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Multivariable Limits and Continuity
Definition Limit of a function of two variables We say that a function $f(x, y)$ approaches the limit $L$ as $(x, y)$ approaches $(a , b)$, and write $$\lim_{(x,y) \to (a,b)} f(x,y) = L$$ if, for every number $\epsilon > 0$, there exists a corresponding number $\delta > 0$ such that for all $(x, y)$ in the domain of $f$, $$\|f(x, y) - L\| < \epsilon$$ whenever $$0<\sqrt{(x-a)^2+(y-b)^2} < \delta.$$ Limit of a function of n variables Given a function $f : D \rightarrow \mathbb{R}$, $D \subseteq \mathbb{R}^n$, we say that the limit of $f (r)$ as $r$ approaches $a$ exists ($\underset{\textbf{r} \to a}{\lim} f(\textbf{r}) = L$), where $\textbf{r}$ is a $n$ dimensional vector and has value $L$ if and only if for every real number $\epsilon > 0$ there exists a real number $\delta > 0$ such that $$\|f(\textbf{r})-L\| <\epsilon$$ whenever $$0< \|\|\textbf{r}-a\|\| < \delta$$We then write $$\lim_{ \textbf{r} \to a} f(\textbf{r}) = L$$ -3
Motivation In the $17^{th}$ century, mathematicians were eagerly interested in the study of the motion of stars and planets, the motion for objects on or close by to the earth. This inquiry involved not only the speed of the object but also its direction of motion at any instant, which was along a tangent line to the path of motion. For finding the speed of a moving object and also the tangent to a curve, the concept of a limit is very important. Using limits we can explain how a function varies. Some functions vary continuously; that is a little change in $x$ produces a little change in $f(x)$ while some other functions can have values that jump randomly, or tend to increase or decrease without bound. Hence, to distinguish between these behaviours precisely the notion of a limit is very useful. -3
Bird's Eye View When we write $\underset{(x,y)\to (a,b)}{\lim} f(x,y) = L$, what we mean is that the limiting value of $ f (x, y) $, as $ (x, y)$ approaches $(a, b)$, is $L$. In simpler terms, if my $f(x,y)$ is arbitrarily close to $L$, then $(x, y)$ will lie in the neighbourhood of $(a,b)$ While computing $\underset{(x,y)\to (a,b)}{\lim} f(x,y)$ it is crucial to note that $ (x, y)$ is never equal to $(a, b)$, the reason being that the function may not be defined at the point $(a,b)$. This allows us to calculate limits, if they exist, of a function $f$,defined on $D\subset\mathbb{R}^{2}$, at some point $(a,b)\in\mathbb{R}^{2}\setminus D $ There are various notations for expressing the limits: $\underset{(x,y)\to (a,b)}{\lim} f(x,y) = L$. $\lim_{x \to a, y \to b} f(x,y) = L$. $ f (x, y) $ approaches $L$ as $ (x, y)$ approaches $(a, b)$. -3
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Video 1: Epsilon-Delta definition of limits of Multivariable Functions
Context of the Definition Let $f$ be a function of two variables, $x$ and $y$. The limit of $f(x,y)$ as $(x,y)$ approaches $(a,b)$ is $L$, $$\lim_{(x,y) \to (a,b)} f(x,y) = L$$ if $\forall \epsilon > 0$ there $\exists \delta > 0$, such that $\|f(x,y)-L\|<\epsilon$, whenever $0< \sqrt{(x-a)^2 + (y-b)^2} < \delta$. This means that for each $\epsilon > 0$, there exists a $\delta > 0$, such that for all points $(x,y)$ in a $\delta$ disk around $(a,b)$, except possibly for $(a,b)$ itself, the value of $f(x,y)$ is no more than $\epsilon$ away from $L$. 0
