. Preliminary Notions - cis.umac.mo

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.. Preliminary Notions.

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...1 A ball in space (R3) of radius r, centered at P(a, b, c) in R3 is given by{ (x, y, z) ∈ R3 | (x − a)2 + (y − b)2 + (z − c)2 < r2 }, which includes allthe points X(x, y, z) in space with distance ∥PX∥ less than r.

...2 A disc in plane (R2) of radius r, centered at P(a, b) in R2 is given by{ (x, y) ∈ R2 | (x − a)2 + (y − b)2 < r2 }, which includes all the pointsX(x, y) in space with distance ∥PX∥ less than r.

...3 In general, one has B(P, r) = { X(x1, · · · , xn) ∈ Rn | ∥PX∥ < r }.

...4 A sphere of radius r centered P consists of all the point X such that thedistance ∥PX∥ = r.

Matb 210/Math200 in 2013-2014

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A region (or domain resp.) or , denoted by R, is a subset of R2 (or R3 resp., inwhich any two points in R or D are joined by a path inside R (or D resp). Inmost cases, one can think R or D as a disc or a ball resp.

.Interior points and Boundary points in a domain (region)..

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...1 A point P of the region R is called an interior point of the region R, if wecan find a ball centered at the point P with positive radius so that the balllies completely inside the region R.

...2 A point P is called a boundary point of the region R, if for any ball centeredat the point P, we can find two other distinct points in that ball: one is in Rand the other is not in R.

...3 Given a point p ∈ Rn and a region R ⊂ Rn, p is called an accumulationpoint of D if for any positive δ > 0, one can find another pointq ∈ D ∩ B(p, δ) \ {p}.Remark. Accumulation point plays an important role in defining theimportant concept of limit.

Matb 210/Math200 in 2013-2014

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.Bounded, Closed and Open Region in Rn..

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...1 A region R is called bounded if the region R lies completely inside a ballcentered at origin with a (sufficiently large) positive radius.

...2 A region R is called open if all the points of the region is an interior point.

...3 A region R is called closed if its complement Rn \ R is open.

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Example Let a, b, c be positive numbers. (i) The boxT = { (x, y, z) | |x| ≤ a, |y| ≤ b, |z| ≤ c } is a closed set in R3. Any point on the6 faces of the box T is a boundary point of T.(ii) S = { (x, y, z) | |x| < a, |y| < b, |z| < c }, which is the subset of T with all 6faces deleted, is an open set in space.

Proof. (ii) For any point P(x0, y0, z0) in S, B(P, δ) ⊂ S, whereδ = 1

2 min{ |x0 − a|, |x0 + a|, |y0 − b|, |y0 + b|, |z0 − c|, |z0 + c| }. Note thatδ > 0 as (x0, y0, z0) does not lie in any of the 6 faces.

Matb 210/Math200 in 2013-2014

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.. Functions of several variables.

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Example. There is a box with length x, width y and height z, then one considertwo geometric quantities:its volume V(x, y, z) = xyz, which is a function of 3 variables of the dimensionsof the box.its surface area S(x, y, z) = 2(xy + yz + zx), which is also a function of 3variables of the dimensions of the box.n

Remark. Not all the functions are given by polynomials in the x, y, z..

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The reason why we need to study functions of several variables is that mostquantity we want to measure or to study are the outcome which can beaffected by more than a single factor.

Matb 210/Math200 in 2013-2014

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.. Functions of several variables:.Domain of a function...

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Definition. Given a function f (x, y) of two variables, the domain Dom(f ) off (x, y) is the set of points (x, y) in which one can evaluate for f ,and obtain afinite value in R. i.e. Domain(f ) = { (x, y, ) ∈ R2 | f (x, y) is a finite number }.The similar definition can be worked out for functions of 3 variables.

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......Example. Define f (x, y) = xy . Determine its domain.

