Limits

Section 2.1 Limits

Before we really start, let’s recall some useful notation.

Definition 2.1.1.

$\bbbn$ is the set $\{1,2,3,\cdots\}$ of all natural numbers.
$\bbbr$ is the set of all real numbers.
$\in$ is read “is an element of”.
$\notin$ is read “is not an element of”.
$\Set{A}{B}$ is read “the set of all $A$ such that $B$”
If $S$ is a set and $T$ is a subset of $S\text{,}$ then $S\setminus T$ is $\Set{x\in S}{x\notin T}\text{,}$ the set $S$ with the elements of $T$ removed. In particular, if $S$ is a set and $a$ is an element of $S\text{,}$ then $S\setminus\{a\}=\Set{x\in S}{x\ne\ a}$ is the set $S$ with the element $a$ removed.
If $n$ is a natural number, $\bbbr^n$ is used for both the set of $n$-component vectors $\llt x_1,x_2,\cdots,x_n \rgt$ and the set of points $(x_1,x_2,\cdots,x_n)$ with $n$ coordinates.
If $S$ and $T$ are sets, then $f:S\rightarrow T$ means that $f$ is a function which assigns to each element of $S$ an element of $T\text{.}$ The set $S$ is called the domain of $f\text{.}$
\begin{align*} [a,b]=\Set{x\in\bbbr}{a\le x\le b} \amp\amp (a,b]=\Set{x\in\bbbr}{a \lt x\le b}\\ [a,b)=\Set{x\in\bbbr}{a\le x \lt b}\amp\amp (a,b)=\Set{x\in\bbbr}{a \lt x \lt b} \end{align*}

The definition of the limit of a function of more than one variable looks just like the definition

Definition 1.3.3 in the CLP-1 text.

of the limit of a function of one variable. Very roughly speaking

\begin{gather*} \lim_{\vx\to \va} f(\vx) = \vL \end{gather*}

if $f(\vx)$ approaches $\vL$ whenever $\vx$ approaches $\va\text{.}$ Here is a more careful definition of limit.

Definition 2.1.2. Limit.

Let

$m$ and $n$ be natural numbers
²
In this text, we will interested in $m,n\in\big\{1,2,3\big\}\text{,}$ but the definition works for all natural numbers $m,n\text{.}$
$\displaystyle \va\in \bbbr^m$
the function $f(\vx)$ be defined for all $\vx$ near
³
To be precise, there is a number $r\gt 0$ such that $f(\vx)$ is defined for all $\vx$ obeying $|\vx-\va|\lt r\text{.}$
$\va$ and take values in $\bbbr^n$
$\displaystyle \vL\in\bbbr^n$

We write

\begin{gather*} \lim_{\vx\to \va} f(\vx) = \vL \end{gather*}

⁴

There is a precise, formal version of this definition that looks just like Definition 1.7.1 of the CLP-1 text.

the value of the function $f(\vx)$ is sure to be arbitrarily close to $\vL$ whenever the value of $\vx$ is close enough to $\va\text{,}$ without

⁵

You may find the condition “without being exactly $\va$” a little strange, but there is a good reason for it, which we have already seen in Calculus I. In the definition $f'(x) = \lim\limits_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}\text{,}$ the function whose limit is being taken, namely $\frac{f(x)-f(a)}{x-a}\text{,}$ is not defined at all at $x=a\text{.}$ This will again happen when we define derivatives of functions of more than one variable.

being exactly $\va\text{.}$

Now that we have extended the definition of limit, we can extend the definition of continuity.

Definition 2.1.3. Continuity.

Let

$m$ and $n$ be natural numbers
$\displaystyle \va\in\bbbr^m$
the function $f(\vx)$ be defined for all $\vx$ near $\va$ and take values in $\bbbr^n$

The function $f$ is continuous at a point $\va$ if
\begin{equation*} \lim_{\vx\to \va} f(\vx) = f(\va) \end{equation*}
The function $f$ is continuous on a set $D$ if it is continuous at every point of $D\text{.}$

Here are a few very simple examples. There will be some more substantial examples later — after, as we did in the CLP-1 text, we build some tools that can be used to build complicated limits from simpler ones.

Example 2.1.4.

