Lesson: Limits and Continuit of Functions of several Variables Lesson Developer: Kapil Kumar Department/College: Assistant Professor, Department of Mathematics, A.R.S.D. College, Universit of Delhi Institute of Lifelong Learning, Universit of Delhi pg.
Table of Contents Chapter: Limits and Continuit of Functions of several Variables : Learning outcomes : Introduction 3: Functions of several variables 4: Limits of functions of several variables o 4.. Non-eistence of it o 4.. Determining the simultaneous its b changing to polar coordinates 5: Algebra of its 6: Repeated its or iterative its 7: Two-path test for non-eistence of a it 8: Continuit at a point o 8.. definition of continuit of a function at a point Eercises Summar References. Learning outcomes: After studing this chapter ou should be able to understand the Functions of several variables Limits of functions of several variables Algebra of its Repeated its or iterative its Two-path test for non-eistence of a it Continuit at a point Institute of Lifelong Learning, Universit of Delhi pg.
. Introduction: In studing a real world phenomenon and applications in geometr, applied mathematics, engineering and natural science, a quantit being investigated usuall depends on two or more independent variables. Therefore we need to etend the basic ideas of the calculus of functions of a single variable to functions of several variables. In this lesson we will stud the its and continuit for multivariable functions. Although the definitions of the it of a function of two or three variables is similar to the definition of the it of a function of a single variable but with a crucial difference. 3. Functions of Several Variables: Real valued functions of several independent real variables are defined in the same wa as the real valued functions of single variable. The domains of the real valued functions of several variables are the sets of ordered pairs (triples, quadruples, n-tuples) of real numbers and the ranges are subsets of real numbers. For eample:. Consider the function V r h, here V denoted the volume of he clinder, r radius and h height of the clinder. Here V depends on r and h. Thus, r and h are called the independent variables and V is called dependent variable.. The relation Z, between, and z determines a value of z corresponding to ever pair of numbers, which are such that. The region determined b the point (, ) is called the domain of the point (, ). 3. The relation Z e determines a function of two variables (, ); the domain of the function being the whole plane i.e., the set of all the ordered pairs of real numbers. Definition : A variable Z is said to be a function of two variables and, denoted b Z f (, ), if to each pair of values of and (over same domain D) there corresponds a definite value of Z. Here and are called the independent variables and Z is called the dependent variable. Definition : Let D is a set of n-tupple of real numbers (,,..., n). A realvalued function f on D is a rule that assign a unique real number Institute of Lifelong Learning, Universit of Delhi pg. 3
Z f (,,..., n ) To each element in D. The set D is called the domain and the set of z-values taken on b f is the function s range. The,,..., n are called independent variable and the Z is called a function of n independent variables. 4. Limits: The definition of the it of a function of two or three variables is similar to the definition of the it of a function of a single variable but with a crucial difference. A function f (, ) is said to tend to a it as a point (, ) tends to the point ( 0, 0) if for ever arbitraril small positive number, there eists a positive number 0 such that f (, ) whenever 0 0, 0 0 Or 0 0 0 Smbolicall, the it of the function f (, ) at the point ( 0, 0) is denoted b f (, ) (, ) ( 0, 0) Or f (, ) 0 0 Where is called the it (the double it or the simultaneous it) of f when (, ) tends to ( 0, 0) simultaneousl. Value Addition: Note. The definition of it sas that the distance between f (, ) and becomes arbitraril small whenever the distance from (, ) to ( 0, 0) is made sufficientl small (but not 0).. The simultaneous it postulates that b whatever path the point is approached, the function f attains the same it. 3. In general the determination whether a simultaneous it eists or not is a difficult matter but ver often a simple consideration enables us to show that the it does not eist. Institute of Lifelong Learning, Universit of Delhi pg. 4