Solution. One can see that if y ̸= 0, then the fraction xy is a finite quantity, and

if y = 0, then no matter what x is (including x = 0), the expression xy is

meaningless. The domain of f (x, y) is the set { (x, y) | y ̸= 0}= R2\ the x-axis.

Matb 210/Math200 in 2013-2014

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Example. The domain of a polynomial function f (x, y) of 2 variables is theentire xy-plane R2;The domain of a polynomial function f (x, y, z) of 3 variablesis the entire space R3.

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Example. Find the domain of the following functions: (a) z(x, y) =√

y cos x;(b) u(x, y, z) = ln(1 − x2 − y2 − z2) (c) v(x, y) = arctan y

1+x2+y2 .

Solution. (a) The domain of z(x, y) =√

y cos x is given by{ (x, y) ∈ R2 | y cos x ≥ 0 }, which is not easy to describe geometrically.(b) The domain of u(x, y, z) is given by{ (x, y, z) ∈ R3 | 1 − (x2 + y2 + z2) > 0 }, which is the unit open (solid) ball{ (x, y, z) | x2 + y2 + z2 < 1 } in space.(c) The domain of arctan is R, and the denominator of rational functions

y1+x2+y2 does not vanish on the entire xy-plane, hence the domain of v(x, y) is

the entire xy-plane R2.

Matb 210/Math200 in 2013-2014

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.Range of a function...

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Given a function f (x, y) or f (x, y, z) of 2 or 3 variables, the range Ran(f ) of f isthe set of all the values of f (x, y) or f (x, y, z) while (x, y) or (x, y, z) runs throughall the points in the domain of f .

Remark. In general, it is very difficult to determine the range of f ..

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Example. For any non-zero vector (a, b, c) define f (x, y, z) = ax + by + cz.Determine the domain of f and the range of f .

Solution. As f is a polynomial of degree 1, so the domain of f consists of allpoints in R3, i.e. Domain(f ) = R3. To determine the range of f , it is enough tochoose some points P(x, y, z) in Dom(f ) and to show that f (x, y, z) takes everyreal number. For this, since (a, b, c) ̸= (0, 0, 0), we know that one of a, b, c isnon-zero, without loss of generality, one may assume that a ̸= 0, otherwiseconsider the other two. For any given ℓ ∈ R, one knows ℓ

a ∈ R, sof (ℓ/a, 0, 0) = ℓ/a × a = ℓ. Hence the set of all real numbers lies in range of f ,i.e. range of f = R..

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Remark. From now on, one will write z = f (x, y) to represent a function,meaning z is an arbitrary value in the range of f .Matb 210/Math200 in 2013-2014

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.Level Curves of function f (x, y)..

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Given a function f (x, y) of two variables, and a fixed real number c, the levelcurve Lc of a function z = f (x, y) at value c is given by the set of points (x, y) inxy-plane such that the value of f (x, y) equals to the given constant c. Innotation, Lc = { (x, y) | f (x, y) = c }.

Example. Define f (x, y) = x2 + y2, and c = 1, then L1 =

{ (x, y) | x2 + y2 = 1 } which is the unit circle centeredat (0, 0) in xy-plane. If one changes the constant c from 0to 1, one obtain a family of concentric circle of radius

√c,

centered at the origin (0, 0).Remark. Level curve Lc for a fixed c is a set of points in xy-plane. Given a nicefunction z = f (x, y), the level curve Lc looks like a curve in xy-plane.

Matb 210/Math200 in 2013-2014

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Example 11. Sketch the level curves of the function

f (x, y) =34

y2 +124

y3 − 132

y4 − x2.

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Solution. Here we use the color to rep-resent the value of f as shown in thescale appeared in right hand side of thefigure.

Matb 210/Math200 in 2013-2014

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.Graph G of a function z = f (x, y)..

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Definition. Let function z = f (x, y), the graph of the function f is given by theset of all points (x, y, f (x, y)) in space R3, where the point (x, y) is in thedomain of the function f . In notation,G = G(f ) = { (x, y, f (x, y) ) ∈ R3 | (x, y) is in domain of f }.

