.[br][br][br][br][u]Remark[/u]: Similar theorem also holds for the limits of functions of three variables.[br][br][br]

[u]Properties of limits[/u][br][br]Limits of functions of two variables have the following properties: Let [math]L[/math] and [math]M[/math] be real numbers and suppose that [math]\lim_{(x,y)\to (a,b)}f(x,y)=L[/math] and [math]\lim_{(x,y)\to (a,b)}g(x,y)=M[/math]. Assume [math]c[/math] is a constant, and [math]m[/math] and [math]n[/math] are integers.[br][br][list][*][math]\lim_{(x,y)\to (a,b)}(f(x,y)\pm g(x,y))=L\pm M[/math][/*][br][*][math]\lim_{(x,y)\to (a,b)}cf(x,y)=cL[/math][/*][br][*][math]\lim_{(x,y)\to (a,b)}f(x,y)g(x,y)=LM[/math][/*][br][*][math]\lim_{(x,y)\to (a,b)}\left(\frac{f(x,y)}{g(x,y)}\right)=\frac LM[/math], provided that [math]M\ne 0[/math][/*][br][*][math]\lim_{(x,y)\to (a,b)}(f(x,y))^n=L^n[/math][/*][br][*]If [math]m[/math] and [math]n[/math] has no common factors and [math]n\ne 0[/math], then [math]\lim_{(x,y)\to (a,b)}(f(x,y))^{\frac mn}=L^{\frac mn}[/math], where we assume [math]L>0[/math] if [math]n[/math] is even.[/*][/list][br][br][br][u]Example[/u]:[br][br]Find [math]\lim_{(x,y)\to (2,8)} (2x^3y-(xy)^{\frac 14})[/math].[br][br][math]\lim_{(x,y)\to (2,8)} (2x^3y-(xy)^{\frac 14})=\lim_{(x,y)\to (2,8)} 2x^3y - \lim_{(x,y)\to (2,8)}(xy)^{\frac 14}[/math][br][math]=2(\lim_{(x,y)\to (2,8)}x)^3(\lim_{(x,y)\to (2,8)}y)-((\lim_{(x,y)\to (2,8)}x)(\lim_{(x,y)\to (2,8)}y))^{\frac 14}=2\cdot 2^3\cdot 8 -(2\cdot 8)^{\frac 14}=126[/math][br][br]([u]Note[/u]: The above steps are just for demonstrating the use of the limit properties. In practice, we rarely write down all these steps. We plug [math]x=2[/math] and [math]y=8[/math] into the expression and compute instead.)[br][br][br][u]Example[/u]:[br][br]Find [math]\lim_{(x,y)\to (4,1)}\frac{xy-4y^2}{\sqrt{x}-2\sqrt{y}}[/math][br][br]In this example, you cannot just plug in the values of [math]x[/math] and [math]y[/math] into the expression and compute the limit because you will get "[math]\frac 00[/math]", which is undefined. Instead, you need to simplify the expression as follows:[br][br][math]\lim_{(x,y)\to (4,1)}\frac{xy-4y^2}{\sqrt{x}-2\sqrt{y}}=\lim_{(x,y)\to (4,1)}\frac{(xy-4y^2)(\sqrt{x}+2\sqrt{y})}{(\sqrt{x}-2\sqrt{y})(\sqrt{x}+2\sqrt{y})}[/math][br][math]=\lim_{(x,y)\to (4,1)} \frac{y(x-4y)(\sqrt{x}+2\sqrt{y})}{x-4y}=\lim_{(x,y)\to (4,1)} y(\sqrt{x}+2\sqrt{y}) = 4[/math][br][br][br][u]Example[/u]:[br][br]Find [math]\lim_{(x,y)\to (0,0)} \frac{x^2y}{x^2+y^2}[/math][br][br]For limits such that [math](x,y)\to (0,0)[/math], a nice trick using polar coordinates may work:[br][br]Recall that any point [math](x,y)[/math] can be expressed in terms of polar coordinates - [math]r=\sqrt{x^2+y^2}[/math], [math]r\cos\theta=x[/math], and [math]r\sin\theta=y[/math]. Moreover, [math](x,y)\to (0,0)[/math] if and only if [math]r\to 0^+[/math]. Then we have [br][br][math]\frac{x^2y}{x^2+y^2}=\frac{r^2\cos^2\theta\cdot r\sin\theta}{r^2}=r\cos^2\theta\sin\theta[/math][br][br]which implies that we have the following inequalities:[br][br][math]-r\leq \frac{x^2y}{x^2+y^2} \leq r[/math][br][br]Using squeeze theorem, we have[br][br][math]\lim_{r\to 0^+} -r \leq \lim_{(x,y)\to (0,0)}\frac{x^2y}{x^2+y^2}\leq \lim_{r\to 0^+}r[/math][br][math]\implies \lim_{(x,y)\to (0,0)}\frac{x^2y}{x^2+y^2}=0[/math][br][br][br][br]

[b][size=150]Continuity[/size][/b][br][br]The definition of continuity of functions of two/three variables is very straightforward. A function [math]f(x,y)[/math] is [b]continuous[/b] at [math](a,b)[/math] if [math]\lim_{(x,y)\to (a,b)}f(x,y)=f(a,b)[/math]. [br][br]Similarly, [math]g(x,y,z)[/math] is [b]continuous[/b] at [math](a,b,c)[/math] if [math]\lim_{(x,y,z)\to (a,b,c)}g(x,y,z)=g(a,b,c)[/math].[br][br][br]By the properties of limits, we can easily see that any polynomial functions in two/three variables are continuous everywhere.[br][br]The following are some ways to make new continuous functions from old ones:[br][br]If [math]g(x)[/math] is continuous and [math]h(y)[/math] is continuous, then [math]f(x,y)=g(x)h(y)[/math] is a continuous function of two variables.[br][br][u]Example[/u]: [math]f(x,y)=x^3\cos y[/math] is continuous. ([math]g(x)=x^3, \ h(y)=\cos y[/math])[br][br][br][br]If [math]h(x,y)[/math] is a continuous function of two variables and [math]g(u)[/math] is a continuous function, then [math]f(x,y)=g(h(x,y))[/math] is continuous.[br][br][u]Example[/u]: [math]f(x,y)=e^{x^3\cos y}[/math] is continuous. ([math]h(x,y)=x^3\cos y, \ g(u)=e^u[/math])[br][br][br][br]If [math]f(x,y)[/math] is a continuous function of two variables and [math]\vec{r}=\langle x(t),y(t)\rangle[/math] is a continuous vector-valued function, then [math]g(t)=f(x(t),y(t))[/math] is continuous.[br][br][u]Example[/u]: [math]g(t)=e^{(\cos^3 t)\cos(\sin t)}[/math] is continuous. ([math]f(x,y)=e^{x^3\cos y}, \ \vec{r}=\langle \cos t, \sin t\rangle[/math])[br][br][br][br][br]