Multivariable limits

We already know how limits can help us find the value of a function whenever the independent variable approaches a certain value. However, when we have multivariable functions, we face the problem of approaching more than one variable at a time. Multivariable limits help us examine the behavior of a multivariable function as all variables approach a certain set of values. In this topic, we will take a look at the mathematical definition of multivariable limits, some examples, and the uniqueness theorem for multivariable limits.

Limit points

Using the concept of norm, we can express the distance between two vectors $\vec{x}$ and $\vec{y}$ as:

$$d(\vec{x},\vec{y})=\|\vec{y} - \vec{x}\|$$One-dimensional limits use the concept of open intervals to get arbitrarily close to a number. We can extend this idea using open n-balls. An n-ball centered at $\vec{x}_0$ of radius $r$ can be defined as the set of all points with distance less than $r$ from the center:

$$B(\vec{x}_0, r) = \vec{x} \in \mathbb{R}^n \mid d(\vec{x}_0,\vec{x}) \lt r \qquad ; \quad d(\vec{x}_0,\vec{x}) = \|\vec{x} - \vec{x}_0\| \lt r \quad ; \quad r \in \mathbb{R}^+$$We can see how the n-ball becomes an interval for $n=1$, a disk bounded by a circle for $n=2$, and a ball bounded by a sphere for $n=3$.

For any set $A \subseteq \mathbb{R}^n$, we call $\vec{x}_0$ a limit point (or accumulation point) if any open n-ball centered at it contains elements of $A$ different from $\vec{x}_0$, with $\vec{x}_0 \in A$:

Definition of a multivariable limit

A limit of a function $f: Df \subseteq \mathbb{R}^n \to Rf \subseteq \mathbb{R}^m$ for an accumulation point $\vec{x}_0 \in \mathbb{R}^n$ of $Df$ can be defined as:

$$\lim\limits_{\vec{x} \to \vec{x}_0} f(\vec{x}) = \vec{L} \mid \vec{L} \in \mathbb{R}^m \qquad ; \quad \| f(\vec{x}) - \vec{L} \| \lt \varepsilon \iff 0 \lt \| \vec{x} - \vec{x}_0 \| \lt \delta$$

As $d(\vec{x}_0,\vec{x}) \to 0 \implies d(\vec{L},f(\vec{x})) \to 0$

Then, as $\vec{x} \to \vec{x}_0 \implies f(\vec{x}) \to \vec{L}$

A more concrete case can be illustrated with a scalar field $f: \mathbb{R}^2 \to \mathbb{R}$ such that $z=f(x,y)$; the limit $L$ exists inside an open disk (n-ball for $n=2$) of radius $\varepsilon$ for an open disk of radius $\delta$ around an accumulation point $(x,y)=(a,b)$:

Thus, for any $\varepsilon \gt 0$, there exists a $\delta \gt 0$ such that for all $(x,y)$ that satisfy $0 \lt \sqrt{(x-a)^2 + (y-b)^2} \lt \delta$, we have $|z-L|< \varepsilon$.

The usual properties of limits can also be extended to the multivariable case. Suppose we have two functions $f(x,y)$ and $g(x,y)$, then we can say:

1. $\lim\limits_{(x,y) \to (x_0,y_0)} [f(x,y)+g(x,y)] = \lim\limits_{(x,y) \to (x_0,y_0)} f(x,y) + \lim\limits_{(x,y) \to (x_0,y_0)} g(x,y)$
2. $\lim\limits_{(x,y) \to (x_0,y_0)} [f(x,y)-g(x,y)] = \lim\limits_{(x,y) \to (x_0,y_0)} f(x,y) - \lim\limits_{(x,y) \to (x_0,y_0)} g(x,y)$
3. $\lim\limits_{(x,y) \to (x_0,y_0)} [f(x,y) \cdot g(x,y)] = \lim\limits_{(x,y) \to (x_0,y_0)} f(x,y) \cdot \lim\limits_{(x,y) \to (x_0,y_0)} g(x,y)$
4. $\lim\limits_{(x,y) \to (x_0,y_0)} \frac{f(x,y)}{g(x,y)} = \frac{\lim\limits_{(x,y) \to (x_0,y_0)} f(x,y)}{\lim\limits_{(x,y) \to (x_0,y_0)} g(x,y)}$; where $\lim\limits_{(x,y) \to (x_0,y_0)} g(x,y) \neq 0$
5. $\lim\limits_{(x,y) \to (x_0,y_0)} [c \cdot f(x,y)] = c \cdot \lim\limits_{(x,y) \to (x_0,y_0)} f(x,y)$; where $c$ is a constant
6. $\lim\limits_{(x,y) \to (x_0,y_0)} [f(x,y)]^n = [\lim\limits_{(x,y) \to (x_0,y_0)} f(x,y)]^n$
7. $\lim\limits_{(x,y) \to (x_0,y_0)} \sqrt[n]{f(x,y)} = \sqrt[n]{\lim\limits_{(x,y) \to (x_0,y_0)} f(x,y)}$

Also, we have:

1. $\lim\limits_{(x,y) \to (x_0,y_0)} h(x) = \lim\limits_{x \to x_0} h(x)$
2. $\lim\limits_{(x,y) \to (x_0,y_0)} h(y) = \lim\limits_{y \to y_0} h(y)$

Examples, examples!