In general, the graph G is a surface in space..

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Example. For any (a, b) ̸= (0, 0), one define f (x, y) = ax + by. Determine thegraph G(f ) of f .

Solution. The graph G of f is a plane in space. For any point P(x, y, z) in thegraph G of f , we know that z = f (x, y) = ax + by. Rewrite it as ax + by − z = 0,then this is an equation of the plane S : ax + by − z = 0 in R3, and hence thegraph of f is a plane. Indeed, For any point P(x, y, z) in the plane S, one canimmediately check that z = ax + by = f (x, y), so P lies in the graph G(f ) of f .

Matb 210/Math200 in 2013-2014

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......Example. Describe the graph G(f ) of the function f (x, y) = x2 + y2.

Solution. Let P(x, y, z) be any point of G(f ). Then we have z = x2 + y2, whichwe have just considered in the definition of level curve. Since the domain of fis R2, so the graph of f is the set of all points of the form (x, y, f (x, y) ). In fact,if we use the cylindrical coordinates, i.e using polar coordinates to describe thexy-plane, and no changes in z-coordinates. In this case,z = x2 + y2 = (r cos θ)2 + (r sin θ)2 = r2. It means that the graph of z = x2 + y2

is a surface of revolution about z-axis. First draw a parabola curve C : z = x2

on xz-plane, and then rotate the curve C in space about z-axis.

Matb 210/Math200 in 2013-2014

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Example 11. Study the graph of the functionf (x, y) = 3

4 y2 + 124 y3 − 1

32 y4 − x2.

Solution. Though the function f (x, y) is a polynomial function, but its graph isnot so easy to visualize by hand. A computer plot of the graph is given at page856 of our textbook. We need more tools in fact!Now we will discuss the y-section of the graph, i.e. one fix the value of y = a,and allow the variable x to change. Think of cutting the graph by using a knifealong the plane y = a (compare with 0 · x + 1 · y + 0 · z = a).f (x, a) = 3

4 a2 + 124 a3 − 1

32 a4 − x2, which represents a family of parabolasz = k − x2 with coefficients k changing with respect to y = a.

Hence the graph of f is symmetric about yz-plane, i.e. (x, y, z) in the graph if and only if(−x, y, z) is also in the graph.

Matb 210/Math200 in 2013-2014

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Remark. 1. The concepts x-section and y-section are important, as they play asignificant role in multiple integral which will be treated in ch. 14.2. Next we will discuss the x-section of the graph by fixing the value of x = b.Then f (b, y) = 3

4 y2 + 124 y3 − 1

32 y4 − b2 is a degree 4 polynomial in y withnegative highest coefficient, which makes the thing complicated. One caneasily show by means of derivative that there are some local maximum values.However these local maxima only happen in one-direction, and we will laterlocate the global maxima or minima by means by means of derivative test,which is more complicated.

Matb 210/Math200 in 2013-2014

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.Level Surface Sc of a function w = f (x, y, z)..

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Definition. Given a function f (x, y, z) of 3 variables, and a fixed number c ∈ R,the level surface Sc of a function w = f (x, y, z) at value c is a set in R3,consisting of all the points (x, y, z) in domain of f such that the value f (x, y, z)equals to the given constant c. In notation, Sc = { (x, y, z) ∈ R3 | f (x, y, z) = c }.

In general, the graph G is a surface in space R3. And it is difficult to visualize agraph, though the function is given by a polynomial..

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Example. Prove that the level surface of f (x, y, z) = x2 + y2 + z2 at c = 1 is justthe sphere of radius 1, centered at (0, 0, 0).

Proof. Let P(x, y, z) be any point in the level surface of Sc of the function f , thenx2 + y2 + z2 = f (x, y, z) = 1, so ∥−→OP∥ = 1, i.e. P lies on the unit sphere S,centered at O(0, 0, 0). On the contrary, for any point Q(x, y, z) lying in the unitsphere S, it follows from the definition of sphere S that x2 + y2 + z2 = 1, andhence that f (Q) = x2 + y2 + z2 = 1, then Q is in the level surface Sc of f .

