Lesson -1: Successive Differentiation

Transcription

1 Lesson -1: Successive Differentiation The process of differentiating a given function again and again is called as Successive Differentiation and the results of such differentiation are called successive derivatives. The higher order differential coefficients will occur more frequently in spreading a function all fields of scientific and engineering applications. Notations: Higher Derivatives of y = f(x) (i) n th order derivative: (ii) f, f, f,...., n th order derivative : (iii) D y, D 2 y, D 3 y,., n th order derivative : D n y (iv) y', y'', y''',......, n th order derivative : y (n) Solved Examples : 1. If y = sin(sin x), prove that y 2 (x) + tan x y 1 + y cos 2 x = 0 Solution: Differentiating y = sin(sin x).. (1) with respect to x, we get y 1 = cos (sin x) cos x.(2) Again differentiating (2) w.r.to x, we find y 2 = -[sin x cos (sin x) + cos 2 x sin (sin x)] = -[tan x cos x cos (sin x) + cos 2 x sin(sin x)] = -[tan x y 1 + y cos 2 x], using (1) Or, y 2 + tan x y 1 + y cos 2 x = 0 2., show that 2y 1 y 3 = 3 y 2 2 Solution: y 1 = = y 2 = -2( ad-bc ).c / (cx + d) 3, y 3 = 6 ( ad bc.c 2 / (cx + d) 4 Now, 2y 1 y 3 = 2. = 12 1

2 = 3 [ - ] 2 = 3 y Find the n th derivative of y = + Solution: y 1 = = - + y 2 = (-1) 2 y 3 = (-1) 3 + (-1) 2 + Differentiating (n-3) times w.r.to x, we find y n = (-1) n + (-1) (n-1) + = (-1) n [ - + ] Some Standard Results: (i) D n (ax + b) m = m (m - 1) (m - 2).(m n + 1) (ax + b) m - n a n (ii) D n [ ] = (iii) D n log (ax + b) = (iv) D n = m n a mx (log a) n Proof: Put a mx = y, taking logarithm, mx log a = log y, differentiating w.r.to x y 1 = m log a or, y 1 = y m log a, y 2 = y 1 m log a = (m log a ) 2 y, y 3 = (m log a ) 3 y Proceeding likewise 2

3 y n = (m log a ) n y = m n ( log a ) n a mx (v) (vi) D n ( e mx ) = m n e mx D n sin(ax + b) = a n sin Proof: y 1 = a cos (ax+b) = a sin etc.. (vii) D n cos(ax + b) = a n cos (viii) D n [ e ax sin( bx+c) ] = (a 2 + b 2 ) n/2 e ax sin { n tan -1 (b/a) + bx + c} Proof: Put y = e ax sin (bx+c) y 1 = e a x {a sin (bx+c) + b cos (bx+c). Let a = r cos α and b = r sin α, thus y 1 = e ax r sin (α+bx+c) y 2 = r e ax {a sin ( α + bx + c) + b cos (α + bx + c) }= r 2 e ax sin ( 2α + bx + c) y 3 = r 3 e ax sin ( 3α + bx + c) and in general y n = r n e ax sin ( nα + bx + c), but r = (a 2 + b 2 ) 1/2 and hence the proof follows. (ix) D n [ e ax cos (bx+c)] = (a 2 + b 2 ) n/2 e ax cos { n tan -1 (b/a) + bx + c} Problems Set for Practice (i) If y = e ax sin bx, prove that y 2 2a y 1 + ( a 2 + b 2 ) y = 0. (ii) If y = log{ x + (1+x 2 ) }, prove that (1+x 2 ) y 2 + x y 1 = 0. (iii) If y = tan -1 (sinh x), prove that y 2 + y 2 1 tan y = 0. (iv) Find y 2, when x = a cos 3 θ and y = b sin 3 θ. (v) If xy = e x + b e -x, prove that x y y 1 x y = 0. (vi) Find the n th derivative of, 3 Leibnitz s Theorem Leibnitz s theorem is useful in the calculation of n th derivatives of the product of two functions. Statement of the theorem: If u and v are functions of x, then n C 1 + n C n C r Solved Examples:

4 1. Find the n th derivative of the function x 2 log 3x. Solution. We take u = log 3x and v = x 2, then v 1 = 2x, v 2 = 2 and v 3, v 4 etc.. are all zero. By Leibnitz s theorem, ( x 2 log 3x.) n = ( log 3x) n x 2 + n C 1 ( log 3x) n-1 2x + n C 2 ( log 3x) n-2 2 = (-1) n-1 + = [ (n - 1)(n - 2) x 2 - n(n - 2) 2 x 2 + n(n - 1) x 2 ] = [ (n - 1)(n - 2) x 2 - n(n - 3) x 2 ] 2. If y = x = 0. Also find y n (0). Solution. y 1 = =... (i) or, (1-x 2 2 ) y 1 = m 2 y 2. Differentiating again, (1-x 2 ) 2 y 1 y 2 2x y 2 1 2m 2 y y 1 = 0 or, (1-x 2 ) y 2 x y 1 m 2 y = (ii) Differentiating n- times following Leibnitz s theorem, (1-x 2 ) y n+2 + n y n+1 (-2x) + i.e, (1-x 2 ) y n (iii) At x=0 y(0) =, y 1 (0) = -m (using (i)), y 2 (0) = m 2 ( using(ii) ). Using (iii) we obtain y 3 (0) = -m (1+m 2 ) y 4 (0) = -m 2 (2 2 +m 2 ) y 5 (0) = (3 2 + m 2 ) y 3 (0) = - m (1 + m 2 ) (3 2 + m 2 ) y 6 (0) = (4 2 + m 2 ) y 4 (0) = - m 2 (2 2 + m 2 ) (4 2 + m 2 ). Hence in general, y n (0) = - m (1 + m 2 ) (3 2 + m 2 ) {(n-2) 2 + m 2 } y n (0) = m 2 (2 2 + m 2 ) (4 2 + m 2 ) {(n-2) 2 + m 2 } for for More clearly, 4

5 y 2k (0) = m 2 (2 2 + m 2 ) (4 2 + m 2 ) {(2k -2) 2 + m 2 } y 2k+1 (0) = - m (1 + m 2 ) (3 2 + m 2 ) {(2k - 1) 2 + m 2 } 3. If x = sin t, y= cos pt, show that (1 x 2 ) y 2 x y 1 + p 2 y = 0. Hence deduce that (1 x 2 ) y n+2 ( 2n + 1) x y n+1 ( n 2 - p 2 ) y n = 0 Solution. i.e, y 1 = So that (1-x 2 ) y p 2 y 2 p 2 = 0. Differentiating again w.r. to x, we obtain (1-x 2 ) 2y 1 y 2-2 x y y y 1 p 2 = 0 or, (1-x 2 ) y 2 - x y 1 + p 2 y = 0 (dividing both sides by 2y 1 ). Now differentiating n more times using Leibnitz s Theorem, we find that (1 x 2 ) y n+2 + n y n+1 (-2x) + Or, (1 x 2 ) y n+2 + (2n + 1) x y n+1 ( n 2-4. If x = tan(log y), then find the value of (1+x 2 ) y n+1 + (2nx 1) y n + n(n-1) y n-1. Solution. Let us take x = tan (log y) i.e, tan -1 x = log y or, y = with respect to x,. Now differentiating y 1 = = i.e, (1 + x 2 ) y 1 y = 0. Apply Leibnitz s theorem in differentiating n- times and get D n y 1 (1 + x 2 ) + n C 1 D n-1 y 1. 2x + n C 2 D n-2 y D n y = 0 Or, (1+x 2 ) y n+1 +2nx y n + n(n-1) y n-1 y n = 0 i.e, (1+x 2 ) y n+1 + (2nx 1) y n + n(n-1) y n-1 = = 2x or, y = [ x + ] m or, [ x - ] m Prove that ( x 2 1) y n+2 + (2n + 1) x y n+1 + ( n 2 m 2 ) y n = 0. 5