If $f(x,y)$ is the constant function which always takes the value $L\text{,}$ then
\begin{equation*} \lim_{(x,y)\to(a,b)} f(x,y) = L \end{equation*}
If $f:\bbbr^2\rightarrow\bbbr^2$ is defined by $f(x,y) = (x,y)\text{,}$ then
\begin{equation*} \lim_{(x,y)\to(a,b)} f(x,y) = (a,b) \end{equation*}
By definition, as $(x,y)$ approaches $(a,b)\text{,}$ $x$ approaches $a$ and $y$ approaches $b\text{,}$ so that if $f:\bbbr^2\rightarrow\bbbr$ is defined by $f(x,y) = x\text{,}$ then
\begin{equation*} \lim_{(x,y)\to(a,b)} f(x,y) = a \end{equation*}
Similarly, if $g:\bbbr^2\rightarrow\bbbr$ is defined by $g(x,y) = y\text{,}$ then
\begin{equation*} \lim_{(x,y)\to(a,b)} g(x,y) = b \end{equation*}

Limits of multivariable functions have much the same computational properties as limits of functions of one variable. The following theorem summarizes a bunch of them. For simplicity, it concerns primarily real-valued functions. That is, functions that output real numbers as opposed to vectors. However it does contain one vector-valued function. The function $\vX$ in the theorem takes as input an $n$-component vector and returns an $m$-component vector. We will not deal with many vector-valued functions here in CLP-3, but we will see a lot in CLP-4.

Theorem 2.1.5. Arithmetic, and Other, Properties of Limits.

Let

$m$ and $n$ be natural numbers
$\va\in \bbbr^m$ and $\vb\in\bbbr^n$
$D$ be a subset of $\bbbr^m$ that contains all $\vx\in\bbbr^m$ that are near $\va$
$\displaystyle c,F,G\in\bbbr$

and

\begin{equation*} f,g:D\setminus\{\va\}\rightarrow\bbbr\qquad \vX:\bbbr^n\setminus\{\vb\}\rightarrow D\setminus\{\va\}\qquad \ga:\bbbr\rightarrow\bbbr \end{equation*}

Assume that

\begin{equation*} \lim_{\vx\to \va} f(\vx) = F\qquad \lim_{\vx\to \va} g(\vx) = G\qquad \lim_{\vy\to \vb} \vX(\vy) = \va\qquad \lim_{t\to F} \ga(t) = \ga(F) \end{equation*}

Then

$\lim\limits_{\vx\rightarrow \va}\big[f(\vx)+g(\vx)\big] =F+G$

$\lim\limits_{\vx\rightarrow \va}\big[f(\vx)-g(\vx)\big] =F-G$
$\lim\limits_{\vx\rightarrow \va} f(\vx)\,g(\vx) =FG$

$\lim\limits_{\vx\rightarrow \va} cf(\vx) =cF$
$\lim\limits_{\vx\rightarrow \va} \frac{f(\vx)}{g(\vx)} =\frac{F}{G} $ if $G\ne 0$
$\displaystyle \lim\limits_{\vy\rightarrow \vb} f\big(\vX(\vy)\big) =F$
$\displaystyle \lim\limits_{\vx\rightarrow \va} \ga\big(f(\vx)\big) =\ga(F)$

This shows that multivariable limits interact very nicely with arithmetic, just as single variable limits did. Also recall, from Theorem 1.6.8 in the CLP-1 text,

Theorem 2.1.6.

The following functions are continuous everywhere in their domains

polynomials, rational functions
roots and powers
trig functions and their inverses
exponential and the logarithm

Example 2.1.7.