Methods of computing the limit 1) Substitute the variables If you want the limit at point $(a, b)$, and the function is continuous at $(a, b)$, then simply substitute the values of $(a, b)$ into the function. This is possible if and only if, you don’t get $\frac{0}{0}$, divide by 0, the square root of a negative number or any indeterminate forms. For example: $\lim_{(x,y) \to (a,b)} e^{x+y^2} = e^{0+0^2} = e^0 = 1$, $\lim_{(x,y) \to (1,2)} x^6y + 2xy = (1)^62 + 2(1)(2) = 2 + 4 = 6$.\(\) 2) Change the co-ordinates or use the Squeeze theorem In the case, when you can't find the limit of a function by simply substituting the variables, without getting $\frac{0}{0}$, dividing by 0, or taking the square root of a negative number, or indeterminate forms. Then, your next best guess is to use the Squeeze Theorem, to trap these indeterminate functions between the easy, nice functions. If those easy, nice functions approach the same limit, then the indeterminate function, trapped between them, must also approach that limit. Example using squeeze theorem: $$\lim_{(x,y) \to (0,0)} x^4 \sin \left( \frac{1}{x^2 + \|y\|} \right)$$ We know that $-1 \leq \sin \left( \frac{1}{x^2 + \|y\|} \right) \leq 1$.By using this, we can make the function easier. By multiplying $x^4$ to the equation, we get: $-x^4 \leq x^4 \sin \left( \frac{1}{x^2 + \|y\|} \right) \leq x^4$. Next, we'll take the limits: $$0 = \lim_{(x,y) \to (0,0)} -x^4 \leq x^4 \sin \left( \frac{1}{x^2 + \|y\|} \right) \leq \lim_{(x,y) \to (0,0)} x^4 = 0$$ So the limit of our example function is stuck between the two limits of the simpler functions, both having limits equal to 0. So by the Squeeze Theorem we get: $$\lim_{(x,y) \to (0,0)} x^4 \sin \left( \frac{1}{x^2 + \|y\|} \right) = 0$$ Sometimes, changing coordinates may be useful. Consider the example below Using polar coordinates, Find $\lim_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2}$. The relationship between a point in rectangular coordinates $(x,y)$ and the polar coordinates $(r, \theta)$ with $r \geq 0$ is $$x = r \cos \theta$$$$y = r \sin\theta$$ from which we can substitute, $$\lim_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2} = \lim_{(x,y) \to (0,0)} \frac{r^3\cos^3\theta + r^3\sin^3 \theta}{r^2\cos^2\theta + r^2\sin^2 \theta}$$ Also saying $(x,y) \to (0,0)$ is equivalent to $r \to 0^+$. Hence, we have: $$= \lim_{r \to 0^+} \frac{r^3(\cos^3\theta + \sin^3 \theta)}{r^2(\cos^2\theta + \sin^2 \theta)} = \lim_{r \to 0^+} \frac{r^3(\cos^3\theta + \sin^3 \theta)}{r^2}$$ $$= \lim_{r \to 0^+} r(\cos^3\theta + \sin^3 \theta) = 0$$ 3) Prove that the limit does not exist If with above two methods, you were not able to find the limits of a function, then you basically want to prove that the limit does not exist. In single variable, you can do this by proving that the limit from the left and the limit from the right are not equal. In multivariable, you just need to prove that the limit isn’t the same for any two directions. Example: $$\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^2} $$ One way to do this is go from the $x$ direction (set $y = 0$ and then find the limit), and then the $y$ direction (set $x = 0$ and then find the limit). $x$ direction: $$\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^2} = \lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + 0 ^2} = \lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 } =1 $$ $y$ direction: $$\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^2} = \lim_{(x,y) \to (0,0)} \frac{0^2}{0^2 + y^2} = \lim_{(x,y) \to (0,0)} \frac{0}{y^2 } =0 $$ The limit in the $x$ direction and the limit in the $y$ direction are not equal, so then the limit does not exist. Methods for proving the limit does not exist 1. Take the limit in the $x$ direction by setting $y = 0$ and the limit in the $y$ direction by setting $x = 0$. 2. Take the limit along the line $y = x$, by setting $y = x$ in the limit. This will make things easier, as somethings will cancel out to. Then also take the limit in the $x$ or $y$ direction and see what happens. 0
