Limits

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We investigate limits of functions of several variables.

Recall that for functions of a single variable, we say that $\Lim {x}{a} f(x) = L$ if the value of $f(x)$ can be made arbitrarily close to $L$ for all $x$ sufficiently close, but not equal to, $x=a$ .

This easily allows us to make a similar definition for functions of several variables.

Suppose that $F$ is a real-valued function of $n$ variables. Assume that the domain of $F$ contains a small $n$ -dimensional ball centered at a point $\pt {a}$ , except, possibly, the point $\pt {a}$ . The limit of $F$ as $\pt {x}$ approaches $\pt {a}$ is $L$ if the value of $F(\pt {x})$ can be made as close as one wishes to $L$ for all $\pt {x}$ in the domain of $F$ sufficiently close, but not equal to, $\pt {a}$ .

When this occurs, we write $\lim _{\pt {x}\to \pt {a}} F(\pt {x}) = L$ .

Suppose that $F$ is a real-valued function of two variables, $\pt {x} = \point {x,y}$ , and $\pt {a} = \point {a,b}$ . Then the statement $\lim _{\pt {x}\to \pt {a}} F(\pt {x}) = L$ can be also written as $\lim _{\point {x,y}\to \point {a,b} }F\left (x,y\right ) = L$

Suppose that $F$ is a function of three variables, $\pt {x} = \point {x,y,z}$ , and $\pt {a} = \point {a,b,c}$ . Similarly, the statement $\lim _{\pt {x}\to \pt {a}} F(\pt {x}) = L$ can be written equivalently as

$\lim _{\point {x,y,z}\to \point {a,b,c}} F\left (x,y,z\right ) = L$

We now focus on functions of two variables, although it is not difficult to state similar results in the more general setting. For the rest of what follows, we will often denote points $\point {x,y}$ by $\pt {x}$ .

While the intuitive idea behind limits seems to remain unchanged, something interesting is worth observing. One of the most important ideas for limits of a function of a single variable is the notion of a sided limit. For functions of a single variable, there were really only two natural ways for $x$ to become close to $a$ ; we could take $x$ to approach the point $a$ from the left or the right. For instance, $\Lim {x}{a^-}f(x)$ tells us to consider the inputs $x<a$ only. In fact, there’s a theorem that guarantees that $\Lim {x}{a} f(x) = L$ if and only if $\Lim {x}{a^-}f(x) =L$ and $\Lim {x}{a^+}f(x) =L$ , meaning that the function must approach the same value as the input approaches $a$ from both the left and the right.

On the other hand, there are now infinitely many ways for $\point {x,y} \to \point {a,b}$ ; we can approach along a straight line path parallel to the $x$ -axis or $y$ -axis, other straight line paths, or even other types of curves.

In order to check whether a limit exists, do we have to verify that the function tends to the same value along infinitely many different paths?

While this may seem problematic, there is some good news; many of the limit laws from before still do hold now.

Limit Laws Let $F$ and $G$ be real-valued functions of two variables, and $b$ , $L$ and $M$ be real numbers, where $\lim _{\pt {x}\to \pt {a}}F(\pt {x}) = L \quad \text {and}\quad \lim _{\pt {x}\to \pt {a}} G(\pt {x}) = M.$ Let $\pt {x} = \point {x_{1},x_{2}}$ , and $\pt {a} = \point {a_{1},a_{2}}$ . We also assume that the domain of $F$ and the domain of $G$ both contain a small disk with the center at $\pt {a}$ , except possibly the point $\pt {a}$ .