4. f (, ) f (, ) f (, ) 0 0 (, ) ( 0, 0 ) 0 0 4.. Non-Eistence of Limit: From the simultaneous it postulates it is ampl clear that if f (, ) and if ( ) is an function such that ( ) 0 when 0 (, ) ( 0, 0). Then f, ( ) 0 functions ( ), ( ), must eist and be equal to. Thus, if we can find two such that the it of f, ( ) and f, ( ) different, then the simultaneous it in question does not eist. Eample : For the function f, (, ) (0, 0). Solution: Let m, then are. Find the it when f (, ) f (, m ) (, ) (0, 0) 0 m 0 0 m m m m m Now, if we take m, then f (, ) f (, m ) (, ) (0, 0) 0 m 0 0 m m m m m Institute of Lifelong Learning, Universit of Delhi pg. 5
Thus, we can find m ( ) and m ( ) such that the it of and f, ( ) f, ( ) question does not eist. are different, then the simultaneous it in Eample : Show that the it f (, ), where (, ) (0, 0) f (, ) 4 does not eist. Solution: First we will approach the it along the line f (, ) f (, m) (, ) (0, 0) 0 3 m 0 m m 0 4 m 0 4 4 m, then Now approach the origin along the curve m and let 0, then we have f (, ) f (m, ) (, ) (0, 0) 0 4 m 0 m 4 4 m 0 m m m Since, two its are different, therefore simultaneous it does not eist. Eample 3: Show that the it (, ) (0, 0) does not eist. Solution: We will approach the origin (0, 0) along the line m, then Institute of Lifelong Learning, Universit of Delhi pg. 6
m m (, ) (0, 0) 0 m 0 m m m Since, the it depends upon m and is different for different values of m, therefore simultaneous it does not eist. 4.. Determining the Simultaneous Limits b Changing to Polar Coordinates: Eample 4: Show that ( ) 0. (, ) (0, 0) Solution: Given that ( ) f (, ) and 0, Now ( ) f (, ) 0 0 ( ) Let r cos and r sin, then we have f (, ) 0 r r cos sin. r (cos sin ) r (cos sin ) cos sin cos r cos sin cos r sin cos r sin 4 4 Institute of Lifelong Learning, Universit of Delhi pg. 7
r cos sin cos r 4 4 Now if and 4 4 or if and Thus, for a given 0, there eists a number, such that ( ) 0 when and. Hence, ( ) 0. (, ) (0, 0) Eample 5: Show that (, ) (0, 0) does not eist. Solution: First we will find the it along the line m, thus we have m m 3 (, ) (0, 0) 0 0 ( m ) 0 ( m ) 3 ( m ) 0 ( m ) Institute of Lifelong Learning, Universit of Delhi pg. 8
Now, we will find the it along the curve 3 m, then 3 ( m ) 3 ( m ) (, ) (0, 0) 0 ( m ) 0 3 m 0 3 ( m ) 3 3 3 ( m ) 0 m 3 ( 0) m m m depends upon m, hence the it does not eist. 5. Algebra of Limits: m 3 Theorem (Properties of its of functions of two variables): If f (, ) and g(, ) be two functions defined on some neighborhood of a point ( 0, 0) such that f (, ) and g(, ) (, ) ( 0, 0 ) (, ) ( 0, 0 ) The following rules hold if, and k are real numbers.. Sum Rule: ( f g)(, ) f (, ) g(, ) (, ) ( 0, 0 ) (, ) ( 0, 0 ) (, ) ( 0, 0 ). Difference Rule: ( f g)(, ) f (, ) g(, ) (, ) ( 0, 0 ) (, ) ( 0, 0 ) (, ) ( 0, 0 ) 3. Product Rule: ( f. g)(, ) f (, ). g(, ) (, ) ( 0, 0 ) (, ) ( 0, 0 ) (, ) ( 0, 0 ) 4. Constant Multiple Rule: Institute of Lifelong Learning, Universit of Delhi pg. 9
( k. f )(, ) k k. f (, ) (, ) ( 0, 0 ) (, ) ( 0, 0 ) 5. Quotient Rule: f f (, ) (, ) ( 0, 0) (, ) provided 0 (, ) ( 0, 0) g g(, ) (, ) ( 0, 0) 6. Power Rule: If r and s are integers with no common factors and s 0, then (, ) ( 0, 0) f s s, provided (, ) r r assume that 0 ) r s is a real number. (If s is even we Eample 6: Appling the definition of it find the it eists. Solution: Given that (, ) (0, 0) 4, if it f (, ) 4 If we take 0, then along the line 0, we have 4 0 0 Similarl, if we take 0, then along the line 0, we have 4 0 0 Therefore, if the it does eists as (, ) (0, 0) then the value of the it must be 0. Now, appling the definition of it, let 0 be given then 4 4 0 Institute of Lifelong Learning, Universit of Delhi pg. 0
4 since 4 4 4 4 Thus, if we choose, then 4 4 0, whenever 0 Thus, 4 0. (, ) (0, 0) Eample 7: Prove that 4. (, ) (, 3) Solution: Using the algebra of its, we have (, ) (, 3) (, ) (, 3) (, ) (, 3) 3 4 Hence, 4. (, ) (, 3) Alternative Method: Using the definition of it method For a given 0, we have 4 Institute of Lifelong Learning, Universit of Delhi pg.