Example: 1. Consider $\lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} $. This gives the $\frac{0}{0}$ form, which means, that the function is undefined at $(1,1)$. We attempt to evaluate the limit by aproaching the point $(1,1)$ along four different paths. $y = x : \lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} = \lim_{(x,y) \to (1,1)} \frac{0}{x-1} = 0$ $y = 2 - x : \lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} = \lim_{(x,y) \to (1,1)} \frac{2(x - 1)}{x-1} = 2$ $y = x^2 : \lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} = \lim_{(x,y) \to (1,1)} \frac{x - x^2)}{x-1} = -1$ $y = 1 : \lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} = \lim_{(x,y) \to (1,1)} \frac{x - 1}{x-1} = 1$ The resulting limiting values found from these following different paths are all different. From this we conclude that limits doesn't exists. 0
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Video 2: Different paths of approach to limit point
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Video 3: Visualizing the limit of a function $f(x,y) = \frac{x-y}{x-1}$ along different paths of approach at point $(1,1)$
2. Consider the function $f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}$. Find $\lim_{(x,y) \to(0,0)} \frac{x^2 - y^2}{x^2 + y^2}$. We will aproach the limit along the path $x = 0$ and $y = 0$. 1. Limit along the path $x = 0$. $$\lim_{(x,y) \to(a,b)} \frac{x^2 - y^2}{x^2 + y^2} = \lim_{(x,y) \to(0,0)} \frac{x^2 - y^2}{x^2 + y^2} \|_{x=0}$$ $$= \lim_{(x,y) \to(0,0)} \frac{- y^2}{ y^2}$$ $$= \lim_{(x,y) \to(0,0)} -1$$ $$= -1$$ 2. Limit along the path $y = 0$. $$\lim_{(x,y) \to(a,b)} \frac{x^2 - y^2}{x^2 + y^2} = \lim_{(x,y) \to(0,0)} \frac{x^2 - y^2}{x^2 + y^2} \|_{y=0}$$ $$\lim_{(x,y) \to(0,0)} \frac{x^2 }{x^2 }$$ $$= \lim_{(x,y) \to(0,0)} 1$$ $$= 1$$ 3. Conclusion: The limits are different, therefore $\lim_{(x,y) \to(0,0)} \frac{x^2 - y^2}{x^2 + y^2}$ does not exist. 0
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Video 4: Visualizing the limit of a function $f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}$ along the path $x=0$ and $y=0$
Continuity of Multivariable Functions Definition A function of many variables $f : D\subseteq\mathbb{R}^{n} \rightarrow \mathbb{R}$ is continuous at a point $r_0 \in D $ if $$f(r_0) = \lim_{\textbf{r}\to r_0} f (\textbf{r})$$ i.e. given any $\epsilon>0$, there exists a $\delta>0$ such that, $$ \lvert f(\textbf{r}) - f(\textbf{r}_{0})\rvert < \epsilon $$ whenever, $$ \lVert \textbf{r} - \textbf{r}_{0} \rVert < \delta$$ 0
Context of the Definition A function $f (x, y)$ is said to be continuous at a point $(a, b)$, if the following results to be true: $(a, b)$ is in the domain of $f$. $\lim_{(x,y)\to(a,b)} f (x, y)$ exists. $\lim_{(x,y)\to(a,b)} f (x, y) = f(a,b)$. If a function $f$ is not continuous at a point $(a, b)$, then it is discontinuous at $(a, b)$. A function $f$ is continuous on a set $D$ if and only if, it is continuous at each point in $D$. Thus, to know whether a function is continuous at a point, we find the limit of the function at the point and simply substitute the point. For the multivariable functions, the following outcomes are true : The polynomial functions are continuous. The rational functions are continuous in their domain. The sum or product of continuous functions, defined on the same domain, is a continuous function. If $f$ is a continuous function and $\alpha\in \mathbb{R} (or \mathbb{C})$ then, $\alpha f$ is a continuous function The quotient of two continuous functions is continuous until and unless the denominator isn't 0. If $f (x, y)$ is continuous and $g (x)$ is defined and continuous on the range of $f$, then $g (f (x, y))$ is also continuous. 0
