2. Real functions of several real variables

Definition 2.0.1 Let M ⊆ Rn be non-empty. By a real function of n real variables defined on M one refers toa relation which assigns to every ordered n-tuple (x1,x2, . . . ,xn) ∈M exactly one real number y ∈ R. We usethe notation:

• y = f (x1, . . . ,xn), • f : M→ R, • f : (x1, . . . ,xn) 7→ y.

The set M is called the domain of f denoted by D( f ).

R The natural domain of a real function f of n real variables is the set of all ordered n-tuples in Rn for which

the expression defining f is well-defined. However, in many cases it may be very difficult (even impossible)to specify the natural domain of f .

� Example 2.1 An example of a real function from Rn to R is,• for n = 1, the function f (x) = x2 +1 with the natural domain D( f ) = R,• for n = 2, the function f (x,y) =

√x+ y with the natural domain D( f ) = {(x,y) ∈ R2 : x+ y≥ 0},

• for n = 3, the function f (x,y,z) = 1√x2+y2+z2−1

with the natural domain D( f ) = {(x,y,z) ∈ R3 : x2 + y2 +

z2 > 1}.�

Definition 2.0.2 Let y = f (x1, . . . ,xn) be a real function of n real variables defined on M ⊆ Rn,M 6= /0. Bythe graph of f one refers to the set

graph( f ) = {(x1, . . . ,xn,y) ∈ Rn+1 : (x1, . . . ,xn) ∈M∧ y = f (x1, . . . ,xn)}.

R In case n = 2, one refers to the graph {(x,y, f (x,y)) ∈ R3 : (x,y) ∈ D( f )} of a real function f of two real

variables as to a surface in R3.

� Example 2.2 The function z = f (x,y) = −x− y+ 1 is a real function of 2 real variables with the naturaldomain D( f ) = R2. The graph of f is the plane −x− y− z+1 = 0 in R3.

�

12 Chapter 2. Real functions of several real variables

Figure 2.1: plane −x− y− z+1 = 0

2.1 Plane sections

Let us restrict our attention only to 3 dimensions: we will deal with real functions of two real variables andrelated surfaces in R3.

Definition 2.1.1 If a plane intersects a solid (a 3-dimensional object), then the region common to the planeand the solid is called a cross-section of the solid. A plane containing a cross-section of the solid may bereferred to as a cutting plane.

The shape of the cross-section of a solid may depend upon the orientation of the cutting plane to the solid.For instance, while all the cross-sections of a ball are disks, the cross-sections of a cube depend on how thecutting plane is related to the cube. If the cutting plane is perpendicular to a line joining the centers of twoopposite faces of the cube, the cross-section will be a square, however, if the cutting plane is perpendicular to adiagonal of the cube joining opposite vertices, the cross-section can be either a point, a triangle or a hexagon.

Definition 2.1.2 If a plane intersects a surface, then the curve common to the plane and the surface is calleda plane section of the surface.

The concept of plane section is related to cross-section. Namely, a plane section is the boundary of across-section of a solid in a cutting plane.

Definition 2.1.3 Consider the surface in R3 defined by a real function of two real variables, i.e., z = f (x,y).The plane sections by cutting planes that are parallel to a coordinate plane (a plane determined by twocoordinate axes) are called level curves or isolines. More specifically, cutting planes with equations of theform z = k, k ∈ R (planes parallel to the xy-plane) produce plane sections that are often called contour lines.

R Level curves are plotted as perpendicular projections into the coordinate planes. In case of contour lines, theprojection is into xy - plane. Thus, one can say that a contour line which is a plane section of the graph of afunction z = f (x,y) by a cutting plane z = z0 is the set

{(x,y) ∈ D( f ) : f (x,y) = z0}.

Let us point out that this means that a contour line of a real function of two real variables is a curve alongwhich the function has a constant value, so that the curve joins points of equal value.

� Example 2.3 Let us sketch few plane sections of the graph of f (x,y) = x2 + y2.Contour lines corresponding to the cutting planes z =−1, z = 0, z = 1, z = 4 are plotted in Figure 2.2. Note

that the contour line corresponding to the cutting plane z =−1 is an empty set.Level curves corresponding to the cutting planes x = 2, x = 4, y =−1, y = 0 are plotted in Figure 2.3.