Matb 210/Math200 in 2013-2014

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Example Given f (x, y, z) = x2 + y2 − z2 defined on the space R3. The figurebelow illustrates the level surfaces Sc = { (x, y, z) | f (x, y, z) = c } of thefunction f (x, y, z), with(i) c = −

√2; (ii) c = −1; (iii) c = 0; (iv) c = 1; and (v) c =

√2.

Matb 210/Math200 in 2013-2014

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.. Limit of the function of several variables

In this section v represents a vector in Rn, sometimes, we call a vector as apoint in the the n-dimensional space..

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...1 Suppose that f is a function defined in Dom(f ) of Rn, this means thatf = f (x1, x2, · · · , xn) is a function of n variables.

...2 Recall a point w in Rn lies in domain Dom(f ) if and only if one canevaluate the expression f (w).

...3 We will encounter the following statement:For any number ε > 0, there exists some δ > 0 satisfying the condition P(depending on the given function f , ε and δ).In fact, this means thatIf someone (your teacher) gives you a positive number ε, then you (as astudent)need to find another positive number δ depending of ε (by variousmeans, theorem or given assumption), so that no matter what happens,the condition P always holds for this δ.

Matb 210/Math200 in 2013-2014

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.. Limit of the function of several variables.

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Definition. Suppose that a scalar function f (v) is defined in the domain(f ) inRn and p be an accumulation point in D. We say that the limit of the functionf (v) as the point v in Rn approaches to p is equal to ℓ, ( denoted bylim

v−>pf (v) = ℓ,) if

for any given positive number ε > 0,there exists a positive number δ > 0 such that

the following condition holds:For any point w (not equal to p) in Rn lying in the domain of f andthe distance between the point w and p is less than δ, then

the distance between f (w) and ℓ is less than ε, i.e. |f (w)− ℓ| < ε.

∀ε > 0 ∃δ > 0 such that for any w( ̸= p) in the ball B(p, δ)∩ domain (f ),one has f (w) lies in the ball B( ℓ, ε) ⊂ R, i.e. |f (w)− ℓ| < ε.

Matb 210/Math200 in 2013-2014

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Example. Show that the limit of the function f (x, y) = 2x + 3y is equal to 0when (x, y) approaches to (0, 0) i.e. lim

(x,y)→(0,0)(2x + 3y) = 0.

Solution. Domain of f is R2. For any given ε > 0, we need to chooseδ = ???,such that

for any (x, y) ∈ R2 = domain (f ) and (x, y) in B((0, 0), δ) \ {(0, 0)}, i.e.x2 + y2 < δ2, we have|f (x, y)− 0| = |2x + 3y| (by Cauchy-Schwarz ineq |v · w| ≤ ∥v∥ · ∥w∥, )

≤√

22 + 32 ·√

x2 + y2 =√

13 ·√

x2 + y2

<√

13δ < ?? < ε.And hence f (x, y) ∈ B(0, ε) for all (x, y) ∈ R2 ∩ B( (0, 0), δ).For the missing condition on δ, one can choose δ = ε√

13or even smaller δ = ε

4which will fill up the gap in the inequality ?? above..

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Remark. One can use Cauchy-Schwarz inequality as above to show thatlim

(x,y,z)→(a,b,c)(Ax + By + Cz) = Aa + Bb + Cc.

Matb 210/Math200 in 2013-2014

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Example. (Important) Prove that(i) lim

(x,y,z)→(a,b,c)x = a; (ii) lim

(x,y,z)→(a,b,c)y = b; (iii) lim

(x,y,z)→(a,b,c)z = c;

(iv) lim(x,y,z)→(a,b,c)

k = k, where k is a constant;

(v) lim(x,y,z)→(0,0,0)

√x2 + y2 + z2 = 0.