when and 3 If and 3, then and 3 and 3 3 ecluding,, thus and 3 3 and 3 3 on adding, we have 4 3 43 3 4 3 Now if, it follows that 3 4 3 3 4 3 4 3 If, we take (or, whichever is smaller) 3 when 4, 3 Hence, 4. (, ) (, 3) 6. Repeated Limits or Iterative Limits: If a function f is defined in some neighborhood of (a, b), then the it f (, ) if it eists, is a function of, sa ( ). Now, if the it ( ) 0 eists and is equal to m, then we write 0 Institute of Lifelong Learning, Universit of Delhi pg.
f (, ) m 0 0 and m is called the repeated it of f as 0 and 0. B changing the order of taking the its, we get the other repeated its i.e., the it f (, ) if it eists, is a function of, sa ( ). Now, if the it ( ) 0 0 eists and is equal to m, then we write f (, ) m 0 0 and m is called the repeated it of f as 0 and 0. Value Addition: Note Two repeated its ma or ma not be equal. I.Q. I.Q. Eample 8: Show that for the function its eists at origin but are unequal. Solution: Given that f (, ), both the repeated f (, ) For first repeated it, we have f (, ) 0 0 0 0 0 0 0 0 () 0 Institute of Lifelong Learning, Universit of Delhi pg. 3
for second repeated it, we have f (, ) 0 0 0 0 0 0 0 0 0 ( ) Hence, both the repeated it eist but are unequal. Value Addition: Note If the simultaneous it of a function f(, ) eists at ( 0, 0), then the two repeated its if the eists are necessaril equal but the converse is not true. However if the repeated its are not equal, then the simultaneous it cannot eist. I.Q. 3 Eample9: Show that for the function the its eist. f (, ) and f (, ) (, ) (0, 0) 0 0 sin, if 0 f (, ), both 0, if 0 eists but f (, ) does not 0 0 Solution: Given that sin, if 0 f (, ) 0, if 0 For repeated it f (, ) sin 0 0 0 0 Institute of Lifelong Learning, Universit of Delhi pg. 4
sin 0 0 sin [using algebra of its] 0 0 0 0 0 0 0 Hence, f (, ) eists and equal to 0. 0 0 Now, repeated it f (, ) sin 0 0 0 0 sin [using algebra of its] 0 0 0 0 sin 0 0 Since, sin 0 does not eists. Therefore, f (, ) 0 0 does not eists. For simultaneous it, Let 0 be given, then Institute of Lifelong Learning, Universit of Delhi pg. 5
If sin sin sin since sin and Now if and Then for a given 0, there eists a number 0 such that sin when and sin 0 (, ) (0, 0) Hence, the simultaneous it eist. Eample 0: Show that for the function f (, ), both the repeated its f (, ) and f (, ) 0 0 0 0 simultaneous it Solution: Given that f (, ) (, ) (0, 0) eists and are equal. But the does not eist. f (, ) For repeated it f (, ) 0 0 0 0 Institute of Lifelong Learning, Universit of Delhi pg. 6
0 0 0 0 0 0 Therefore, f (, ) 0 0 eists and equal to 0. Now, repeated it f (, ) 0 0 0 0 0 0 0 0 0 0 Hence, f (, ) eists and equal to 0. 0 0 Hence, both the repeated it eists and are equal to zero. Now, the simultaneous it, f (, ) (, ) (0, 0) (, ) (0, 0) First taking the it along the line m, then we have Institute of Lifelong Learning, Universit of Delhi pg. 7