6 Solution. + = x so, 2 2x + 1 = 0. Solving this quadratic equation we find i.e, = x. Therefore y = (x ) m or, y = (x ) m. Let us consider y = (x ) m Then y 1 = m (x ) m-1 { 1 + } or, = m y. Squaring both sides, we find ( x ) y 1 = m 2 y 2. Again differentiating with respect to x, ( x 2 1 ) 2y 1 y 2 + 2x y m 2 y y 1 = 0. Simplifying, ( x 2 1 ) y 2 + x y 1 - m 2 y = 0 Differentiating n-more times using Leibnitz s theorem, we arrive at ( x 2 1) y n+2 + (2n + 1) x y n+1 + ( n 2 m 2 ) y n = 0. Problems Set for Practice In each of the following problems, apply Leibnitz s theorem to get the results. (i) Show that } = (Take u = and v = log x, then apply Leibnitz s theorem) (ii) If y = a cos (log x) + b sin (log x), prove that x 2 y n+2 + (2n + 1) x y n+1 + (n 2 + 1) y n = 0 (iii) If y = x n log x, show that y n+1 = (iii) y n = D n ( x n log x ), prove that y n = n y n-1 + ( n 1)! Lesson -2: Mean Value Theorems & Expansion of Functions Mean-Value Theorems In this topic we shall discuss a few important theorems of differential calculus- Rolle s Theorem, Lagrange s & Cauchy s Mean-Value Theorem. Expansions of functions by Taylor s and Maclaurin s theorem. 6

7 Rolle s theorem and its application Statement: Let a function f be defined on a closed interval [a,b]. Suppose, further that (i) f is continuous on [a,b] (ii) f is differentiable in the open interval (a,b) and (iii) f(a) = f(b) then, there exists at least one point x=c lying within a < c < b, such that f (c) = 0. Proof. Since f is continuous on [a,b], it must attains it s lub M and glb m there, i.e, there exist two points of [a,b] such that f( Now either M=m or, M I. If M=m, then f(x) must be constant for all x and hence f (x) = 0 at all points of [a,b]. II. Next suppose M Then since f(a) = f(b) and m either M or m, if not both must be different from f(a) or f(b). Then f( f(a) and f( f(b). Thus is neither a nor b. By hypothesis f(x) is differentiable in the open interval (a,b), so f ( ) exists. We shall prove that f ( ) = o. Since f ( ) exists, R f ( ) = L f ( ) = a finite number or, Since f( ) = M ( the greatest value of f(x) in [a,b], f( - f( ) If h > 0, then And if h < 0, then 0 Taking limits h + and h in the two cases, we get R f ( ) and L f ( ). But R f ( ) = L f ( ) =. f ( ). Hence the only possibility is f ( ) = 0. The geometric interpretation of Rolle's Theorem is that if f is a continuous function whose domain is a closed interval and f has tangent lines at every point of its graph except possibly the endpoints, then at least one of those tangent lines is horizontal. Corollary. If a < b are two roots of the equation f(x)=0, then the equation f (x) = 0 will have at least one root between a and b, provided 7