Examples 1. Is $f (x, y) = x^2y+3x^3y^5-x+2y$ continuous at $ (0, 0)$? Where is it continuous? $\rightarrow$ $f (x, y)$ is continuous on $R^2$ being a polynomial function, and also continuous at $ (0, 0)$. 2. Where is $f(x, y) = \frac{3x-2y}{x^2+y^2}$ continuous? $\rightarrow$ As $f(x, y)$ being the quotient of two continuous functions, it is continuous untill its denominator is not equal to 0 that is on $R^{2}\setminus \{ (0, 0)\}$. 3. Where is $f (x, y) = \frac{1}{x^2-y}$ continuous? $\rightarrow$ As $f(x, y)$ being the quotient of two continuous functions, it is continuous untill it's denominator is not 0. The denominator will become $0$ along the parabola $y = x^2$. Therefore, $f(x, y)$ is continuous on {$(x, y) \in R^2 \| y \neq x^2$}. 0
Polynomials are continuous everywhere and the Rational functions are continuous everywhere they are defined. The function $f(x,y)=\frac{3x^2y}{x^2+y^2}$, is not defined at $(0,0)$, and hence not continuous at $(0,0)$. However, we know that $\lim_{(x,y)\to(0,0)}f(x,y)=0$, so we can easily "fix" the problem, by extending the definition of $f$ so that $f(0,0)=0$. Note that in contrast to this example, we cannot fix $f(x,y)=\frac{3x^2y}{x^2+y^2}$ at $(0,0)$ because the limit does not exist. No matter what value we try to assign to $f$ at $(0,0)$ the surface will have a "jump" there. 0
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Video 5: The function $f(x,y) = \frac{3x^2y}{x^2 + y^2}$ is not continuous at point $(0,0)$
Applications The limit concept is essential to understanding the real number system and its distinguishing characteristics. In one sense real numbers can be defined as the numbers that are the limits of convergent sequences of rational numbers. Limits are the method by which the derivative, or rate of change, of a function, is calculated. Limits are also a key to calculate integrals. The integral calculates the entire area of a region by summing up an infinite number of small pieces of it. 0
History In the $3^{rd}$ century B.C, Archimedes of Syracuse first developed the idea of limits to measure curved figures and the volume of a sphere. By carving these figures into small pieces that can be approximated, then increasing the number of pieces, the limit of the sum of pieces can give the desired quantity. Archimedes' thesis, "The Method", was lost until 1906, when mathematicians discovered that Archimedes came on the brink of discovering infinitesimal calculus. As Archimedes's work was unknown until the twentieth century, others developed the modern mathematical concept of limits. Sir Issac Newton and German Gottfried Wilhelm von Leibniz developed independently, the general principles of calculus (of which the idea of limits is a crucial part) within the seventeenth century. 0
Pause and Ponder What is the role of the limits in derivatives of a function? The derivative of a function $f(x)$ at a point $x_0$ is the limit of the difference quotient at $x = x_0$, as the increment $h = x- x_0$ of the independent variable $x$ approaches 0. In mathematical words: $f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0+h) - f(x_0)}{h}$ So derivatives are a kind of special limits, that gives useful information about the functions and their behavior. 0
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Figure 1: Limit Definition of the Derivative [1]
References https://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx http://www.math.ucsd.edu/~jbriones/math20c.15w/handouts/Math20C_Handout5.pdf http://ccjohnson.org/math2060/lectures/Lecture10-LimitsContinuity.pdf https://ksuweb.kennesaw.edu/~plaval/math2203/funcnD_limcont.pdf Book - Thomas’ Calculus Early Transcendentals 0
Figures [1] Reproduced from https://www.acsu.buffalo.edu/~mbschild/images.html 4

