Limits for Scalar Valued Multivariable Functions - Giri-Subrahmanya/temp GitHub Wiki

We saw that there are two ways to approach a point $c$ in $ℝ$.

• From its left
• From its right

Let us focus the number of ways the number of ways to approach a point $(x,y)$ in $ℝ^2$.

Along how many ways can we approach $(1,1)$?

Along $y=x$	Along $y=2-x$	Along $y=x^2-2x+2$	Along $e^{x-1}$	Along $\ln^{x}+1$

We could come up with more curves.

If the domain of a scalar valued multivariable function is $ℝ^2$, we see that there are infinitely many ways to approach a point. We could extend this observation to domain being $ℝ^3$, $ℝ^4$, $ℝ^5$ and so on.

As an extension of the informal description of limit of functions of one variable, the following could be a heuristic about limit of a scalar valued multivariable function.

The limit of a scalar valued multivariable function exists at a point $p=(x_1,x_2,...,x_n)\in$ Domain and is equal to $L\in ℝ$ if the limit at $p$ along every curve passing through $p$ exists and is equal to $L$.

Let us look at this closely by observing the value of the following functions at $(1,1)

• $f(x,y)=x^2+y^2$
• $g(x,y)=\frac{xy}{(x^2-1)(y-1)}$

Along every curve, we can approach $(1,1)$ either by PATH-1 or by PATH-2.

Along $y=x$	Along $y=2-x$	Along $y=x^2-2x+2$	Along $e^{x-1}$	Along $\ln^{x}+1$

• PATH-1 - $x<1$
• PATH-2 - $x>1$

The first column is the value $x$ close to $1$, but less than $1$. Using $x$, the value of $y$ has been computed for different curves. Using these two, the function's value has been computed.

$x$		$y=x$	$f(x,y)$		$y=2-x$	$f(x,y)$		$y=x^2-2x+2$	$f(x,y)$		$y=e^{x-1}$	$f(x,y)$		$y=\ln(x)+1$	$f(x,y)$
0.991		0.991	1.964162		1.009	2.000162		1.000081	1.982243007		0.9910403788	1.964242032		0.9909592553	1.964081246
0.992		0.992	1.968128		1.008	2.000128		1.000064	1.984192004		0.9920319148	1.96819132		0.9919678283	1.968064172
0.993		0.993	1.972098		1.007	2.000098		1.000049	1.986147002		0.9930244429	1.972146544		0.9929753851	1.972049115
0.994		0.994	1.976072		1.006	2.000072		1.000036	1.988108001		0.9940179641	1.976107713		0.9939819277	1.976036073
0.995		0.995	1.98005		1.005	2.00005		1.000025	1.990075001		0.9950124792	1.980074834		0.9949874582	1.980025042
0.996		0.996	1.984032		1.004	2.000032		1.000016	1.992048		0.9960079893	1.984047915		0.9959919786	1.984016021
0.997		0.997	1.988018		1.003	2.000018		1.000009	1.994027		0.9970044955	1.988026964		0.996995491	1.988009009
0.998		0.998	1.992008		1.002	2.000008		1.000004	1.996012		0.9980019987	1.992011989		0.9979979973	1.992004003
0.999		0.999	1.996002		1.001	2.000002		1.000001	1.998003		0.9990004998	1.996002999		0.9989994997	1.996001

The first column is the value $x$ close to $1$, but greater than $1$. Using $x$, the value of $y$ has been computed for different curves. Using these two, the function's value has been computed.