�

2.2 Continuity of real functions of several real variables 13

Figure 2.2: contour lines z = 0, z = 1, z = 4

(a) x = 2, x = 4 (b) y =−1, y = 0

Figure 2.3: level curves

2.2 Continuity of real functions of several real variables

The notion of continuity for real functions of several real variables is a generalization of the same notion for realfunctions of one real variable. Intuitively,

f : M ⊆ Rn→ R is continuous at a ∈M if the values f (x) are "close" to the value f (a)for all points x ∈M which are close to a.

Note that it makes sense to talk about continuity of f only at the points at which f is defined. It is the same asfor real functions of one real variable.

14 Chapter 2. Real functions of several real variables

Definition 2.2.1 Let f : D( f )⊆Rn→R and let a = (a1, . . . ,an)∈D( f ). We say the function f is continuousat the point a if

∀ε > 0 ∃δ > 0 ∀x ∈ Oδ (a)∩D( f ) : f (x) ∈ Oε( f (a)).

We say f is continuous if it is continuous at every point a ∈ D( f ).

The condition on continuity can be formulated in equivalent ways, for instance:• ∀ε > 0 ∃δ > 0 ∀x ∈ Oδ (a)∩D( f ) : | f (x)− f (a)|< ε• ∀ε > 0 ∃δ > 0 ∀x ∈ D( f ) : ρ(x,a)< δ ⇒ | f (x)− f (a)|< ε• ∀Oε( f (a)) ∃Mδ := Oδ (a)∩D( f ) : f (Mδ )⊆Oε( f (a))The continuity of functions can be determined by applying the theorems on the sum, difference, product,

quotient and composition of continuous functions:

Theorem 2.2.1 Let f : D( f )⊆ Rn→ R and g : D(g)⊆ Rn→ R be continuous. Then(i) the functions f ±g and f g are continuous on D( f )∩D(g),

(ii) the function fg is continuous on (D( f )∩D(g))\{x ∈ D(g) : g(x) = 0}.

Theorem 2.2.2 Let f : D( f ) ⊆ R→ R and g : D(g) ⊆ Rn→ R be continuous. Then the function f ◦ g iscontinuous on {x ∈ D(g) : g(x) ∈ D( f )}.

Let us point out the function f in Theorem 2.2.2 is a real function of one real variable.The following proposition together with Theorems 2.2.1, 2.2.2 allows one to prove continuity of real

functions of several variables by using the knowledge on continuity of real functions of one real variable:

Proposition 2.2.3 Let i ∈ {1,2, . . . ,n} be arbitrary. The function f : Rn→ R defined as

f (x1,x2, . . . ,xn) = xi

is continuous on Rn.

Proof. Let i∈ {1,2, . . . ,n} be arbitrary. Consider the function f (x1,x2, . . . ,xn) = xi. We will prove the continuityof f by proving the continuity of f at every point a ∈ Rn.

Consider arbitrary a ∈ Rn and let us show that

∀ε > 0 ∃δ > 0 : ρ(x,a)< δ ⇒ | f (x)− f (a)|< ε.

Let ε > 0 be arbitrary and let δ be chosen as δ := ε . Then, if

ρ(x,a) =√

(x1−a1)2 +(x2−a2)2 + · · ·+(xn−an)2 < δ = ε,

then obviously (because 0≤ (x j−a j)2 for all j = 1, . . . ,n)

| f (x)− f (a)|= |xi−ai|=√

(xi−ai)2 ≤√

(x1−a1)2 +(x2−a2)2 + · · ·+(xn−an)2 < ε.

Thus, f is continuous at a and because a was chosen arbitrarily from Rn, f is continuous on Rn. �

� Example 2.4 Let us prove the function f (x,y) = x+ y is continuous on R2.The functions f1(x,y) = x and f2(x,y) = y are continuous according to Proposition 2.2.3. From Theorem

2.2.1, the sum f1(x,y)+ f2(x,y), which is the function f , is a continuous function on R2. �

� Example 2.5 Let us prove the function f (x,y) = arctan(x−y)x2+y2 is continuous on its natural domain R2 \{(0,0)}.