1. Let

$$f\left(x,y\right)=\begin{pmatrix} {\frac{\sin\left(x\right)}{x}}\\ {\frac{x-1}{y+1}}\\ \end{pmatrix}$$

and

$$\lim \limits_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}f\left(x,y\right)=\overline{L}$$

Then,

$$\lim \limits_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}f\left(x,y\right)=\vec{L}\,\Rightarrow\,\lim\limits_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}\begin{pmatrix} {\frac{\sin\left(x\right)}{x}}\\ {\frac{x-1}{y+1}}\\ \end{pmatrix}=\begin{pmatrix} {\lim \limits_{\begin{pmatrix} {x,y}\\ \end{pmatrix}\to \begin{pmatrix} {0,0}\\ \end{pmatrix}}\frac{\sin\left(x\right)}{x}}\\ {\lim \limits_{\begin{pmatrix} {x,y}\\ \end{pmatrix}\to \begin{pmatrix} {0,0}\\ \end{pmatrix}}\frac{x-1}{y+1}}\\ \end{pmatrix}=\vec{L}$$Also,

$$\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}\frac{\sin\left(x\right)}{x}=\lim_{x\to 0}\frac{\sin\left(x\right)}{x}=1 \\ \lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}\frac{x-1}{y+1}=-1$$Thus,

$${\vec{L}=\begin{pmatrix} {1}\\ {-1}\\ \end{pmatrix}}$$

2. Let

$$f(x,y)=3x^2 +xy-5$$And

$$\lim \limits_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {-2}\\ {5}\\ \end{pmatrix}}f\left(x,y\right)=L$$Then

$$\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {-2}\\ {5}\\ \end{pmatrix}}f\left(x,y\right)=\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {-2}\\ {5}\\ \end{pmatrix}}3x^{2}+xy-5=3\left(\lim_{x\to -2}x\right)^{2}+\left(\lim_{x\to -2}x\right)·\left(\lim_{x\to 5}y\right)-5$$So,

$$L=12-10-5=-3$$

3. Let

$$f\left(x,y\right)=\frac{x^{2}-y^{2}}{x^{2}+2y^{2}}$$And

$$\lim \limits_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}f\left(x,y\right)=L$$Then

$$\lim \limits_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}\frac{x^{2}-y^{2}}{x^{2}+2y^{2}}=\frac{0}{0}$$If we evaluate the variables separately, we might be able to solve this indeterminate form:

$$\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}\frac{x^{2}-y^{2}}{x^{2}+2y^{2}}=\lim_{y\to 0}\left[\lim_{x\to 0}\left(\frac{x^{2}-y^{2}}{x^{2}+2y^{2}}\right)\right]=\lim_{y\to 0}\frac{-y^{2}}{2y^{2}}=-\frac{1}{2}$$$$\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {0}\\ {0}\\ \end{pmatrix}}\frac{x^{2}-y^{2}}{x^{2}+2y^{2}}=\lim_{x\to 0}\left[\lim_{y\to 0}\left(\frac{x^{2}-y^{2}}{x^{2}+2y^{2}}\right)\right]=\lim_{x\to 0}\frac{x^{2}}{x^{2}}=1$$The good news is that the limit is no longer indeterminate; the bad news is that now we have two answers! But what does this mean?

Uniqueness of Limits

With single-variable limits, we say that in order for the limit to exist, its value has to be the same whether we approach the limit point from the left or from the right. When we take the limit of a multivariable function, we can approach the limit point from a multitude of directions – even in the two-variable case, we already have infinite ways of approaching it!

For example:

• Iterated limits:

$$\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {x_{0}}\\ {y_{0}}\\ \end{pmatrix}}f\left(x,y\right)=\begin{cases} {\lim \limits_{y\to y_{0}}\left[\lim \limits_{x\to x_{0}}f\left(x,y\right)\right]=L_{a}}&{}\\ {\lim \limits_{x\to x_{0}}\left[\lim \limits_{y\to y_{0}}f\left(x,y\right)\right]=L_{b}}&{}\\ \end{cases}$$

• Straight line:

$$\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {x_{0}}\\ {y_{0}}\\ \end{pmatrix}}f\left(x,y\right)\Rightarrow\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {x_{0}}\\ {y_{0}+m\left(x-x_{0}\right)}\\ \end{pmatrix}}f\left(x,y\right)=\lim_{x\to x_{0}}\left(\lim_{y\to y_{0}+m\left(x-x_{0}\right)}f\left(x,y\right)\right)=L_{c}$$

• Parabola:

$$\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {x_{0}}\\ {y_{0}}\\ \end{pmatrix}}f\left(x,y\right)\Rightarrow\lim_{\begin{pmatrix} {x}\\ {y}\\ \end{pmatrix}\to \begin{pmatrix} {x_{0}}\\ {y_{0}+a\left(x-x_{0}\right)^{2}}\\ \end{pmatrix}}f\left(x,y\right)=\lim_{x\to x_{0}}\left(\lim_{y\to y_{0}+a\left(x-x_{0}\right)^{2}}f\left(x,y\right)\right)=L_{d}$$

The uniqueness theorem tells us that if a limit exists, its value is unique and independent of the path taken to approach the limit point. If the limit is path-dependent, it does not exist.

Then, if the limit $L$ exists,

$$L=L_a=L_b=L_c=L_d= \dots$$So, we can confidently say that the last limit from the previous section (Example 3) does not exist.

Conclusion

To sum up, in this topic we have learned that:

• We can extend single-variable concepts like the absolute value, open intervals, and the definition of limit itself to multivariable functions, keeping the same underlying logic.
• For a multivariable limit to exist, the definition must hold true, and, when evaluated, its value must be the same regardless of the path taken.