Solution. (iv) is the easiest one. Define r(x, y, z) = k, as|r(x, y, z)− k| = |k − k| = 0, so for any given ε > 0, one can choose δ = ε > 0.The detail can be filled as in (i) proved below.(v) For any ε > 0, choose δ = ε, then for any (x, y, z) ∈ B( (0, 0, 0), δ ), we have√(x − 0)2 + (y − 0)2 + (z − 0)2 < δ ♡, it follows from the condition ♡ above

that |√

x2 + y2 + z2 − 0| =√

x2 + y2 + z2 < δ = ε. Solim

(x,y,z)→(a,b,c)

√x2 + y2 + z2 = 0.

Matb 210/Math200 in 2013-2014

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Example. (Important) Prove that (i) lim(x,y,z)→(a,b,c)

x = a; (ii) lim(x,y,z)→(a,b,c)

y = b;

(iii) lim(x,y,z)→(a,b,c)

z = c; (iv) lim(x,y,z)→(a,b,c)

k = k, where k is a constant;

(v) lim(x,y,z)→(0,0,0)

√x2 + y2 + z2 = 0.

Solution. (i) Let f (x, y, z) = x, and its domain is R3. For any given ε > 0, onechoose δ = ε > 0, such that for any (x, y, z) ∈ domain (f ) = R3, with(x, y, z) ∈ B( (a, b, c), δ) \ {(a, b, c)}, then |f (x, y, z)− a| = |x − a| =√(x − a)2 ≤

√(x − a)2 + (y − b)2 + (z − c)2 < δ = ε. (ii) and (iii) can be

proved in a similar way.

Matb 210/Math200 in 2013-2014

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.Laws of Limit..

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Let c be a constant, and suppose that f (v) and g(v) are two functions definedon the same domain D such that lim

v→pf (v) = A and lim

v→pg(v) = B exist and both

are finite, then we have...1 Sum and Difference Rules: lim

v→p( f (v) + g(v) ) = A + B, and

limv→p

( f (v)− g(v) ) = A − B;

...2 Product Rule: limv→p

( f (v) · g(v) ) = A · B;

...3 Scalar Multiplication Rule: limv→p

( c · f (v) ) = c · A;

...4 Quotient Rule: limv→p

( f (v)g(v) ) =

AB , provided that B ̸= 0.

Remarks. (i) The following examples are the applications of these 4 rules.(ii) In Quotient Rule above, the condition B ̸= 0 is necessary for the statement.

Matb 210/Math200 in 2013-2014

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...1 The limit of polynomial function p(x, y) in two variables, or p(x, y, z) inthree variables, at any point s is equal to the function value p(s).

...2 Suppose that f (v) = p(v)q(v) is a rational function, i.e. p(v) and q(v) are

polynomials, then the limit of f (v) at any point s is equal to the functionvalue f (s), provided that q(s) is not zero

Proof (a) Let s = (a, b, c) and v = (x, y, z) ∈ R3, the case of two variables canbe treated similarly. Any polynomial is a sum of monomials of the formAxnymzk, it follows from laws of product and scalar multiplication that

lim(x,y,z)→(a,b,c)

Axnymzk = Aanbmck. Then it follows from the law of addition that

lim(x,y,z)→(a,b,c)

f (x, y, z) = f (a, b, c) for any polynomial f in x, y, z. (b) As f (v) = p(v)q(v)

where p and q are polynomial in n-vector v, suppose that q(s) ̸= 0, it follows

from the law of quotient that lim(x,y,z)→(a,b,c)

f (x, y, z) =lim

(x,y,z)→(a,b,c)p(x,y,z)

lim(x,y,z)→(a,b,c)

q(x,y,z) =p(a,b,c)q(a,b,c) .

Matb 210/Math200 in 2013-2014

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.Sandwich Theorem of Limit..