Multivariable Limits and Continuity

Definition

Limit of a function of two variables

We say that a function $f(x, y)$ approaches the limit $L$ as $(x, y)$ approaches $(a , b)$, and write $$\lim_{(x,y) \to (a,b)} f(x,y) = L$$ if, for every number $\epsilon > 0$, there exists a corresponding number $\delta > 0$ such that for all $(x, y)$ in the domain of $f$, $$|f(x, y) - L| < \epsilon$$ whenever $$0<\sqrt{(x-a)^2+(y-b)^2} < \delta.$$

Limit of a function of n variables

Given a function $f : D \rightarrow \mathbb{R}$, $D \subseteq \mathbb{R}^n$, we say that the limit of $f (r)$ as $r$ approaches $a$ exists ($\underset{\textbf{r} \to a}{\lim} f(\textbf{r}) = L$), where $\textbf{r}$ is a $n$ dimensional vector and has value $L$ if and only if for every real number $\epsilon > 0$ there exists a real number $\delta > 0$ such that $$|f(\textbf{r})-L| <\epsilon$$ whenever $$0< ||\textbf{r}-a|| < \delta$$We then write $$\lim_{ \textbf{r} \to a} f(\textbf{r}) = L$$

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Motivation

In the $17^{th}$ century, mathematicians were eagerly interested in the study of the motion of stars and planets, the motion for objects on or close by to the earth. This inquiry involved not only the speed of the object but also its direction of motion at any instant, which was along a tangent line to the path of motion.

For finding the speed of a moving object and also the tangent to a curve, the concept of a limit is very important. Using limits we can explain how a function varies. Some functions vary continuously; that is a little change in $x$ produces a little change in $f(x)$ while some other functions can have values that jump randomly, or tend to increase or decrease without bound. Hence, to distinguish between these behaviours precisely the notion of a limit is very useful.

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Bird's Eye View

When we write $\underset{(x,y)\to (a,b)}{\lim} f(x,y) = L$, what we mean is that the limiting value of $ f (x, y) $, as $ (x, y)$ approaches $(a, b)$, is $L$. In simpler terms, if my $f(x,y)$ is arbitrarily close to $L$, then $(x, y)$ will lie in the neighbourhood of $(a,b)$

While computing $\underset{(x,y)\to (a,b)}{\lim} f(x,y)$ it is crucial to note that $ (x, y)$ is never equal to $(a, b)$, the reason being that the function may not be defined at the point $(a,b)$. This allows us to calculate limits, if they exist, of a function $f$,defined on $D\subset\mathbb{R}^{2}$, at some point $(a,b)\in\mathbb{R}^{2}\setminus D $

There are various notations for expressing the limits:

$\underset{(x,y)\to (a,b)}{\lim} f(x,y) = L$.
$\lim_{x \to a, y \to b} f(x,y) = L$.
$ f (x, y) $ approaches $L$ as $ (x, y)$ approaches $(a, b)$.

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Video 1: Epsilon-Delta definition of limits of Multivariable Functions

Context of the Definition

Let $f$ be a function of two variables, $x$ and $y$. The limit of $f(x,y)$ as $(x,y)$ approaches $(a,b)$ is $L$, $$\lim_{(x,y) \to (a,b)} f(x,y) = L$$ if $\forall \epsilon > 0$ there $\exists \delta > 0$, such that $|f(x,y)-L|<\epsilon$, whenever $0< \sqrt{(x-a)^2 + (y-b)^2} < \delta$.

This means that for each $\epsilon > 0$, there exists a $\delta > 0$, such that for all points $(x,y)$ in a $\delta$ disk around $(a,b)$, except possibly for $(a,b)$ itself, the value of $f(x,y)$ is no more than $\epsilon$ away from $L$.

Methods of computing the limit

1) Substitute the variables

If you want the limit at point $(a, b)$, and the function is continuous at $(a, b)$, then simply substitute the values of $(a, b)$ into the function. This is possible if and only if, you don’t get $\frac{0}{0}$, divide by 0, the square root of a negative number or any indeterminate forms.