Constant Law: $\lim _{\pt {x}\to \pt {a}} b = b$
Identity Law: $\lim _{\pt {x}\to \pt {a}} x_i = a_i$ , where $i=1,2.$
Sum/Difference Law: $\lim _{\pt {x}\to \pt {a}}\big (F(\pt {x})\pm G(\pt {x})\big ) = L\pm M$
Scalar Multiple Law: $\lim _{\pt {x}\to \pt {a}} b\cdot F(\pt {x}) = bL$
Product Law: $\lim _{\pt {x}\to \pt {a}} \left (F(\pt {x})\cdot G(\pt {x})\right ) = LM$
Quotient Law: $\lim _{\pt {x}\to \pt {a}} \frac {F(\pt {x})}{G(\pt {x})} = \frac {L}{M}$ , if $M\neq 0$

In practice, this allows us to compute many limits in a similar fashion as before.

Compute $\Lim {\point {x,y}}{\point {1,2}} \frac {2x+4y}{x-3y}$ .

We show how the above properties are used quite explicitly. $\begin{align*} \Lim{\point{x,y}}{\point{1,2}} \frac{2x+4y}{x-3y} & = \frac{\Lim{\point{x,y}}{\point{1,2}}(2x+4y)}{\Lim{\point{x,y}}{\point{1,2}}(x-3y)} \textrm{ by the quotient law. } \\ &= \frac{\Lim{\point{x,y}}{\point{1,2}}2x+\Lim{\point{x,y}}{\point{1,2}} 4y}{\Lim{\point{x,y}}{\point{1,2}}x-\Lim{\point{x,y}}{\point{1,2}}3y} \textrm{ by the sum/difference law. } \\ &= \frac{2\left[\Lim{\point{x,y}}{\point{1,2}}x\right]+4\left[\Lim{\point{x,y}}{\point{1,2}} y\right]}{\left[\Lim{\point{x,y}}{\point{1,2}}x\right]-3\left[\Lim{\point{x,y}}{\point{1,2}}y\right]} \textrm{ by the scalar multiple law. } \\ &= \frac{2[1]+4[2]}{[1]-3[2]} \textrm{ by the identity law. } \\ &= -2 \end{align*}$

Essentially, the above laws allow us to evaluate limits by directly substituting values into the given function, provided the end result is a constant. Henceforth, when a limit can be evaluated by direct substitution, we will not show the details.

As it turns out, another old technique works well too.

Compute $\lim _{(x,y)\to (0,0) }\frac {x^3+xy^2}{x^2+y^2}$ .

To compute this limit, note that direct substitution leads us to the indeterminate form $\frac {0}{0}$ . $\begin{align*} \lim_{\point{x,y}\to\point{0,0}}\frac{x^3+xy^2}{x^2+y^2} &=\lim_{\point{x,y}\to\point{0,0}} \frac{\answer[given]{x}\left(x^2+y^2\right)}{\answer[given]{x^2+y^2}}\\ &=\lim_{\point{x,y}\to\point{0,0}} x\\ &=\answer[given]{0} \end{align*}$

What allows us to perform the cancellation of the common factors of $x^2+y^2$ ? Note that when determining whether a limit exists or not, we must look near the point $\point {0,0}$ , but not at the point $\point {0,0}$ . No matter how close a point $\point {x,y}$ is to the point $\point {0,0}$ , as long as $\point {x,y} \neq \point {0,0}$ , then $x^2+y^2 \neq 0$ . So, this cancellation is valid.

When limits don’t exist

Unfortunately, there are difficulties that arise now that did not before when we have to handle indeterminate forms. Since limits exist only when the function tends to the same value along every path, we can use this to show that some limits do not exist.

If it is possible to arrive at different limiting values by approaching along different paths, the limit does not exist.

This is analogous to the left and right hand limits of single variable functions not being equal, implying that the limit does not exist.

Suppose that $F(x,y) = \begin {cases} 2x+2y , & x \geq 3 \\ 3x-y, & x<3 \end {cases}$ .

Determine whether $\Lim {\point {x,y}}{\point {2,2}} F(x,y)$ exists.

We need to determine which piece of the function should be used for each limit.

For $\Lim {\point {x,y}}{\point {2,2}} F(x,y)$ , note that any path that approaches $\point {2,2}$ must eventually lie in the portion of the domain of $F(x,y)$ where $F(x,y) = 3x-y$ . The image below shows the domain of $F(x,y)$ and the formulas used to evaluate it for each $(x,y)$ in the $(x,y)$ -plane.