8 (i) (ii) f(x) is continuous on [a,b] and f is differentiable in (a,b). If f(x) be a polynomial, the conditions (i) and (ii) are satisfied. Hence Between any two roots of a polynomial f(x) lies at least one zero of the polynomial f (x) Example 6 Verify Rolle s Theorem for f(x) = on [0,4]. Solution. Here f(x) is continuous on [0,4] and f (x) = = = f(x) is differentiable on [0,4]. So by Rolle s theorem f (x) should have at least one zero within (0,4), i.e, = 0 for some x (0,4). Equating = 0, we find x = -2. Here x = - 2 lies within (0,4). Mean-Value Theorem: Lagrange s Form. Statement. If a function f is a) f is continuous on [a,b] b) f is differentiable in (a,b) then there exists at least one value of x, say c, such that = f (c, for a < c Geometrically, this is equivalent to stating that the tangent line to the graph of f at c parallel to the chord joining the points (a, f(a)) and (b, f(b) ). Example 7. Verify Lagrange s Mean-Value Theorem for the function f(x) = 2x 2 7x 10 over ( 2, 5 ) and find c of the Lagrange s Mean-Value Theorem. Solution. Here a=2 and b=5, hence by Lagrange s Mean-Value Theorem f (c or, f (c = (i) 8

9 Again f (x = 4x 7, so f (c = 4c 7 (ii) Combining (i) and (ii) we get 4c 7 = 7 so, c = and lies within (2,5). Example 8 Prove that if <, 0 < a < b Hence show that < < Solution. We take f(x) = so that ( = f(x) satisfies all the conditions of Lagrange s Mean-Value Theorem, so or, 1 + a 2 < 1 + c 2 < 1 + b 2 or, = for some c, a < c < b or, < < or, < < Now let a =1 and b =, then < or, < < Example 9 Estimate Solution. We take f(x) =, x [27,28]. So that f (x) =. Then using Lagrange s Mean-Value Theorem = f, x 0 (27,28) i.e, f(28) = f(27) + f (x 0 ) = + = 3 +, 27 < x 0 <28 < 3 + = 3 +. Thus, < 3 9

10 Cauchy s Mean-Value Theorem. Statement. If f(x) and g(x) are continuous in [a,b] and differentiable in (a,b) and g (x) any x in (a,b), then there exists at least one point x= c in (a,b) such that for = We consider the function F(x) = f(x) - g(x) Here g(b) g(a) 0 since g (c), because otherwise g(b) = g(a) and by Rolle s theorem g (c) = 0. Now F(x) is continuous in [a,b] since f(x) and g(x) are continuous in [a,b]. Again since f (x) and g (x) exist in (a,b), F (x) exists in (a,b) and that F (x) = f (x) - g (x) Clearly, F(a) = F(b). Thus F(x) satisfies all the conditions of Rolle s theorem in the interval [a,b]. Therefore there should have at least one point c of x between a and b, such that F (c) = 0 i.e, 0 = f (c) - g (c) or, = Example 10. If in the Cauchy s Mean-Value Theorem we take f(x) = e x and g(x) = e -x, then prove that c is the arithmetic mean between a and b. Solution. f(x) and g(x) satisfies the conditions of Cauchy s Mean-Value Theorem, so there exists at least one point x=c in (a,b) such that =, a < c < b or, = i.e, c = Example11. Using Cauchy Mean Value Theorem, show that 1 - < cos x for x Solution. Applying Cauchy s Mean-Value Theorem f(x) = 1 cos x and g(x) = on the interval [0,x]. We get = < 1 for some c. On simplification proof follows. Problems Set for Practice 10

11 1.For each of the following, verify that the hypotheses of Rolle's Theorem are satisfied on the given interval. Then find all value(s) of c in that interval that satisfy the conclusion of the theorem. (i) f(x) = -x 2-4x 11 on [0,4] (ii) f(x) = - sin x on [0,2 ] 2. Use the mean value theorem (MVT) to establish the following inequalities. (i) e x > 1 + x for x R (ii) < log x < x 1 for x > 1. 3.Does there exist a differentiable function f:[0,2] R satisfying f(0) = -1, f(2) = - 4 and f (x) < 2 for x [0,2]? Generalized Mean-Value Theorem I. Taylor s Theorem with Lagrange s Form of Remainder: Statement. Let f(x) be a function defined in the closed interval [a,a+h] such that (i) (n-1) th derivative f (n-1) is continuous on [a,a+h] and (ii) n th derivative f (n) exists in (a,a+h). Then there exists at least one number, where 0 < < 1 such that f(a+h) = f(a) + h f (a) + f (a) f (n-1) (a) + R n (1) where R n = f (n) (a + is called the Lagrange s Form of Remainder after n terms. II. Taylor s Theorem with Cauchy s Form of Remainder: Statement. Statement. Let f(x) be a function defined in the closed interval [a,a+h] such that (i) (n- 1) th derivative f (n-1) is continuous on [a,a+h] and (ii) n th derivative f (n) exists in (a,a+h). Then there exists at least one number, where 0 < < 1 such that f(a+h) = f(a) + h f (a) + f (a) f (n-1) (a) + R n (2) where R n = f (n) (a + is called the Cauchy s Form of Remainder after n terms. Note1.If we take n=1 in (1), Taylor s theorem reduces to Lagrange s Mean-Value Theorem. 11