The continuity of the functions (x,y) 7→ (x− y) and (x,y) 7→ (x2 + y2) follows from Theorem 2.2.1 andProposition 2.2.3. The continuity of the function (x,y) 7→ arctan(x−y) follows from Theorem 2.2.2. Finally, thecontinuity of the function f given as a fraction of two continuous functions follows from Theorem 2.2.1. �

2.3 Limits of real functions of several real variables 15

R If f : R→ R is continuous on (a,b) ⊆ R and g : R→ R is continuous on (c,d) ⊆ R, then the functionsf (x)± g(y), f (x)g(y) are continuous on (a,b)× (c,d) and the function f (x)g(y) is continuous on ((a,b)×(c,d))\{(x,y) ∈ (a,b)× (c,d) : g(y) = 0}.

For example, the function f (x,y) = x3

arcsin(1+y) is continuous on R× ((−2,−1)∪ (−1,0)) because thefunction x3 is continuous on R, the function arcsin(1+ y) is continuous on (−2,0) and arcsin(1+ y) = 0for y =−1.

One of the important features of continuous real functions of one real variable on a closed interval is thatthey attain their minimum and maximum on this interval. One refers to this fact as extreme value theorem andits generalization to real functions of several real variables is stated below.

Definition 2.2.2 Consider a function f : M ⊆ Rn→ R. We say the function f has(i) a global maximum at a ∈M if f (x)≤ f (a) for all x ∈M,

(ii) a global minimum at b ∈M if f (b)≤ f (x) for all x ∈M.

Theorem 2.2.4 Let M ⊆ Rn be a closed and bounded set and let a function f : M→ R be continuous on M.Then there exists a,b ∈M such that f has a global maximum at a and a global minimum at b.

2.3 Limits of real functions of several real variablesThe limit of a function f at a point a gives us an information about the function values of f in a neighbourhoodof a. Namely, the notation

limx→a

f (x) = L

means that the function values f (x) are "close" to the limit L for the points x "close" to the limiting point a.

Definition 2.3.1 Let a ∈ Rn and let f : M ⊆ Rn→ R be such that Pδ (a)∩M 6= /0 for every δ > 0. We saythat the function f has the limit L at the point a, denoted as lim

x→af (x) = L, if

∀Oε(L) ∃Pδ (a) ∀x ∈M∩Pδ (a) : f (x) ∈ Oε(L).

R Let us point out that f does not need to be defined at a as well as it does not need to be defined at all pointsof Pδ (a).

The definition of limx→a

f (x) = L can be stated in several equivalent ways, for instance

∀ε > 0 ∃δ > 0 ∀x ∈M : 0 < ρ(x,a)< δ ⇒ | f (x)−L|< ε.

Theorem 2.3.1 A function f : Rn→ R has at a point a ∈ Rn at most one limit.

Proof. The proof is identical to the proof of the statement for n = 1. �

Theorem 2.3.2 Let M ⊆ Rn be an open set and let f : M ⊆ Rn→ R. Then the function f is continuous ata ∈M if and only if lim

x→af (x) = f (a).

The theorem above makes it easy to compute the limits of real functions of several real variables at the pointsat which these functions are continuous. We will not deal with more complicated situations where the limitswould be computed at the points at which the functions are not continuous or not defined.

However, note that the theorems on the limits of sum, difference, product, quotient and composition offunctions are still valid in case of real functions of several real variables. There is no theorem such as l’Hospitalrule for real functions of several real variables.

� Example 2.6

16 Chapter 2. Real functions of several real variables

• lim(x,y)→(1,2)

(x2 + y) = 3

• lim(x,y)→(0,0)

x4− y4

x2 + y2= lim

(x,y)→(0,0)(x2− y2) = 0

�

� Example 2.7 Let us compute the limit

lim(x,y)→(0,0)

xyx2 + y2

. (2.1)

The notation (x,y)→ (0,0) means that the point (x,y) is getting closer to the point (0,0) in the metric ρ .One specific way of getting closer to (0,0) is to be moving on a line passing through (0,0). Such line has aparametrization x = t, y = kt, t ∈ (−∞,∞) where k ∈ R. By the substitution of these expressions for x and y inthe limit (2.1) one derives the limit in one variable t:

limt→0

t k tt2 +(kt)2

= limt→0

k1+ k2

=k

1+ k2. (2.2)

Note that if the limit (2.1) exists, then it equals the limit (2.2). However, the opposite implication does not holdin general.