We see that for any point $(x,y)$ sufficiently close to $(2,2)$ the following holds

$\Lim {\point {x,y}}{\point {2,2}} F(x,y) = \Lim {\point {x,y}}{\point {2,2}} (3x-y) = 4.$

Determine whether $\Lim {\point {x,y}}{\point {3,2}} F(x,y)$ exists.

For $\Lim {\point {x,y}}{\point {3,2}} F(x,y)$ , note that it is possible to approach $\point {3,2}$ along a path for which $F(x,y) = 3x-y$ by taking $x \to 3^-$ or $F(x,y) = 2x+2y$ by taking $x \to 3^+$ .

Along the path $x>3, y=2$ , we have $F(x,y) = 2x+2y$ , so if $\Lim {\point {x,y}}{\point {3,2}} F(x,y)$ exists, we must have $\Lim {\point {x,y}}{\point {3,2}} F(x,y) = 10$ .
Along the path $x<3, y=2$ , we have $F(x,y) = 3x-y$ , so if $\Lim {\point {x,y}}{\point {3,2}} F(x,y)$ exists, we must have $\Lim {\point {x,y}}{\point {3,2}} F(x,y) = 7$ .

Thus, $\Lim {\point {x,y}}{\point {2,2}} F(x,y)$ does not exist.

Now, let’s consider an example in which we do not have a piecewise function.

Determine whether $\lim _{\point {x,y}\to \point {0,0}} \frac {3xy}{x^2+y^2}$ exists.

Let’s find two different paths in the domain along which $F(x,y)=\frac {3xy}{x^2+y^2}$ approaches different values as $\point {x,y} \to \point {0,0}$ .

If we approach $\point {0,0}$ along the line $y=0$ in the $xy$ -plane, we have $F(x,y) = F(x,0) = \frac {3x(0)}{x^2+(0)^2} = 0, x \neq 0.$ Note that this tells us that if the limit exists, it must be $0$ .
If we approach $\point {0,0}$ along the line $y=x$ in the $xy$ -plane, we have $F(x,y) = F(x,x) = \frac {3x(x)}{x^2+(x)^2} = \frac {3x^2}{2x^2} = \frac {3}{2}, x \neq 0.$ Note that this tells us that if the limit exists, it must be $\frac {3}{2}$ .

Since $F(x,y)=\frac {3xy}{x^2+y^2}$ approaches different values along different paths,

$\lim _{\point {x,y}\to \point {0,0}} \frac {3xy}{x^2+y^2}$ does not exist.

As it turns out, there are nice ways to think about the above result both geometrically and analytically.

A geometric viewpoint of analyzing along level curves

Note that for the curve $x=0$ in the domain of the function $F(x,y)$ , we found $F(x,y) = 0$ for every $(x,y)$ . This means that the curve $x=0$ is part of the level curve associated to $z=0$ of $F(x,y)$ . Similarly, for the curve $y=x$ in the domain of the function $F(x,y)$ , we found $F(x,y) = \frac {3}{2}$ for every $(x,y)$ . This means that the curve $y=x$ is part of the level curve associated to $z= \frac {3}{2}$ of $F(x,y)$ .

These level curves share $(0,0)$ as a boundary point; had either been defined there, they would intersect. Since there are different $z$ -values associated to each level curve, it is impossible that $\Lim {(x,y)}{(0,0)} F(x,y)$ exists. This notion can be generalized.

If two different level curves of a function share a common boundary point, then the limit at that boundary point does not exist.