12 III. Maclaurin s Theorem Statement. Let f(x) be a function defined in the closed interval [0,x] such that (i) f (n-1) is continuous on [0, x] and (ii) f (n) exists in (0, x) then there exists at least one number, where 0 < < 1 such that f(x) = f(0) + x f (0) + f (0) f (n-1) (0) + R n where R n = f (n) (, 0 < < 1 [Lagrange s Form] and R n = f (n) ( 0 < < 1 [Cauchy s Form] Note2. Putting a = 0 and h = x in (1) we get Maclaurin s Theorem with Lagrange s Form of Remainder and the same substitution in (2) gives Maclaurin s Theorem with Cauchy s Form of Remainder. Some Useful Limits (a) = 0 for x < 1. (b) = 0 for and for x > 1. (c) = 0 for all values of x (d) Example12. Find the Maclaurin s theorem with Lagrange s form of remainder for f(x) = sin x. Solution. f (n) (x) = = sin and f (2n) (x) = (-1) n sin x, f (2n+1) (x) = (-1) n sin f (n) (0) = sin. Therefore f(0) = 0, f (0) = 1, f (0) = 0,....., f (2n) (0) = 0 and f (2n+1) (0) = (-1) n. Substituting these values in the Maclaurin s theorem with Lagrange s form of remainder, i.e, f(x) = f(0) + x f (0) + f (0) (-1) n-1 f (2n-1) (0) + f (2n) ( ), 0 < < 1. We find 12

13 sin x = 0 + x (-1) n-1 + (-1) n sin 0 < < 1 Example13. Verify Maclaurin s theorem for f(x) = remainder upto three terms when x = 1. with Lagrange s form of Solution. Here we use the formula, f(x) = f(0) + x f (0) + f (0) + f (n) ( 0 < < 1 At x=1 f(1) = f(0) + f (0) + f (0) + f (n) ( or, 0 = 1 + (-1) + (-1) (-1) + (-1) (-1) (-1) Simplifying, = -. = - Here, lies within (0,1), hence Maclaurin s theorem with Lagrange s form of remainder is verified. Example14. Express log (1+x) with Maclaurin s theorem with Lagrange s form of remainder. Solution. f(x) = log (1+x). f (x) =, f (x) = -,...., f (n) (x) = (-1) n-1. Substituting these in the Maclaurin s theorem with Lagrange s form of remainder, we get log (1+x) = log 1 + x.1 + (-1) (-1) n (-1) n-1, 0 < < 1 = x (-1) n-2 + (-1) n-1, 0 < < 1 Example15. Expand f(x) = (a+x) n, n Solution. Maclaurin s theorem with Lagrange s form of remainder f(x) = f(0) + x f (0) f(x) + f (0) f (n-1) (0) + f (n) (, 0 < <1 we find, (a+x) n = a n + x. n a n-1 +.n(n-1) a n n(n-1)(n-2)..{n-(n-2)}a +. n! Example16. Express the polynomial 2x 3 + 7x 2 + x 6 in powers of (x-1). 13