In this example we evaluate

\begin{equation*} \lim_{(x,y)\rightarrow (2,3)} \frac{x+\sin y}{x^2y^2+1} \end{equation*}

as a typical application of Theorem 2.1.5. Here “$\eqover{a}$” means that part (a) of Theorem 2.1.5 justifies that equality. Start by computing separately the limits of the numerator and denominator.

\begin{align*} &\lim_{(x,y)\rightarrow (2,3)} \big(x+\sin y\big)\ \eqover{a} \lim_{(x,y)\rightarrow (2,3)} x +\lim_{(x,y)\rightarrow (2,3)}\sin y\\ &\hskip0.5in\eqover{e} \lim_{(x,y)\rightarrow (2,3)} x +\sin\Big(\lim_{(x,y)\rightarrow (2,3)} y\Big)\\ &\hskip0.5in=\ 2+\sin 3\\ &\lim_{(x,y)\rightarrow (2,3)} \big(x^2y^2+1\big)\ \eqover{a}\ \lim_{(x,y)\rightarrow (2,3)} x^2y^2 +\lim_{(x,y)\rightarrow (2,3)}1\\ &\hskip0.5in\eqover{b}\ \Big(\lim_{(x,y)\rightarrow (2,3)} x\Big) \Big(\lim_{(x,y)\rightarrow (2,3)} x\Big) \Big(\lim_{(x,y)\rightarrow (2,3)} y\Big) \Big(\lim_{(x,y)\rightarrow (2,3)} y\Big)+1\\ &\hskip0.5in=\ 2^23^2+1 \end{align*}

Since the limit of the denominator is nonzero, we can simply divide.

\begin{align*} \lim_{(x,y)\rightarrow (2,3)} \frac{x+\sin y}{x^2y^2+1}\ &\eqover{c}\ \frac {\lim\limits_{(x,y)\rightarrow (2,3)}(x+\sin y)} {\lim\limits_{(x,y)\rightarrow (2,3)}(x^2y^2+1)}\\ &=\ \frac{2+\sin 3}{37} \end{align*}

Here we have used that $\sin x$ is a continuous function.

While the CLP-1 text’s Definition 1.3.3 of the limit of a function of one variable, and our Definition 2.1.2 of the limit of a multivariable function look virtually identical, there is a substantial practical difference between the two. In dimension one, you can approach a point from the left or from the right and that’s it. There are only two possible directions of approach. In two or more dimensions there is “much more room” and there are infinitely many possible types of approach. One can even spiral in to a point. See the middle and right hand figures below.

The next few examples illustrate the impact that the“extra room” in dimensions greater than one has on limits.

Example 2.1.8.

As a second example, we consider $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{x^2y}{x^2+y^2}\text{.}$ In this example, both the numerator, $x^2y\text{,}$ and the denominator, $x^2+y^2\text{,}$ tend to zero as $(x,y)$ approaches $(0,0)\text{,}$ so we have to be more careful.

A good way to see the behaviour of a function $f(x,y)$ when $(x,y)$ is close to $(0,0)$ is to switch to the polar coordinates, $r,\theta\text{,}$ that are defined by

\begin{align*} x&=r\cos\theta\\ y&=r\sin\theta \end{align*}

The points $(x,y)$ that are close to $(0,0)$ are those with small $r\text{,}$ regardless of what $\theta$ is. Recall that $\lim\limits_{{(x,y)\rightarrow (0,0)}} f(x,y)=L$ when $f(x,y)$ approaches $L$ as $(x,y)$ approaches $(0,0)\text{.}$ Substituting $x=r\cos\theta\text{,}$ $y=r\sin\theta$ into that statement turns it into the statement that $\lim\limits_{{(x,y)\rightarrow (0,0)}} f(x,y)=L$ when $f(r\cos\theta,r\sin\theta)$ approaches $L$ as $r$ approaches $0\text{.}$ For our current example

\begin{gather*} \frac{x^2y}{x^2+y^2} =\frac{(r\cos\theta)^2(r\sin\theta)}{r^2} =r\cos^2\theta\sin\theta \end{gather*}

As $\big|r\cos^2\theta\sin\theta\big|\le r$ tends to $0$ as $r$ tends to $0$ (regardless of what $\theta$ does as $r$ tends to $0$) we have

\begin{equation*} \lim\limits_{(x,y)\rightarrow (0,0)}\frac{x^2y}{x^2+y^2}=0 \end{equation*}

Example 2.1.9.