For example:

$\lim_{(x,y) \to (a,b)} e^{x+y^2} = e^{0+0^2} = e^0 = 1$,
$\lim_{(x,y) \to (1,2)} x^6y + 2xy = (1)^62 + 2(1)(2) = 2 + 4 = 6$.

2) Change the co-ordinates or use the Squeeze theorem

In the case, when you can't find the limit of a function by simply substituting the variables, without getting $\frac{0}{0}$, dividing by 0, or taking the square root of a negative number, or indeterminate forms. Then, your next best guess is to use the Squeeze Theorem, to trap these indeterminate functions between the easy, nice functions. If those easy, nice functions approach the same limit, then the indeterminate function, trapped between them, must also approach that limit.

Example using squeeze theorem:

$$\lim_{(x,y) \to (0,0)} x^4 \sin \left( \frac{1}{x^2 + |y|} \right)$$

We know that $-1 \leq \sin \left( \frac{1}{x^2 + |y|} \right) \leq 1$.By using this, we can make the function easier.

By multiplying $x^4$ to the equation, we get: $-x^4 \leq x^4 \sin \left( \frac{1}{x^2 + |y|} \right) \leq x^4$.

Next, we'll take the limits:

$$0 = \lim_{(x,y) \to (0,0)} -x^4 \leq x^4 \sin \left( \frac{1}{x^2 + |y|} \right) \leq \lim_{(x,y) \to (0,0)} x^4 = 0$$

So the limit of our example function is stuck between the two limits of the simpler functions, both having limits equal to 0. So by the Squeeze Theorem we get:

$$\lim_{(x,y) \to (0,0)} x^4 \sin \left( \frac{1}{x^2 + |y|} \right) = 0$$

Sometimes, changing coordinates may be useful. Consider the example below

Using polar coordinates, Find $\lim_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2}$.
The relationship between a point in rectangular coordinates $(x,y)$ and the polar coordinates $(r, \theta)$ with $r \geq 0$ is

$$x = r \cos \theta$$$$y = r \sin\theta$$

from which we can substitute, $$\lim_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2} = \lim_{(x,y) \to (0,0)} \frac{r^3\cos^3\theta + r^3\sin^3 \theta}{r^2\cos^2\theta + r^2\sin^2 \theta}$$

Also saying $(x,y) \to (0,0)$ is equivalent to $r \to 0^+$. Hence, we have:

$$= \lim_{r \to 0^+} \frac{r^3(\cos^3\theta + \sin^3 \theta)}{r^2(\cos^2\theta + \sin^2 \theta)} = \lim_{r \to 0^+} \frac{r^3(\cos^3\theta + \sin^3 \theta)}{r^2}$$

$$= \lim_{r \to 0^+} r(\cos^3\theta + \sin^3 \theta) = 0$$

3) Prove that the limit does not exist

If with above two methods, you were not able to find the limits of a function, then you basically want to prove that the limit does not exist. In single variable, you can do this by proving that the limit from the left and the limit from the right are not equal. In multivariable, you just need to prove that the limit isn’t the same for any two directions.

Example:

$$\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^2} $$

One way to do this is go from the $x$ direction (set $y = 0$ and then find the limit), and then the $y$ direction (set $x = 0$ and then find the limit).

$x$ direction:

$$\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^2} = \lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + 0 ^2} = \lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 } =1 $$

$y$ direction:

$$\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^2} = \lim_{(x,y) \to (0,0)} \frac{0^2}{0^2 + y^2} = \lim_{(x,y) \to (0,0)} \frac{0}{y^2 } =0 $$

The limit in the $x$ direction and the limit in the $y$ direction are not equal, so then the limit does not exist.

Methods for proving the limit does not exist

1. Take the limit in the $x$ direction by setting $y = 0$ and the limit in the $y$ direction by setting $x = 0$.
2. Take the limit along the line $y = x$, by setting $y = x$ in the limit. This will make things easier, as somethings will cancel out to. Then also take the limit in the $x$ or $y$ direction and see what happens.