For different k’s in (2.2) we obtain different limits. This means that by coming with (x,y) to the point (0,0)along different lines will lead us to different values of the limit lim

(x,y)→(0,0)

xyx2 + y2

which is in contradiction with

Theorem 2.3.1. Hence, the limit (2.1) does not exist. �

� Example 2.8 Let us compute the limit

lim(x,y)→(0,0)

x3

x2 + y2. (2.3)

By considering the lines x = t, y = kt, t ∈ (−∞,∞) where k ∈ R in (2.3), we compute the limit

limt→0

t3

t2 +(kt)2= lim

t→0

t1+ k2

= 0.

Hence, 0 is a candidate to be the value of (2.3). To prove this let us check the definition:

∀ε > 0 ∃δ > 0 ∀(x,y) ∈Pδ ((0,0)) :x3

x2 + y2∈ Oε(0).

So, for every ε > 0 we need to find δ > 0 such that the following holds:

0 <√

x2 + y2 < δ ⇒∣∣∣∣ x3x2 + y2

∣∣∣∣< ε. (2.4)Let us estimate

∣∣∣ x3x2+y2 ∣∣∣ by using the quantity√x2 + y2 :∣∣∣∣ x3x2 + y2∣∣∣∣= |x|x2x2 + y2 ≤ |x|(x2 + y2)x2 + y2 = |x|=√x2 ≤√x2 + y2.

Then, for every ε > 0 we can take δ = ε to derive (2.4). �

� Example 2.9 Consider the function f (x,y) = x2+y22x and compute lim

(x,y)→(0,0)f (x,y).

Let us plot the contour lines of f corresponding to the cutting planes z =±1,±2,±3, see Figure 2.4.One can see that for 0 6= z0 ∈ R the contour line is given as

x2−2xz0 + y2 = 0, i.e. (x− z0)2 + y2 = z20,

which is the circle with the center (z0,0) and the radius |z0|. Note that the line x = 0 is omitted from the contourplot.

Because in arbitrarily small neighbourhood of (0,0) the function f attains all real non-zero values, lim(x,y)→(0,0)

f (x,y)

does not exist. However, note that for (x,y) getting close to (0,0) along the lines one obtains the limit

limt→0

t2 + k2t2

2t= lim

t→0(12

t(1+ k2)) = 0 for all k ∈ R. �

2.3 Limits of real functions of several real variables 17

Figure 2.4: contour lines z =±1,±2,±3

Very useful approach to compute limits of real functions of two real variables is to rewrite these functions inpolar coordinates. Let us recall a point (x,y) ∈ R2 can be specified by its distance r from the point (0,0) (radiusof a circle centered at (0,0)) and by the angle ϕ between the x-axis and the line passing through the points (0,0)and (x,y). (r,ϕ), where r ∈ [0,∞) and ϕ ∈ [0,2π), are then referred to as the polar coordinates of (x,y). Notethat

x = r cos(ϕ), y = r sin(ϕ).

� Example 2.10 Let us use the polar coordinates to compute the limit

lim(x,y)→(0,0)

√x2 + y2

2x.

Since (x,y) = (r cos(ϕ),r sin(ϕ)) for r ∈ [0,∞) and ϕ ∈ [0,2π), one obtains x2+y2 = r2. Because (x,y)→ (0,0)means that ρ((x,y),(0,0)) =

√x2 + y2 = r, one can rewrite the limit as

limt→0

r2r cos(ϕ)

=1

2cos(ϕ).

Because the limit depends on the angle which specifies the direction in which (x,y) moves towards (0,0), the

limit lim(x,y)→(0,0)

√x2 + y2

2xdoes not exist. �

� Example 2.11

lim(x,y)→(0,0)

x2 yx2 + y2

= limr→0

r3 cos2(ϕ)sin(ϕ)r2

= limr→0

(r cos2(ϕ)sin(ϕ)) = 0

�

18 Chapter 2. Real functions of several real variables

2.4 Exercises1. Determine the domain of the function f and sketch the obtained set:

(a) f (x,y) =1

ln(x+2)(b) f (x,y) = ln(x− y)

(c) f (x,y) =√

xy

(d) f (x,y) =√

1− x2 +√

1− y2

(e) f (x,y) =

√y2

x

(f) f (x,y) =

√1− x2− y

x2 + y2

(g) f (x,y) =1

x−1arccos(x2 + y2)