$x$		$y=x$	$f(x,y)$		$y=2-x$	$f(x,y)$		$y=x^2-2x+2$	$f(x,y)$		$y=e^{x-1}$	$f(x,y)$		$y=\ln(x)+1$	$f(x,y)$
1.009		1.009	2.036162		0.991	2.000162		1.000081	2.018243007		1.009040622	2.036243976		1.008959741	2.03608076
1.008		1.008	2.032128		0.992	2.000128		1.000064	2.016192004		1.008032086	2.032192685		1.00796817	2.032063831
1.007		1.007	2.028098		0.993	2.000098		1.000049	2.014147002		1.007024557	2.028147459		1.006975614	2.028048887
1.006		1.006	2.024072		0.994	2.000072		1.000036	2.012108001		1.006018036	2.024108289		1.005982072	2.024035929
1.005		1.005	2.02005		0.995	2.00005		1.000025	2.010075001		1.005012521	2.020075167		1.004987542	2.020024959
1.004		1.004	2.016032		0.996	2.000032		1.000016	2.008048		1.004008011	2.016048086		1.003992021	2.016015979
1.003		1.003	2.012018		0.997	2.000018		1.000009	2.006027		1.003004505	2.012027036		1.002995509	2.012008991
1.002		1.002	2.008008		0.998	2.000008		1.000004	2.004012		1.002002001	2.008012011		1.001998003	2.008003997
1.001		1.001	2.004002		0.999	2.000002		1.000001	2.002003		1.0010005	2.004003001		1.0009995	2.004001

Notice that $g(1,1)$ is not defined

The first column is the value $x$ close to $1$, but less than $1$. Using $x$, the value of $y$ has been computed for different curves. Using these two, the function's value has been computed.

$x$		$y=x$	$g(x,y)$		$y=2-x$	$g(x,y)$		$y=x^2-2x+2$	$g(x,y)$		$y=e^{x-1}$	$g(x,y)$		$y=\ln(x)+1$	$g(x,y)$
0.991		0.991	6089.631738		1.009	-6200.240589		1.000081	-682825.9892		0.9910403788	6117.325429		0.9909592553	6061.937859
0.992		0.992	7718.875502		1.008	-7843.373494		1.000064	-972702.8112		0.9920319148	7750.041499		0.9919678283	7687.709338
0.993		0.993	10097.06421		1.007	-10239.4196		1.000049	-1452677.165		0.9930244429	10132.69458		0.9929753851	10061.4337
0.994		0.994	13764.01427		1.006	-13930.17943		1.000036	-2307932.557		0.9940179641	13805.5971		0.9939819277	13722.43131
0.995		0.995	19850.12531		1.005	-20049.62406		1.000025	-3990074.687		0.9950124792	19900.04156		0.9949874582	19800.20896
0.996		0.996	31062.62525		1.004	-31312.12425		1.000016	-7796968.437		0.9960079893	31125.04158		0.9959919786	31000.20883
0.997		0.997	55305.68074		1.003	-55638.51333		1.000009	-18490865.43		0.9970044955	55388.93049		0.996995491	55222.43093
0.998		0.998	124625.1251		1.002	-125124.6246		1.000004	-62437687.19		0.9980019987	124750.0416		0.9979979973	124500.2086
0.999		0.999	499250.1251		1.001	-500249.6248		1.000001	-499750374.7		0.9990004998	499500.0416		0.9989994997	499000.2085

The first column is the value $x$ close to $1$, but greater than $1$. Using $x$, the value of $y$ has been computed for different curves. Using these two, the function's value has been computed.

$x$		$y=x$	$g(x,y)$		$y=2-x$	$g(x,y)$		$y=x^2-2x+2$	$g(x,y)$		$y=e^{x-1}$	$g(x,y)$		$y=\ln(x)+1$	$g(x,y)$
1.009		1.009	6256.29728		0.991	-6144.688408		1.000081	688999.4538		1.009040622	6228.436915		1.008959741	6284.157831
1.008		1.008	7906.374502		0.992	-7780.876494		1.000064	980515.9363		1.008032086	7875.041833		1.00796817	7937.707338
1.007		1.007	10311.34905		0.993	-10167.99365		1.000049	1462881.871		1.007024557	10275.55202		1.006975614	10347.14624
1.006		1.006	14014.01352		0.994	-13846.84834		1.000036	2321822.071		1.006018036	13972.26401		1.005982072	14055.76314
1.005		1.005	20150.12469		0.995	-19949.62594		1.000025	4010075.312		1.005012521	20100.04177		1.004987542	20200.20771
1.004		1.004	31437.62475		0.996	-31187.12575		1.000016	7828219.062		1.004008011	31375.04175		1.003992021	31500.20783
1.003		1.003	55805.68037		0.997	-55471.84778		1.000009	18546421.61		1.003004505	55722.26395		1.002995509	55889.09685
1.002		1.002	125375.1249		0.998	-124874.6254		1.000004	62562687.81		1.002002001	125250.0417		1.001998003	125500.2081
1.001		1.001	500750.1249		0.999	-499749.6252		1.000001	500250375.4		1.0010005	500500.0417		1.0009995	501000.2082