As a third example, we consider $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{x^2-y^2}{x^2+y^2}\text{.}$ Once again, the best way to see the behaviour of $f(x,y)=\frac{x^2-y^2}{x^2+y^2}$ for $(x,y)$ close to $(0,0)$ is to switch to polar coordinates.

\begin{equation*} f(x,y)=\frac{x^2-y^2}{x^2+y^2}=\frac{(r\cos\theta)^2-(r\sin\theta)^2}{r^2} =\cos^2\theta-\sin^2\theta =\cos(2\theta) \end{equation*}

Note that, this time, $f$ is independent of $r$ but does depend on $\theta\text{.}$ Here is a greatly magnified sketch of a number of level curves for $f(x,y)\text{.}$

Observe that

as $(x,y)$ approaches $(0,0)$ along the ray with $2\theta =30^\circ\text{,}$ $f(x,y)$ approaches the value $\frac{\sqrt{3}}{2}$ (and in fact $f(x,y)$ takes the value $\cos(30^\circ)=\frac{\sqrt{3}}{2}$ at every point of that ray)
as $(x,y)$ approaches $(0,0)$ along the ray with $2\theta =60^\circ\text{,}$ $f(x,y)$ approaches the value $\frac{1}{2}$ (and in fact $f(x,y)$ takes the value $\cos(60^\circ)=\frac{1}{2}$ at every point of that ray)
as $(x,y)$ approaches $(0,0)$ along the ray with $2\theta =90^\circ\text{,}$ $f(x,y)$ approaches the value $0$ (and in fact $f(x,y)$ takes the value $\cos(90^\circ)=0$ at every point of that ray)
and so on

So there is not a single number $L$ such that $f(x,y)$ approaches $L$ as $r=|(x,y)|\rightarrow 0\text{,}$ no matter what the direction of approach is. The limit $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{x^2-y^2}{x^2+y^2}$ does not exist.

Here is another way to come to the same conclusion.

Pick any really small positive number. We’ll use $10^{-137}$ as an example.
Pick any real number $F$ between $-1$ and $1\text{.}$ We’ll use $F=\frac{\sqrt{3}}{2}$ as an example.
Looking at the sketch above, we see that $f(x,y)$ takes the value $F$ along an entire ray $\theta={\rm const}\text{,}$ $r\gt 0\text{.}$ In the case $F=\frac{\sqrt{3}}{2}\text{,}$ the ray is $2\theta=30^{\circ}\text{,}$ $r\gt 0\text{.}$ In particular, because the ray extends all the way to $(0,0)\text{,}$ $f$ takes the value $F$ for some $(x,y)$ obeying $|(x,y)|\lt 10^{-137}\text{.}$
That is true regardless of which really small number you picked. So $f(x,y)=\frac{x^2-y^2}{x^2+y^2}$ does not approach any single value as $r=|(x,y)|$ approaches $0$ and we conclude that $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{x^2-y^2}{x^2+y^2}$ does not exist.

Subsection 2.1.1 Optional — A Nasty Limit That Doesn’t Exist

Example 2.1.10.

In this example we study the behaviour of the function

\begin{equation*} f(x,y)=\begin{cases} \frac{(2x-y)^2}{x-y} & \text{if } x\ne y\\ 0 & \text{if } x=y \end{cases} \end{equation*}

as $(x,y)\rightarrow (0,0)\text{.}$ Here is a graph of the level curve, $f(x,y)=-3\text{,}$ for this function.

Here is a larger graph of level curves, $f(x,y)=c\text{,}$ for various values of the constant $c\text{.}$

As before, it helps to convert to polar coordinates — it is a good approach

⁶

Not just a pun.

. In polar coordinates

\begin{equation*} f(r\cos\theta,r\sin\theta) =\begin{cases} r\frac{(2\cos\theta-\sin\theta)^2}{\cos\theta-\sin\theta} & \text{if } \cos\theta\ne\sin\theta \\ 0 & \text{if } \cos\theta=\sin\theta \end{cases} \end{equation*}

If we approach the origin along any fixed ray $\theta=\text{const}\text{,}$ then $f(r\cos\theta,r\sin\theta)$ is the constant $\frac{(2\cos\theta-\sin\theta)^2}{\cos\theta-\sin\theta}$ (or $0$ if $\cos\theta=\sin\theta$) times $r$ and so approaches zero as $r$ approaches zero. You can see this in the figure below, which shows the level curves again, with the rays $\theta=\frac{1}{8}\pi$ and $\theta=\frac{3}{16}\pi$ superimposed.