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Suppose that f (v) and g(v) are function defined on the same domain D in Rn,such that 0 ≤ |f (v)| ≤ g(v) for all v ∈ D(⊂ Rn).If lim

v→pg(v) = 0, then lim

v→pf (v) = 0.

Solution. For any given ε > 0, it follows from limv→p

g(v) = 0 that there exists

δ > 0 such that for any point w ∈ B(p, δ) ∩ D and w ̸= p, one has

|g(w)| = |g(w)− 0| < ε. ♡

In particular, it follows from ♡ that for the same δ > 0, one has

|f (w)− 0| = |f (w)| < |g(w)| < ε,

for any w ∈ B(p, δ) ∩ D \ {p} i.e. limv→p

f (v) = 0.

Matb 210/Math200 in 2013-2014

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......Example. Evaluate the limit lim

(x,y)→(0,0)

xy(x + y)x2 + y2 .

Solution. Let f (x, y) =xy(x + y)x2 + y2 . We apply the first principle to show that

lim(x,y)→(0,0)

f (x, y) = 0. Observe that f (x, 0) = f (0, y) = 0 for all non-zero x and y.

By using completing square, one has x2 + y2 ≥ 2|xy| > 0, for all(x, y) ∈ D = R2\ both coordinate axes,For any (x, y) ∈ D, one has

0 ≤∣∣∣∣xy(x + y)

x2 + y2

∣∣∣∣ ≤ |xy(x + y)|2|xy| =

12|x + y| ≤ 1

2(|x|+ |y|)

≤ 12(√

x2 + y2 +√

x2 + y2) =√

x2 + y2,

and hence it follows from the sandwich theorem of limit thatlim

(x,y)→(0,0)

xy(x+y)x2+y2 = 0.

Matb 210/Math200 in 2013-2014

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Example. Let f be a scalar function defined in some domain D in Rn withrange R ⊂ R, and lim

v→pf (v) = ℓ. Suppose that g : R → R is continuous scalar

function defined on R. Prove that(i) the composite function g ◦ f is defined on D;(ii) lim

v→pg ◦ f (v) = g(ℓ).

Solution. (i) This follows easily by directly plugging into the composite functiong ◦ f . (ii) For any p = (a, b, c) ∈ D3 = domain (f ), and any given ε > 0, it followsfrom the continuity of g at ℓ, i.e. lim

w→ℓg(w) = g(ℓ) thatthere exists δ′ > 0 such

that for any w ∈ R ∩ B(ℓ, δ′) we have g(w) ∈ B(g(ℓ), ε), i.e.|g(w)− g(ℓ)| < ε ♠. Then it follows from lim

v→pf (v) = ℓ and δ′ > 0 that there

exists δ > 0 such that for any v = (x, y, z) ∈ D ∩ B(p, δ) \ {p}, one has|f (v)− ℓ| < δ′.In this case, f (v) ∈ B(ℓ, δ′), and hence it follows from ♠ that|g(f (v))− g(ℓ)| < ε for all v ∈ (x, y, z) ∈ D ∩ B(p, δ) \ {p}. In particular, wehave lim

v→pg ◦ f (v) = g(ℓ).

Matb 210/Math200 in 2013-2014

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Proposition. Suppose that the limit lim(x,y)→(a,b)

f (x, y) = ℓ as (x, y) approaches to

(a, b), and let C : x = x(t), y = y(t) for any continuous curve passing through(a, b) at t = 0. Prove that the directional limit g(t) = f (x(t), y(t)) of the functionf along the curve C at t = 0 is given by ℓ, i.e. lim

t→0g(t) = ℓ.

Proof. For any ε > 0, there exists δ > 0 such that for any(x, y) ∈domain(f ) ∩ B((a, b), δ) \ {(a, b)}, one has |f (x, y)− ℓ| < ε. ♠ Asδ > 0, it follows from the continuity of the curve C, there exists δ′ > 0 such thatfor any t ∈ (0 − δ′, 0 + δ′) and t ̸= 0 we have

√(x(t)− a)2 + (y(t)2 − b)2 < δ.