Example:

1. Consider $\lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} $.

This gives the $\frac{0}{0}$ form, which means, that the function is undefined at $(1,1)$. We attempt to evaluate the limit by aproaching the point $(1,1)$ along four different paths.

$y = x : \lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} = \lim_{(x,y) \to (1,1)} \frac{0}{x-1} = 0$

$y = 2 - x : \lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} = \lim_{(x,y) \to (1,1)} \frac{2(x - 1)}{x-1} = 2$

$y = x^2 : \lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} = \lim_{(x,y) \to (1,1)} \frac{x - x^2)}{x-1} = -1$

$y = 1 : \lim_{(x,y) \to (1,1)} \frac{x - y}{x-1} = \lim_{(x,y) \to (1,1)} \frac{x - 1}{x-1} = 1$

The resulting limiting values found from these following different paths are all different. From this we conclude that limits doesn't exists.

Video 2: Different paths of approach to limit point

Video 3: Visualizing the limit of a function $f(x,y) = \frac{x-y}{x-1}$ along different paths of approach at point $(1,1)$

2. Consider the function $f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}$. Find $\lim_{(x,y) \to(0,0)} \frac{x^2 - y^2}{x^2 + y^2}$.

We will aproach the limit along the path $x = 0$ and $y = 0$.

1. Limit along the path $x = 0$.

$$\lim_{(x,y) \to(a,b)} \frac{x^2 - y^2}{x^2 + y^2} = \lim_{(x,y) \to(0,0)} \frac{x^2 - y^2}{x^2 + y^2} |_{x=0}$$

$$= \lim_{(x,y) \to(0,0)} \frac{- y^2}{ y^2}$$

$$= \lim_{(x,y) \to(0,0)} -1$$

$$= -1$$

2. Limit along the path $y = 0$.

$$\lim_{(x,y) \to(a,b)} \frac{x^2 - y^2}{x^2 + y^2} = \lim_{(x,y) \to(0,0)} \frac{x^2 - y^2}{x^2 + y^2} |_{y=0}$$

$$\lim_{(x,y) \to(0,0)} \frac{x^2 }{x^2 }$$

$$= \lim_{(x,y) \to(0,0)} 1$$

$$= 1$$

3. Conclusion: The limits are different, therefore $\lim_{(x,y) \to(0,0)} \frac{x^2 - y^2}{x^2 + y^2}$ does not exist.

Video 4: Visualizing the limit of a function $f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}$ along the path $x=0$ and $y=0$

Continuity of Multivariable Functions

Definition

A function of many variables $f : D\subseteq\mathbb{R}^{n} \rightarrow \mathbb{R}$ is continuous at a point $r_0 \in D $ if $$f(r_0) = \lim_{\textbf{r}\to r_0} f (\textbf{r})$$ i.e. given any $\epsilon>0$, there exists a $\delta>0$ such that, $$ \lvert f(\textbf{r}) - f(\textbf{r}_{0})\rvert < \epsilon $$ whenever, $$ \lVert \textbf{r} - \textbf{r}_{0} \rVert < \delta$$

Context of the Definition

A function $f (x, y)$ is said to be continuous at a point $(a, b)$, if the following results to be true:

$(a, b)$ is in the domain of $f$.
$\lim_{(x,y)\to(a,b)} f (x, y)$ exists.
$\lim_{(x,y)\to(a,b)} f (x, y) = f(a,b)$.

If a function $f$ is not continuous at a point $(a, b)$, then it is discontinuous at $(a, b)$.

A function $f$ is continuous on a set $D$ if and only if, it is continuous at each point in $D$.

Thus, to know whether a function is continuous at a point, we find the limit of the function at the point and simply substitute the point.

For the multivariable functions, the following outcomes are true :

The polynomial functions are continuous.
The rational functions are continuous in their domain.
The sum or product of continuous functions, defined on the same domain, is a continuous function.
If $f$ is a continuous function and $\alpha\in \mathbb{R} (or \mathbb{C})$ then, $\alpha f$ is a continuous function
The quotient of two continuous functions is continuous until and unless the denominator isn't 0.
If $f (x, y)$ is continuous and $g (x)$ is defined and continuous on the range of $f$, then $g (f (x, y))$ is also continuous.