2. Determine the domain of f :(a) f (x,y) = (ex+y,

√x− y, log(y−|3x|))

(b) f (x,y) =(√−y+ x2−1, y√

2(1−x2)+y

)3. Sketch the graph of the function f :

(a) f (x,y) = 1− x(b) f (x,y) = 2x+ y−1(c) f (x,y) = y2

(d) f (x,y) =−√

x2 + y2 +1

(e) f (x,y) = 1− x2− y2

(f) f (x,y) =1

x2 + y2(g) f (x,y) = xy(h) f (x,y) =

√1− x2− y2

4. Sketch the domain of f (x,y) = arccos(|x− y|) and its π2 - contour line.5. Sketch the domain of f and determine its properties. Into the same figure plot the contour line z = a (or

the contour line passing through the point A ∈ D( f )):(a) f (x,y) =

√xy, a = 2, A = [1,0],

(b) f (x,y) = arccos(xy), a = π2 , A = [−12 ,−1],

(c) f (x,y) =

√x+1− (y−1)2

x, a ∈ {1,2, 12},

(d) f (x,y) = arcsin(|xy2−1|+1), a ∈ {π3 ,π2 },

(e) f (x,y) =1

x(y+2), A ∈ {[1,−1], [−1,1]},

(f) f (x,y) =√

sin(x2 + y2), a = 1,

(g) f (x,y) = ln(

x4− x2− y2

), a = 0.

6. Determine D( f ), properties of D( f ) and the plane sections corresponding to the specified cutting planes:(a) f (x,y) =

√3√

x− 4√

xarcsin(xy), cutting planes: z = 0, x = 8,

(b) f (x,y) =y2

x, cutting planes z = 2, x = 2, y = 2.

7. Prove that the constant function f (x,y) = 5 is continuous on R2.8. Justify whether the function f is continuous:

(a) f (x,y) =√

x2 + y2

(b) f (x,y) =1y2

(c) f (x,y) = sgn(xy)

(d) f (x,y) =x+ y

x

9. Plot the points (2, π2 ), (1,−π2 ) and (3,5) given in polar coordinates.

10. Describe the set of points lying on(a) a line segment parallel to x-axis,(b) a line segment parallel to y-axis,(c) a circle centered at origin,(d) a semicircle centered at origin.

by using polar coordinates.11. Compute lim

(x,y)→(0,0)

yx2 + y

.

2.5 Answers 19

2.5 AnswersExercise 1.:

(a) D( f ) = {(x,y) ∈ R2|x+2 > 0∧ x+2 6= 1} (b) D( f ) = {(x,y) ∈ R2|x− y > 0}

(c) D( f ) = {(x,y) ∈ R2|x/y≥ 0∧ y 6= 0}(d) D( f ) = {(x,y) ∈ R2|1− x2 ≥ 0∧1− y2 ≥ 0}= 〈−1,1〉2

(e) D( f ) = {(x,y) ∈ R2|y2/x≥ 0∧ x 6= 0} (f) D( f ) = {(x,y) ∈ R2 \{(0,0)}|1− x2− y≥ 0}

(g) D( f ) = (x,y) ∈ R2|x−1 6= 0∧−1≤ x2 + y2 ≤ 1

Exercise 2.:(a) D( f ) = {(x,y) ∈ R2|x− y≥ 0∧ y−|3x|> 0}(b) D( f ) = {(x,y) ∈ R2|x2−1≥ y∧2(1− x2)+ y > 0}

20 Chapter 2. Real functions of several real variables

Exercise 3.:

(a) (b)

(c) (d)

(e)(f)

(g)(h)

2.5 Answers 21

Exercise 4.:

Figure 2.7: D( f ) = {(x,y) ∈ R2|−1≤ |x− y| ≤ 1}, π/2-contour line is y = x

Exercise 5.:

(a) D( f ) = {(x,y) ∈ R2|xy≥ 0} is open, path-connected,it is neither closed, nor bounded, nor convex; 2-contourline is y = 4/x; 0-contour line (since f (1,0) = 0) is{(x,y) ∈ D( f )|xy = 0} = {(x,y) ∈ D( f )|x = 0∨ y = 0}which is union of x-axis and y-axis