In particular, the point (x(t), y(t)) on the curve C lies in the ball B((a, b), δ), andhence it follows from ♠ that |g(t)− ℓ| = |f (x(t), y(t))− ℓ| < ε for anyt ∈ (0 − δ′, 0 + δ′) \ {0}, so lim

t→0g(t) = ℓ.

Remark. One can use the result of this proposition to disprove the limit ofcertain function does not exist.

Matb 210/Math200 in 2013-2014

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Example. Discuss the limit of the following function f (x, y) at (0, 0) such thatf (x, y) = xy

x2+y2 if (x, y) ̸= (0, 0), and f (0, 0) = 0.

Solution. The limit does not exists at (0, 0). The reason is that one can obtaintwo different directional limits namely 1

2 and − 12 along two different lines y = x

and y = −x respectively. Indeed. along the line y = x, one can parameterizethe line ℓ : x(t) = t, and y(t) = t for t ∈ R, then f (t, t) = t·t

t2+t2 = 12 , for all t ∈ R,

and hence the directional limit of f along the line ℓ at t = 0 is limt→0

f (t, t) = 12 .

Similarly along the line ℓ′ : x(t) = t and y(t) = −t for t ∈ R, then thedirectional limit of f along the line ℓ′ at t = 0 is lim

t→0f (t,−t) = − 1

2 . As these two

limits are not the same, so the original limit lim(x,y)→(0,0)

f (x, y) does not exist.

Matb 210/Math200 in 2013-2014

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Example. Determine the limit of f (x, y) =2x2y

x4 + y2 exist as (x, y) approaches to

(0, 0).

Solution. Let r(t) = (t, t2), which traces out thered curve. For any t ∈ R \ {0}, one has f ◦ r(t) =f (t, t2) = 2t2·t2

t4+t4 = 1, and hence the limit of f alongthe curve r(t) is 2. However, one can check thatf (0, t) = 0 for all non-zero t, so the limit of f alongthe x-axis is 0. As these two limits along differentpaths are not the same, so the limit of f (x, y) doesnot exist as (x, y) approaches to (0, 0).

Matb 210/Math200 in 2013-2014

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Example. Define f (x, y) = sin(xy)xy if xy ̸= 0 and f (x, y) = 1 otherwise. Prove

that f (x, y) is continuous at (0, 0).

Solution. Note that Dom(f ) = R2. For any ε > 0,

it follows from limt→0

sin tt

= 1 and the definition

of limit that there exists δ′ > 0 such that for allt ∈ (−δ′, δ′) \ {0}, we have

∣∣∣ sin tt − 1

∣∣∣ < ε ♡.

Let δ =√

δ′, which is also positive. Then for any (x, y) ∈ B((0, 0), δ) \ {(0, 0)},

we have√

x2 + y2 < δ, and hence |xy| < x2+y2

2 < 12 δ2 < 1

2 δ′ < δ′, so it follows

from ♡ that |f (x, y)− f (0, 0)| ≤∣∣∣∣ sin xy

xy− 1

∣∣∣∣ < ε for all

(x, y) ∈ B((0, 0), δ) \ {(0, 0)}. In particular, we have lim(x,y)→(0,0)

f (x, y) = f (0, 0),

so f is continuous at (0, 0).

Matb 210/Math200 in 2013-2014

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Example. Determine the values of the following limits, if exist:

lim(x,y)→(0,0)

√x2+y2+1−1

x2+y2 .

Proof. First simplify the function by rationalizing the numerator, for any

(x, y) ̸= (0, 0),

√x2 + y2 + 1 − 1

x2 + y2

= x2+y2

(√

x2+y2+1+1)(x2+y2)= 1√

x2+y2+1+1. It follows from the laws of limits that

lim(x,y)→(0,0)

√x2 + y2 + 1 − 1

x2 + y2 = lim(x,y)→(0,0)

1√x2 + y2 + 1 + 1

=1

lim(x,y)→(0,0)

(√

x2 + y2 + 1 + 1)by quotient law of limit

=1√

lim(x,y)→(0,0)

(x2 + y2 + 1) + 1by composition of continuous function

= 1√1+1

= 12 .

