Multivariable Limits and Continuity |
---|
Definition
Limit of a function of two variablesWe 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 variablesGiven 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 |
MotivationIn 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 ViewWhen 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:
-3 |
-2 |
Video 1: Epsilon-Delta definition of limits of Multivariable Functions |
Context of the DefinitionLet $f$ 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 limit1) Substitute the variablesIf 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:
2) Change the co-ordinates or use the Squeeze theoremIn 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 $\lim_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2}$. $$x = r \cos \theta$$$$y = r \sin\theta$$ 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 existIf 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 exist1. Take the limit in the $x$ direction by setting $y = 0$ and the limit in the $y$ direction by setting $x = 0$. 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 |
0 |
Video 2: Different paths of approach to limit point |
0 |
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 |
0 |
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 FunctionsDefinitionA 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 DefinitionA function $f (x, y)$ is said to be continuous at a point $(a, b)$, if the following results to be true:
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 :
0 |
Examples1. 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? 3. Where is $f (x, y) = \frac{1}{x^2-y}$ continuous? 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 |
0 |
Video 5: The function $f(x,y) = \frac{3x^2y}{x^2 + y^2}$ is not continuous at point $(0,0)$ |
Applications
0 |
HistoryIn 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. 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 PonderWhat 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 |
0 |
Figure 1: Limit Definition of the Derivative [1] |
References
0 |
Figures[1] Reproduced from https://www.acsu.buffalo.edu/~mbschild/images.html
4 |
Contributor: |
Mentor & Editor: |
Verified by: |
Approved On: |
The work under this website is licenced under a Creative Commons Attribution-Share Alike 4.0 International License CC BY-SA