If you move towards the origin on either of those rays, you first cross the $f=3$ level curve, then the $f=2$ level curve, then the $f=1$ level curve, then the $f=\frac{1}{2}$ level curve, and so on.

That $f(x,y)\rightarrow 0$ as $(x,y)\rightarrow (0,0)$ along any fixed ray is suggestive, but does not imply that the limit exists and is zero. Recall that to have $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0\text{,}$ we need $f(x,y)\rightarrow 0$ no matter how $(x,y)\rightarrow (0,0)\text{.}$ It is not sufficient to check only straight line approaches.

In fact, the limit of $f(x,y)$ as $(x,y)\rightarrow (0,0)$ does not exist. A good way to see this is to observe that if you fix any $r \gt 0\text{,}$ no matter how small, $f(x,y)$ takes all values from $-\infty$ to $+\infty$ on the circle $x^2+y^2=r^2\text{.}$ You can see this in the figure below, which shows the level curves yet again, with a circle $x^2+y^2=r^2$ superimposed. For every single $-\infty \lt c \lt \infty\text{,}$ the level curve $f(x,y)=c$ crosses the circle.

Consequently there is no one number $L$ such that $f(x,y)$ is close to $L$ whenever $(x,y)$ is sufficiently close to $(0,0)\text{.}$ The limit $\lim_{(x,y)\rightarrow(0,0)} f(x,y)$ does not exist.

Another way to see that $f(x,y)$ does not have any limit as $(x,y)\rightarrow (0,0)$ is to show that $f(x,y)$ does not have a limit as $(x,y)$ approaches $(0,0)$ along some specific curve. This can be done by picking a curve that makes the denominator, $x-y\text{,}$ tend to zero very quickly. One such curve is $x-y=x^3$ or, equivalently, $y=x-x^3\text{.}$ Along this curve, for $x\ne 0\text{,}$

\begin{align*} f(x,x-x^3)&=\frac{{(2x-x+x^3)}^2}{x-x+x^3} =\frac{{(x+x^3)}^2}{x^3}\\ &=\frac{{(1+x^2)}^2}{x}\longrightarrow \begin{cases}+\infty & \text{as $x\rightarrow 0$ with } x \gt 0 \\ -\infty & \text{as $x\rightarrow 0$ with } x \lt 0 \end{cases} \end{align*}

The choice of the specific power $x^3$ is not important. Any power $x^p$ with $p \gt 2$ will have the same effect.

If we send $(x,y)$ to $(0,0)$ along the curve $x-y=ax^2$ or, equivalently, $y=x-ax^2\text{,}$ where $a$ is a nonzero constant,

\begin{align*} \lim_{x\rightarrow 0}f(x,x-ax^2) &=\lim_{x\rightarrow 0}\frac{{(2x-x+ax^2)}^2}{x-x+ax^2} =\lim_{x\rightarrow 0}\frac{{(x+ax^2)}^2}{ax^2}\\ &=\lim_{x\rightarrow 0}\frac{{(1+ax)}^2}{a} =\frac{1}{a} \end{align*}

This limit depends on the choice of the constant $a\text{.}$ Once again, this proves that $f(x,y)$ does not have a limit as $(x,y)\rightarrow (0,0)\text{.}$

Exercises 2.1.2 Exercises

Exercise Group.

Exercises — Stage 1

1.

Suppose $f(x,y)$ is a function such that $\lim\limits_{(x,y)\to(0,0)}f(x,y)=10\text{.}$

True or false: $|f(0.1,0.1)-10| \lt |f(0.2,0.2)-10|$

2.

A millstone pounds wheat into flour. The wheat sits in a basin, and the millstone pounds up and down.

Samples of wheat are taken from various places along the basin. Their diameters are measured and their position on the basin is recorded.

Consider this claim: “As the particles get very close to the millstone, the diameters of the particles approach 50 $\mu$m.” In this context, describe the variables below from Definition 2.1.2.

$\displaystyle \mathbf x$
$\displaystyle \mathbf a$
$\displaystyle \mathbf L$

3.

Let $f(x,y)=\dfrac{x^2}{x^2+y^2}\text{.}$

Find a ray approaching the origin along which $f(x,y)=1\text{.}$
Find a ray approaching the origin along which $f(x,y)=0\text{.}$
What does the above work show about a limit of $f(x,y)\text{?}$

4.