An analytic viewpoint of analyzing along level curves

In the first approach to this problem, along the path $y=x$ , we saw that there was convenient cancellation that occurred near, but not at $x=0$ . We note that this is not the only path along which this happens; evaluating $\lim _{\point {x,y}\to \point {0,0}} \frac {3xy}{x^2+y^2}$ along the lines $y=mx$ produces a similar result. $\begin{align*} F(x,y) = F(x,mx) &= \frac{3x\left(\answer[given]{m x}\right)}{x^2+\left(\answer[given]{mx}\right)^2} \\ &= \frac{3mx^2}{x^2(m^2+1)}\\ &= \frac{3m}{m^2+1}, \qquad x \neq 0\\ \end{align*}$

Note that this “cancellation” is the algebraic consequence of analyzing along a level curve; in order for the function to be constant, there can be no explicit dependence on $x$ and $y$ . However, the function $F(x,y)$ tends to a different value as $(x,y) \to (0,0)$ along $y=mx$ for each choice of $m$ , so the limit $\lim _{\point {x,y}\to \point {0,0}} \frac {3xy}{x^2+y^2}$ cannot exist.

Note that we can recover our previous results from this exercise. Along $y=0$ , we have $m=0$ , and $\lim _{x\to 0}\frac {3m}{m^2+1} = \answer [given]{0}.$ Now along $y=x$ , $m=1$ , and $\lim _{x\to 0}\frac {3m}{m^2+1} = \answer [given]{3/2}.$ Since we find differing limiting values when computing the limit along different paths, we must conclude that the limit does not exist. The advantage here is that we’ve found a whole class of paths along which the function tends to different values. This occurs because the lines $y=mx$ are all parts of different level curves of $F(x,y)$ .

We finish by presenting you with a plot of $F$ :

We conclude this section with an example that illustrates the necessity to analyze a function along every path in its domain in order to conclude that a limit exists.

Show $\lim _{\point {x,y}\to \point {0,0}} \frac {\sin (xy)}{x+y}$ does not exist by finding the limit along the path $y=-\sin (x)$ .

Let $F\point {x,y} = \frac {\sin (xy)}{x+y}.$ We will show that $\lim _{\point {x,y}\to \point {0,0}} F\point {x,y}$ does not exist by finding the limit along the path $y=-\sin (x)$ . First, however, let’s try the same trick that worked before and see what happens. Consider the limits found along the lines $y=mx$ as done above. $\begin{align*} \lim_{\point{x,mx}\to\point{0,0}} \frac{\sin\big(x(\answer[given]{mx})\big)}{x+\answer[given]{mx}} &= \lim_{x\to 0} \frac{\sin (mx^2)}{x(m+1)} \\ &= \lim_{x\to 0} \frac{\sin(mx^2)}{x}\cdot\answer[given]{\frac{1}{m+1}}. \end{align*}$

By applying L’Hôpital’s Rule, we can show this limit is $0$ except when $m=-1$ , that is, along the line $y=-x$ . This line is not in the domain of $F$ , so we have found the following fact: along every line $y=mx$ in the domain of $F$ , $\lim _{\point {x,y}\to \point {0,0}} F\point {x,y}=0.$ Now consider the limit along the path $y=-\sin (x)$ :

$\begin{align*} \lim_{\point{x,-\sin(x)}\to\point{0,0}} \frac{\sin\big(-x\sin(x)\big)}{x-\sin(x)} &= \lim_{x\to0} \frac{\sin\big(-x\sin(x)\big)}{x-\sin(x)} \end{align*}$

Now apply L’Hôpital’s Rule twice to find a limit is of the form $\relax \boldsymbol {\tfrac {\#}{0}}$ . Hence the limit does not exist. Step back and consider what we have just discovered.

Along any line $y=mx$ in the domain of the $F\point {x,y}$ , the limit is $0$ .
However, along the path $y=-\sin (x)$ , which lies in the domain of the $F\point {x,y}$ for all $x\neq 0$ , the limit does not exist.

Since the limit is not the same along every path to $(0,0)$ , we say $\lim _{\point {x,y}\to \point {0,0}}\frac {\sin (xy)}{x+y}$ does not exist. We finish by presenting you with a plot of $F$ :