14 Solution. We use Taylor s theorem with Lagrange s form of remainder taking a=1 and h = x-1 and obtain f(x) = f(1) + f (1) + f (1) f (n-1) (1) + f (n) (a+, 0 < < 1 = ( ) + (x-1) ( ) + ( ) +.12 = (x-1) + 13 (x-1) (x-1) 3 Taylor s Series Statement.Let f(x), f (x), f (x), , f (n) (x) exist finitely however large n may be in any interval (x- enclosing the point x and let R n 0 as n. Then Taylor s series of finite form can be extended to an infinite series of the form f(x + h) = f(x) + h f (x) + f (x) f (n) (x) +...., (3) Note. Taylor s series expansion of f(x) in the neighbourhood of x=a : Taking x=a and h = x-a in (3), we get f(x) = f(a) + (x-a) f (a) + f (a) f (n) (a) +....,. (4) Maclaurin s Series Statement. Let f(x), f (x), f (x), , f (n) (x) exist finitely however large n may be in any interval (- and R n 0 as n. Then Maclaurin s series of finite form can be extended to an infinite series of the form f(x) = f(0) + x f (0) + f (0) f (n) (0) ,. [Putting a=0 in (4)] (5) The infinite series (5) is the expansion of f(x) in the neighbourhood of the point x=0. Example16. Expand the function f(x) = e x in the form of Maclaurin s series in the neighbourhood of the point x=0. Solution. Here Lagrange s remainder after n terms Rn =, ( 0 < < 1) 0 as n. Then Maclaurin s series for e x is given by 14

15 f(x) = f(0) + x f (0) + f (0) f (n) (0) i.e, e x = 1 + x Example17. Maclaurin s series for sin x and cos x Solution. (i) f(x) = sin x = sin ( + = 0, since and Now using f(x) = f(0) + x f (0) + f (0) f (n) (0) , we find sin x = x (-1) n (ii)f(x) = cos x = cos ( + = 0, since and cos x = (-1) n Example18. Maclaurin s series for log (1+x). Solution. log (1+x) =, = - ( ) 2, , = (-1) n-1 (n-1)!( ) n. Therefore using (5) log (1+x) = x (-1) n Example19.(a) Find the Maclaurin s series for the function f(x) = (1+x) m where m is not necessarily an integer and hence show that the formula for the binomial series works for nonintegral exponents as well. (b) Use your answer to find the expansion of up to the term in x 6. Solution. f(x) = (1+x) m, so f(0) = 1, f (x) = m(1+x) m-1 f (0) = m. Likewise f (0) = m(m-1),.., f (k) (0) = m (m-1) (m-2).... {m-(k-1)} and so 15

16 (1+x) m = 1 + mx + x 2 + x x k This is an infinite series. If m is a positive integer the series will stop when K = m and will agree with the standard binomial expansion. (b) f(x) = = (1-x 2 ) -1/2 = 1+ ( = 1 + x 2 + x Lesson -3: Reduction Formulae I. I n = = = = - = - = - - (n-1) = - I n-2 - (n - 1) I n Or, { 1 + (n-1)}i n = - I n-2 I n = I n 2 (6) I. J n = = +, using (6) = 0 + J n-2 16

17 = J n-2 II. I n = = = = = I n-2 ( n 1 ) I n I n = I n 2 (7) II. J n = = + = J n-2 Example 20. Using reduction formula, evaluate Solution. We know that J n = = J n-2. Therefore J 7 = = J 5 = = = III. = [ sinx = I m,n = 17

18 = = dx = + = + = + = + I m,n-2 - I m,n Or, (1 + + I m,n = + I m,n-2 Or, I m,n = + I m,n-2 Alternately, = + I m,n-2 (8) Writing, I m,n = and proceeding as earlier, we can show that I m,n = - + I m-2,n (9) III. J m,n = = [ + J m,n-2, using (8) = J m,n-2 (10a) Also using (9), J m,n = = [ - + J m-2,n 18