Let $f(x,y)=x^2-y^2$

Express the function in terms of the polar coordinates $r$ and $\theta\text{,}$ and simplify.
Suppose $(x,y)$ is a distance of 1 from the origin. What are the largest and smallest values of $f(x,y)\text{?}$
Let $r \gt 0\text{.}$ Suppose $(x,y)$ is a distance of $r$ from the origin. What are the largest and smallest values of $f(x,y)\text{?}$
Let $\epsilon \gt 0\text{.}$ Find a positive value of $r$ that guarantees $|f(x,y)| \lt \epsilon$ whenever $(x,y)$ is at most $r$ units from the origin.
What did you just show?

5.

Suppose $f(x,y)$ is a polynomial. Evaluate $\lim\limits_{(x,y)\to(a,b)}f(x,y)\text{,}$ where $(a,b)\in\mathbb R^2\text{.}$

Exercise Group.

Exercises — Stage 2

6.

Evaluate, if possible,

$\displaystyle \ds\lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)$
$\displaystyle \ds\lim_{(x,y)\rightarrow(0,0)}\ \frac{x}{x^2+y^2}$
$\displaystyle \ds\lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2}{x^2+y^2}$
$\displaystyle \ds\lim_{(x,y)\rightarrow(0,0)}\ \frac{x^3}{x^2+y^2}$
$\displaystyle \ds\lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2y^2}{x^2+y^4}$
$\displaystyle \ds\lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy}$

7. (✳).

Find the limit: $\ds \lim_{(x,y)\to(0,0)}\frac{x^8+y^8}{x^4+y^4}\text{.}$
Prove that the following limit does not exist: $\ds \lim_{(x,y)\to(0,0)}\frac{xy^5}{x^8+y^{10}}\text{.}$

8. (✳).

Evaluate each of the following limits or show that it does not exist.

$\displaystyle \ds\lim_{(x,y)\rightarrow (0,0)}\frac{x^3-y^3}{x^2+y^2}$
$\displaystyle \ds\lim_{(x,y)\rightarrow (0,0)}\frac{x^2-y^4}{x^2+y^4}$

Exercise Group.

Exercises — Stage 3

9. (✳).

Evaluate each of the following limits or show that it does not exist.

$\displaystyle \ds\lim_{(x,y)\rightarrow (0,0)}\frac{2x^2 + x^2y - y^2x + 2y^2}{x^2 + y^2}$
$\displaystyle \ds\lim_{(x,y)\rightarrow(0,1)} \frac{x^2y^2 -2 x^2y + x^2} {(x^2 + y^2-2y+1)^2}$

10.

Define, for all $(x,y)\ne(0,0)\text{,}$ $f(x,y)=\frac{x^2y}{x^4+y^2}\text{.}$

Let $0\le \theta \lt 2\pi\text{.}$ Compute $\ds\lim_{r\rightarrow 0^+}f(r\cos\theta,r\sin\theta)\text{.}$
Compute $\ds\lim_{x\rightarrow 0}f(x,x^2)\text{.}$
Does $\ds\lim_{(x,y)\rightarrow (0,0)}f(x,y)$ exist?

11. (✳).

Compute the following limits or explain why they do not exist.

$\displaystyle \ds\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{x^2+y^2}$
$\displaystyle \ds\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(xy)}{x^2+y^2}$
$\displaystyle \ds\lim_{(x,y)\rightarrow(-1,1)}\frac{x^2+2xy^2+y^4}{1+y^4}$
$\displaystyle \ds\lim_{(x,y)\rightarrow(0,0)}|y|^x$

12.

Evaluate each of the following limits or show that it does not exist.

$\displaystyle \ds\lim_{(x,y)\rightarrow (0,0)}\begin{cases} \frac{x^2}{y-x} &\text{if }y\ne x \\ 0 & \text{if }y=x \end{cases}$
$\displaystyle \ds\lim_{(x,y)\rightarrow (0,0)}\begin{cases} \frac{x^8}{y-x} &\text{if }y\ne x \\ 0 & \text{if }y=x \end{cases}$

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