f (, ) f (, m) (, ) (0, 0) 0. m 0 m m 0 ( m ) m 0 ( m ) m ( m ) Since, the it depends upon m therefore, Simultaneous it does not eist. I.Q. 4 7. Two-Path Test for Non-Eistence of a Limit: If a function f(, ) has different its along two different paths as (, ) approaches ( 0, 0), then f (, ) does not eist. (, ) ( 0, 0) Eample : Show that the following functions are discontinuous at the origin (I) (II) 4 4 f (, ), (, ) (0, 0), f (0, 0) 0 4 4 3 f (, ), (, ) (0, 0), f (0, 0) 0 6 Solution: (I) Given that f (, ) 4 4 4 4 Simultaneous it along the line f (, ) f (, m) (, ) (0, 0) 0 0 m m 4 4 4 4 4 4 m, we have Institute of Lifelong Learning, Universit of Delhi pg. 8
( m ) 4 4 0 4 ( 4 m ) 4 ( m ) 0 ( 4 m ) 4 ( m ) 4 ( m ) Since the it depends on m, therefore simultaneous it does not eist. (II) Given that f (, ) 3 6 Simultaneous it along the line f (, ) f (, m) (, ) (0, 0) 0 m 0 m 6 6 m 4 3 0 ( 6 4 m ) 3 0 ( 6 4 m ) 0 ( 0) 0 m m, we have Now, taking the simultaneous it along the curve 3, we have 3 f (, ) f (, ) (, ) (0, 0) 0 0 6 6 0 6 6 Institute of Lifelong Learning, Universit of Delhi pg. 9
0 Since the simultaneous it along two different paths is different, therefore simultaneous it does not eist. Hence, the function f(, ) is discontinuous at (0, 0). 8. Continuit at a Point: A function f(, ) is said to be continuous at a point ( 0, 0) of its domain of definition, if (I) f (, ) is defined at ( 0, 0). (II) f (, ) (, ) (0, 0) eists (III) f (, ) f (, ) (, ) ( 0, 0) 0 0 8.. definition of continuit of a function at a point: A function f (, ) is said to be continuous at a point ( 0, 0) of its domain of definition if for a given 0, there eists a 0 such that f (, ) f (, ) whenever, 0 0 0 0 Value Addition: Note. The point at which a function f(, ) is not continuous is called the point of discontinuous or the function is said to be discontinuous at that point.. A function f(, ) is said to be continuous if it is continuous at ever point of its domain. I.Q. 5 I.Q. 6 Institute of Lifelong Learning, Universit of Delhi pg. 0
Eample : Show that the following functions are continuous at the origin (I) (II), (, ) (0, 0) f (, ) 0, (, ) (0, 0), (, ) (0, 0) f (, ) 0, (, ) (0, 0) Solution: (I) Given that f (, ), (, ) (0, 0) Now let 0 be given, then f (, ) 0 0 Let r cos, rsin, then r cos r sin r (cos sin ) r cos sin r 6 4 r cos sin r 4 ( ), If Institute of Lifelong Learning, Universit of Delhi pg.
or if 4 4, Thus, for a given 0, there eists 4 0 such that 0 whenever, Hence, 0 (, ) (0, 0) also f (0, 0) 0 Thus, 0 f (0, 0) (, ) (0, 0) Hence, f (, ) is continuous at (0, 0). (II) Given that f (, ), (, ) (0, 0) Now let 0 be given, then f (, ) 0 0 Let r cos, rsin, then Institute of Lifelong Learning, Universit of Delhi pg.