• From PATH-1, along all the five curves, the function's value is approaching $2$ as $(x,y)$ approaches $(1,1)$.
• From PATH-2, along all the five curves, the function's value is approaching $2$ as $(x,y)$ approaches $(1,1)$.
• These observations seems to suggest that the limit of $f(x,y)$ as $(x,y)\rightarrow (1,1)$ exists and is equal to $2$.
• However, these observations are not enough to say that the limit of $f(x,y)$ as $(x,y)\rightarrow (1,1)$ exists.
• We would have to check the same for every curve passing through $(1,1)$ to arrive at a conclusion. We could come up with as many curves as we want. Therefore, this is not a good approach. However, this provides an idea of the concept of limits for multivariable functions.

• From both the paths, along all the five curves, the function's value is not approaching any particular value.
• A small change in $(x,y)$ is changing the value of the function drastically.
• From PATH-1, along $y=x^2-2x+2$, for example, $g(0.998,1.000004)=-62437687.19$ and $g(0.999,1.000001)=-499750374.7$. This clearly shows the drastic change. A similar observation can be made for the function's value along other curves.
• Along $y=x^2-2x+2$, as $(x,y)$ approaches $(1,1)$ from PATH-1, the function's value is approaching $\infty$, whereas, as $(x,y)$ approaches $(1,1)$ along PATH-2, the function's value is approaching $-\infty$.
• These observations are enough to say that the limit of $g(x,y)$ as $(x,y)\rightarrow (1,1)$ does not exist. Later, we shall see a formal way to arrive at this conclusion.

As mentioned in the introductory section, the aim of these notes is to provide intuition about Multivariable Calculus. Refer to the lecture slides for the definition of limit. The above explanations were hopefully clear enough to understand the concept of limit.

However, we shall look at the some rules which has to be followed for finding the limit of a function.

1. If $\lim\limits_{x \rightarrow a} f(x)=F$, $\lim\limits_{x\rightarrow a} g(x)=G$ and $c\in ℝ$, then $\lim\limits_{x\rightarrow a} (xf+g)(x)=xF+G$.

2. If $\lim\limits_{x \rightarrow a} f(x)=F$, $\lim\limits_{x\rightarrow a} g(x)=G$, then $\lim\limits_{x\rightarrow a} (fg)(x)=FG$.

3. If $\lim\limits_{x \rightarrow a} f(x)=F$, $\lim\limits_{x\rightarrow a} g(x)=G\ne 0$, then the function $\frac{f}{g}$ is defined in at least a small interval around $a$ and $\lim\limits_{x\rightarrow a} \frac{f}{g}(x)=\frac{F}{G}$.

4. Composition $:$ Suppose $f$ is a scalar-valued multivariable function and g is a function of one variable such that the composition $g\circ f$ is well-defined. If $\lim\limits_{x\rightarrow a} f(x)=F$, $\lim\limits_{x->F} g(x)=L$, then $(g\circ f)(x)=L$.

5. The Sandwich Principle $:$ If $\lim\limits_{x\rightarrow a} f(x)=L$, $\lim\limits_{x\rightarrow a}g(x)=L$, and $h(x)$ is a function such that $f(x)≤h(x)≤g(x)$, then $\lim\limits_{x\rightarrow a}h(x)=L$