Matb 210/Math200 in 2013-2014

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Example. The limit lim(x,y)→(0,0)

e−1

x2+y2 cos(x2 + y2) =

A. 0 B. −1 C. 1 D. e−1 E. does not exist.

Solution. First observe the following inequality holds

ex ≥ 1 + x for any x ≥ 0.

Then it follows that∣∣∣∣e −1x2+y2 cos(x2 + y2)− 0

∣∣∣∣ ≤ 1

e1

x2+y2≤ 1

1 + 1x2+y2

=x2 + y2

1 + x2 + y2 ≤ x2 + y2. It

follows from the sandwich theorem and rules of limit thatlim

(x,y)→(0,0)e

−1x2+y2 cos(x2 + y2) = 0.

Matb 210/Math200 in 2013-2014

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......Example. Determine all the points (x, y) at which f (x, y) = 1x−y has no limit.

Solution. First domain D = Dom(f ) = { (x, y) ∈ R2 | x − y ̸= 0 }= R2 \ { (t, t) | t ∈ R }. At every point (a, b) in D, one knows that a ̸= b, andhence one can find a circular disc B with center (a, b) and radius r = 1

3 |a − b|.We are going to prove that all the point (x, y) of the disc B completely lies in D.Observe that |x − a| ≤

√(x − a)2 + (y − b)2 ≤ 1

3 |a − b|, and similarly|y − b| ≤ 1

3 |a − b|. By triangle inequality,|x − y| = |(x − a) + (a − b) + (b − y)|

≥ | |(x − a) + (a − b)| − |(b − y)| |≥ | | |(a − b)| − |x − a| | − |(b − y)| |≥ |a − b| − 1

3 |a − b| − 13 |a − b| = 1

3 |a − b| > 0, and hence x ̸= y, i.e. (x, y) lies in

D. So it follows from the rules of limit that lim(x,y)→(a,b)

1x − y

=1

a − b. (Continue ..)

Matb 210/Math200 in 2013-2014

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......Example. Determine all the points (x, y) at which f (x, y) = 1x−y has no limit.

Solution. It remains to prove that limit of the function does not exist at any point(a, a). For any point (a, a), one can check the limit of the function along the lineℓ : y + x = 2a passing through (a, a). Rewrite the equation in parametric form,(x, y) = (a + t, a − t), where t ∈ R, then the limit of the function along the line ℓ

is given by lim(x,y)→(a,b) along ℓ

1x − y

= limt→0

1(a + t)− (a − t)

= limt→0

12t

which is

certainly not finite, and hence it does not exist.

Matb 210/Math200 in 2013-2014

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Example. Sketch the graph of the function f (x, y) = x2yx2+y2 for all (x, y) ̸= (0, 0).

Determine the domain Dom(f ) of f , and the set of points at which f has limit.

Proof. Dom(f ) = { (x, y) | denominator x2 + y2 of f does notvanish } = R2 \ {(0, 0)}. It follows from the laws of limit that

lim(x,y)→(a,b)

f (x, y) = f (a, b) for any point (a, b) ∈ Dom(f ). Notice that (0, 0) is an

accumulation point of Dom(f ) but not in Dom(f ). We claim thatlim

(x,y)→(0,0)f (x, y) = 0. It follows from x2 ≤ x2 + y2, that for any (x, y) ̸= (0, 0),

one has |f (x, y)− 0| = x2

x2+y2 · |y| ≤ |y| ≤√

x2 + y2. From the sandwichtheorem of limit that lim

(x,y)→(0,0)f (x, y) = 0.

Matb 210/Math200 in 2013-2014