(b) D( f ) = {(x,y) ∈ R2| − 1 ≤ xy ≤ 1} is closed, un-bounded, path-connected, it is neither open, nor con-vex; π/2-contour line is {(x,y) ∈ D( f )|xy = 0} ={(x,y) ∈ D( f )|x = 0∨ y = 0}; π/3-contour line (sincef (−1/2,−1) = π/3) is y = 1/(2x)

(c) D( f ) = {(x,y) ∈ R2|(x+ 1− (y− 1)2)/x ≥ 0∧ x 6=0} is unbounded, it is neither open, nor closed, norpath-connected, nor convex; 1-contour line is {(x,y) ∈D( f )|x 6= 0∧ (y = 0∨ y = 2)}; 2-contour line is {(x,y) ∈D( f )|x 6= 0∧ x−1/3 =−1/3(y−1)2}; 1/2-contour lineis {(x,y) ∈ D( f )|x−4/3 = 4/3(y−1)2}

(d) D( f ) = {(x,y) ∈ R2| |xy2 − 1| ≤ 0} is closed, un-bounded, it is neither open, nor path-connected, nor con-vex; π/3-contour line is an empty set since R( f ) = {π/2};π/2-contour line is {(x,y) ∈ D( f )|y =±1/

√x}

22 Chapter 2. Real functions of several real variables

(e) D( f ) = {(x,y) ∈ R2|x 6= 0 ∧ y 6= −2} is open, un-bounded, it is neither closed, nor path-connected, norconvex; 1-contour line (since f (1,−1) = 1) is {(x,y) ∈D( f )|y = 1/x−2}; −1/3-contour line (since f (−1,1) =−1/3) is {(x,y) ∈ D( f )|y =−3/x−2}

(f) D( f ) =⋃

k∈N0{(x,y) ∈ R2|2kπ ≤ x2 + y2 ≤ (2k +

1)π} is closed, unbounded, it is neither open, nor path-connected, nor convex; 1-contour line is

⋃k∈N0{(x,y) ∈

D( f )|x2 + y2 = π/2+2kπ}

(g) D( f ) = {(x,y) ∈ R2|x/(4− x2− y2) > 0∧ 4− x2−y2 6= 0} is open, unbounded, it is neither closed, norpath-connected, nor convex; 0-contour line is {(x,y) ∈D( f )|(x+1/2)2 + y2 = 17/4∧ x 6= 0}

Exercise 6.:(a) D( f ) = {(x,y) ∈ R2|x≥ 1∧−1/x≤ y≤ 1/x} is closed, unbounded, path-connected, it is neither open,

nor convex;(b) D( f ) = {(x,y) ∈ R2|x 6= 0} is open, unbounded, it is neither closed, nor path-connected, nor convex;

plane section corresponding to z = 2 is {(x,y,2) ∈ R3|x = y2/2∧ x 6= 0}; plane section corresponding tox = 2 is {(2,y,z) ∈ R3|z = y2/2}; plane section corresponding to y = 2 is {(x,2,z) ∈ R3|z = 4/x}

Exercise 7.:One needs to prove that f is continuous at every (x0,y0) ∈ R2, i.e. that

∀ε > 0∃δ > 0∀(x,y) ∈ Oδ (x0,y0) : f (x,y) = 5 ∈ Oε(5).

For that it is sufficient to note that 5 ∈ Oε(5) for every ε > 0. Hence, for any ε > 0, δ > 0 can be chosenarbitrarily, e.g. δ = ε , δ = 1, . . . , for the implication (x,y) ∈ Oδ (x0,y0)⇒ f (x,y) = 5 ∈ Oε(5) to be satisfied.

Exercise 8.:

(a) continuous(b) continuous

(c) not continuous(d) continuous

Exercise 9.:

Exercise 10.:

2.5 Answers 23

(a) {(r,ϕ)|ϕ = 0∧ r ∈ 〈0,2〉}(b) {(r,ϕ)|ϕ = π/2∧ r ∈ 〈0,1〉}

(c) {(r,ϕ)|r = 2∧ϕ ∈ 〈0,2π〉}(d) {(r,ϕ)|r = 2∧ϕ ∈ 〈0,π〉}

Exercise 11.:

lim(x,y)→(0,0)

yx2 + y

does not exist since the limit limt→0

−kt2

t2− kt2=

kk−1

, where t was substituted for x and −kt2

was substituted for y in the original limit, varies for different k 6= 1.