19 = J m-2,n (10b) (10a) & (10b) implies J m,n = J m,n-2 = J m-2,n (10c) IV. I m,n = = -- = - = - = I m,n + I m-1,n-1 Or, ( 1 + ) I m,n = + I m-1,n-1 i.e, I m,n = + I m-1,n-1 (12) IV. J m,n = = + J m-1,n-1, using (12) = + J m-1,n-1 (13) Example21. If J m,n =, then prove tha J m,n = [ ] 19

20 Solution. Using (13) J m,n = = + J m-1,n-1 J m,m = + J m-1,m-1 = + [ + J m-2,m-2 ] = J m-3.m-3 ] = J 1,1 ] = J 1,1 = [using J 1,1 = = ] = [ ] Hence proved. V. I n = = = + 2n 20

21 = + 2n = +2n I n 2n a 2 I n+1 Or, (1-2n) I n = 2n a 2 I n+1 Example22. If I n =, show that (n-1) ( I n + I n-2 ) = 2 Solution. I n = = = = = - = = = - I n-2 + = - I n-2 + = 2 - I n-2 i.e, (n-1) ( I n + I n-2 ) = 2 21

22 Module-III Lesson I : Calculus of Functions of Several Variables Examples of functions of two or three variables: (a) z f(x,y) = x + y is defined in the entire x-y plane. (b) z g(x,y) = is defined over the entire x-y plane excluding the only point (0,0). (c) z is defined over the region x+y, which include the points of the line x+y = 1. (d) z of the line x+y = 1. is defined over the region x+y > 1, which does not include points (e) z is defined over the closed circular region x 2 + y 2 (f) z h(x,y) = log (1-x 2 y 2 ) is defined over the region x 2 + y 2 < 1. (g) z = is defined in the shaded region : (h) f(x,y,z) = x 2 y + y 2 z + z 2 x is a function of three variables x, y, z. It is defined for all points (x, y, z) R 3. Limit 22

23 Let z = f(x,y) be given in a domain D, and let (x 1,y 1 ) be a point of D or a boundary point of D. Then the equation (14) means the following: Given any ε > 0, a D and within the neighbourhood of (x 1,y 1 ) of radius, except possibly for (x 1,y 1 ) itself, one has. (15) In other words, if (x, y) is in D and 0 < (x - x 1 ) 2 + (y - y 1 ) 2 < (16) then (15 ) holds. Thus if the variable point (x,y) is sufficiently close to (but not at) its limiting position (x 1,y 1 ), the value of the function f(x,y) is as close as desired to its limiting value A.. Examples on Limit 2 Example23. Given f(p) = f(x,y) =, establish Solution. Let be given. We are to find a such that in some neighbourhood of (0,0) for all points (x,y) < or, Now, clearly x 2 < x 2 + y 2 and y 2 < x 2 + y 2 x 2 y 2 < (x 2 + y 2 ) 2 so that < = < holds if, 0 < 2 i.e when Therefore whenever 23

24 Example24. To prove does not exist. Solution. The domain of f is the whole xy-plane punctured at the origin (0,0). For existence of limit we are to examine the values of f near (0,0). If we allow the limit through the straight line y= mx, we observe that f(x,y) = f(x,mx) = = so, This shows that the limit is different for different lines. Continuity If the point (x 1, y 1 ) is in D and (x 1, y 1 ) (17) Then f(x,y) is said to be continuous at (x 1,y 1 ). If this holds for every point (x 1,y 1 ) of D, then f(x,y) is said to be continuous in D. A function f(x,y) is said to be bounded when (x,y) is restricted to a set E, if there is a number M such that when (x,y) is in E. For example, z = x 2 + y 2 is bounded with M = 2, if Example25. The function defined by f(x,y) = is continuous at (0,0). Solution. For = < <, whenever i.e, x 2 + y 2 < 2. This implies. Therefore = i.e, f(x,y) is continuous at (0,0). 24