Examples

1. Is $f (x, y) = x^2y+3x^3y^5-x+2y$ continuous at $ (0, 0)$? Where is it continuous?

$\rightarrow$ $f (x, y)$ is continuous on $R^2$ being a polynomial function, and also continuous at $ (0, 0)$.

2. Where is $f(x, y) = \frac{3x-2y}{x^2+y^2}$ continuous?
$\rightarrow$ As $f(x, y)$ being the quotient of two continuous functions, it is continuous untill its denominator is not equal to 0 that is on $R^{2}\setminus \{ (0, 0)\}$.

3. Where is $f (x, y) = \frac{1}{x^2-y}$ continuous?
$\rightarrow$ As $f(x, y)$ being the quotient of two continuous functions, it is continuous untill it's denominator is not 0. The denominator will become $0$ along the parabola $y = x^2$. Therefore, $f(x, y)$ is continuous on {$(x, y) \in R^2 | y \neq x^2$}.

Polynomials are continuous everywhere and the Rational functions are continuous everywhere they are defined.

The function $f(x,y)=\frac{3x^2y}{x^2+y^2}$, is not defined at $(0,0)$, and hence not continuous at $(0,0)$. However, we know that $\lim_{(x,y)\to(0,0)}f(x,y)=0$, so we can easily "fix" the problem, by extending the definition of $f$ so that $f(0,0)=0$.

Note that in contrast to this example, we cannot fix $f(x,y)=\frac{3x^2y}{x^2+y^2}$ at $(0,0)$ because the limit does not exist. No matter what value we try to assign to $f$ at $(0,0)$ the surface will have a "jump" there.

Video 5: The function $f(x,y) = \frac{3x^2y}{x^2 + y^2}$ is not continuous at point $(0,0)$

Applications

The limit concept is essential to understanding the real number system and its distinguishing characteristics. In one sense real numbers can be defined as the numbers that are the limits of convergent sequences of rational numbers.
Limits are the method by which the derivative, or rate of change, of a function, is calculated.
Limits are also a key to calculate integrals. The integral calculates the entire area of a region by summing up an infinite number of small pieces of it.

History

In the $3^{rd}$ century B.C, Archimedes of Syracuse first developed the idea of limits to measure curved figures and the volume of a sphere. By carving these figures into small pieces that can be approximated, then increasing the number of pieces, the limit of the sum of pieces can give the desired quantity.
Archimedes' thesis, "The Method", was lost until 1906, when mathematicians discovered that Archimedes came on the brink of discovering infinitesimal calculus.

As Archimedes's work was unknown until the twentieth century, others developed the modern mathematical concept of limits. Sir Issac Newton and German Gottfried Wilhelm von Leibniz developed independently, the general principles of calculus (of which the idea of limits is a crucial part) within the seventeenth century.

Pause and Ponder

What is the role of the limits in derivatives of a function?

The derivative of a function $f(x)$ at a point $x_0$ is the limit of the difference quotient at $x = x_0$, as the increment $h = x- x_0$ of the independent variable $x$ approaches 0. In mathematical words:

$f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0+h) - f(x_0)}{h}$

So derivatives are a kind of special limits, that gives useful information about the functions and their behavior.

Figure 1: Limit Definition of the Derivative [1]

References

https://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx
http://www.math.ucsd.edu/~jbriones/math20c.15w/handouts/Math20C_Handout5.pdf
http://ccjohnson.org/math2060/lectures/Lecture10-LimitsContinuity.pdf
https://ksuweb.kennesaw.edu/~plaval/math2203/funcnD_limcont.pdf
Book - Thomas’ Calculus Early Transcendentals

Figures

[1] Reproduced from https://www.acsu.buffalo.edu/~mbschild/images.html

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