, If r cos r sin r (cos sin ) r r r cos sin r 4 cos sin ( ) or if, Thus, for a given 0, there eists 0 such that 0 whenever, Hence, 0 (, ) (0, 0) also f (0, 0) 0 Thus, 0 f (0, 0) (, ) (0, 0) Hence, f (, ) is continuous at (0, 0). I.Q. 7 I.Q. 8 Institute of Lifelong Learning, Universit of Delhi pg. 3
Eample 3: Show that the function f (, ) is continuous at the origin, where, (, ) (0, 0) f (, ). 0, (, ) (0, 0) Solution: Given that f (, ), (, ) (0, 0) Now let 0 be given, then f (, ) 0 0 Let r cos, rsin, then 3 r (cos sin ) r (cos sin ) 3 r (cos sin ) r r (cos sin ) r r r (cos sin ) r If, cos sin Institute of Lifelong Learning, Universit of Delhi pg. 4
or if, Thus, for a given 0, there eists 0 such that 0 whenever, Hence, 0 (, ) (0, 0) also f (0, 0) 0 Thus, 0 f (0, 0) (, ) (0, 0) Hence, f (, ) is continuous at (0, 0). I.Q. 9 I.Q. 0 Eercise:. Show that the it does not eist in each case (i) (, ) (0, 0) (ii) (, ) (0, 0) 4 (iii) ( ) (, ) (0, 0) (iv) (, ) (0, 0) (v) (, ) (0, 0) 4 4 (vi) (, ) (0, 0) (vii) (, ) (0, 0) (viii) (, ) (0, 0) 4. Show that the it eist in each case (i) (, ) (0, 0) (ii) (, ) (0, 0) 4 4 Institute of Lifelong Learning, Universit of Delhi pg. 5
(iii) (, ) (0, 0) (iv) (, ) (0, 0) 3. Show that the functions are discontinuous at origin. (i) 4 4, (, ) (0, 0) f (, ) 0, (, ) (0, 0) (ii) f (, ) 0,, (iii), (, ) (0, 0) f (, ) 0, (, ) (0, 0) 4. Show that the following functions are continuous at origin (i) (ii), (, ) (0, 0) f (, ) 0, (, ) (0, 0), (, ) (0, 0) f (, ) 0, (, ) (0, 0) 5. Discuss the continuit of the following function at origin (i) (ii) sin sin, (, ) (0, 0) f (, ) 0, (, ) (0, 0), (, ) (0, 0) f (, ) 0, (, ) (0, 0) Institute of Lifelong Learning, Universit of Delhi pg. 6
, (, ) (0, 0) 6. Show that for the function f (, ), both the repeated 0, (, ) (0, 0) its eists at the origin and are equal but the simultaneous its does not eist. 7. Show that for the function f (, ) 0 0 sin, (, ) (0, 0) f (, ) 0, (, ) (0, 0) eist at the origin. eists, but the other repeated it and the double it do not 8. Show that for the following functions both the repeated it eist but the double it does not at origin (i) f (, ) 4 4 (ii) (iii), (, ) (0, 0) f (, ) 0, (, ) (0, 0), (, ) (0, 0) f (, ) 0, (, ) (0, 0) Summar: In this lesson we have emphasized on the followings Functions of several variables Limits of functions of several variables Algebra of its Repeated its or iterative its Two-path test for non-eistence of a it Continuit at a point Institute of Lifelong Learning, Universit of Delhi pg. 7
References:. G. B. Thomas, R. L. Finne: Calculus, Pearson Education, /e, (0). H. Anton, I. Bivens, S. Davis: Calculus, John Wile and Sons, 7/e, (0) 3. Widder: Advanced Calculus, PHI Learning Private Limited. 4. Straffin J., Applications of Calculus, MAA notes number 9, The mathematical association of America (993). Institute of Lifelong Learning, Universit of Delhi pg. 8