UNIVERSIDAD CARLOS III DE MADRID MATHEMATICS II EXERCISES (SOLUTIONS ) CHAPTER : Limits and continuit of functions in R n. -. Sketch the following subsets of R. Sketch their boundar and the interior. Stud whether the following are closed, open, bounded and/or conve. (a) A = {(, ) R : 0 < (, ) (, 3) < }. (b) B = {(, ) R : 3 }. (c) C = {(, ) R : <, }. (d) D = {(, ) R : + < }. (e) E = {(, ) R : <, < /, > 0}. (f) F = {(, ) R : + }. (g) G = {(, ) R : ( ) +, }. (a) The set represents the disk of center C = (, 3) and radius with the center removed. A The function f : R R defined b f(, ) = (, ) (, 3) = ( ) + ( 3) is continuous and the set A ma be written as A = {(, ) R : 0 < f(, ) < } = {(, ) R : f(, ) (0, )} Since, the interval (0, ) R is open, the set A is open. It is also bounded, since it is contained in the disc {(, ) R : (, ) (, 3) < }. In addition, it is not conve since the points P = (, 4) and Q = (, ) belong to A but the conve combination (, 4) + (, ) = (, 3) does not belong to the set A. The interior, boundar and closure of A are represented in the following figure interior boundar closure Note that A A =. This gives another wa to prove that the set A is open. (b) The function f : R R defined b f(, ) = 3 is continuous and the set B ma be written as B = {(, ) R : f(, ) 0} = {(, ) R : f(, ) [0, )} Since, the interval [0, ) R is closed, the set B is closed.
= 3 B The set B is not bounded since, for eample, the points (, 0), (, 0),..., (n, 0),... belong to B and lim (n, 0) = + n Furthermore, it is not conve since the points P = (0, 0) and Q = (, ) belong to B but the conve combination C = P + Q = (, ).0 0.8 Q = 3 0.6 0.4 C 0. P 0. 0.4 0.6 0.8.0 does not belong to B, because it does not satisf the equation 3. The interior of B is the set {(, ) R : < 3 }. The boundar of B is the set (B) = {(, ) R : = 3 }. And the closure of B is the set B = B (B) = {(, ) R : 3 }. Since, B = B, the set is closed. (c) Graphicall, the set C is = = - = The points P and Q in the figure = -
3 Q = P = - = = - belong to (C). Since, P / C, we see that C is not closed and since Q C, we see that C is not open. Graphicall, we see that the set C is conve. An alternative wa to prove this is b noting that the set C is determined b the following linear inequalities >, <,, The interior, boundar and closure of A are represented in the following figure interior boundar closure We see that (C) C, so the set is not open. Furthermore, C C so the set is not closed. (d) The following functions defined from R into R are continuous. f (, ) = f (, ) = + f 3 (, ) = + + f 4 (, ) = + The set D = + = - D = - - = - is defined b D = {(, ) R : f (, ) < 0, f (, ) < 0, f 3 (, ) > 0, f 4 (, ) > 0} so it is open and conve. The set D is bounded because is contained in the disc of center (0, 0) and radius. The interior, boundar and closure of A are represented in the following figure
4 Interior boundar closure Since, (D) D =, the set is open. (e) The graphic representation of E is = = / E X=0 The functions f (, ) = f (, ) = / f 3 (, ) = are defined from R into R and are continuous. The set E is defined b E = {(, ) R : f (, ) < 0, f (, ) < 0, f 3 (, ) > 0} so it is open. The set E is not bounded because the points (n, 0) n =,,... belong to E and lim (n, 0) = lim n = + n n In addition, it is not conve because the points P = (0, 0) and Q = (, 0 8) belong to E but the conve combination R = P + Q = (0 6, 0 4)
5 P R Q E does not belong to E, because it does not satisf the inequalit <. The interior, boundar and closure of E are represented in the following figure Interior boundar closure Since (E) E =, the set is open. (f) Graphicall, the F is = - = - The function f(, ) = defined from R into R is continuous. The set F is F = {(, ) R : f(, ) } so is closed. The set F is not bounded because the points are in E and The figure (n, 0) n =,,... lim (n, 0) = lim n = + n n
6 P R Q shows wh F is not conve. The interior, closure and boundar of F are represented in the following figure = - = - = - Interior = - boundar = - = - closure Since (F ) F, the set F is closed. (g) Graphicall the set G is
7 = The functions f(, ) = ( ) + and g(, ) = defined from R into R are continuous. The set G is G = {(, ) R : f(, ), g(, ) } so it is closed. The set G is bounded because it is contained in the disc of center (, 0) and radius. Further, the set G is conve. The interior, boundar and the closure of G are represented in the following figure Interior boundar closure Since (G) G, the set G is closed. -. Let A be a subset of R. Discuss which of the following assertions are true. (a) Int(A) = A (A). (b) (A) = (R A) = (A C ). (c) (A) is bounded. (d) A is closed if and onl if A C is open. (e) A is bounded if and onl if A C is not bounded. (f) A is closed if and onl if (A) A. (g) A is open if and onl if ( A) A =. (a) Yes, because: Int(A) ε > 0 : B(, ε) A ε > 0 : B(, ε) (R n \ A) = A and A. (b) Yes, because: (R n \ A) = R n \ A R n \ (R n \ A) = R n \ A A = (A). (c) No. Eample: A = {(, ) R : 0}. (d) Yes. B definition. (e) No. Eample: A = {(, ) R : 0}. (f) Yes, because: A is closed R n \ A is open R n \ A = Int(R n \ A). But, from (a) and (b), Int(R n \ A) = (R n \ A) \ (R n \ A) = (R n \ A) \ (A). Therefore A is closed R n \ A = (R n \ A) \ A ( A) A. (g) Yes, because: A is open A = Int(A) A = A \ A A A =. -3. Find the domain of the following functions. (a) f(, ) = ( + ) /. (b) f(, ) =. (c) f(, ) = e e. (d) f(, ) = e. (e) f(, ) = ln( + ).
8 (f) f(, ) = ln( + ). (g) f(,, z) = + z. (h) f(, ) = +. (a) {(, ) R : + }. (b) {(, ) R : 0}. (R ecept the aes). (c) R. (d) R. (e) {(, ) R : + > 0}. (f) R \ {(0, 0)}. (g) {(,, z) R 3 : z > 0}. (h) {(, ) R : }. -4. Find the range of the following functions. (a) f(, ) = ( + + ) /. (b) f(, ) = +. (c) f(, ) = +. (d) f(, ) = ln( + ). (e) f(, ) = ln( + + ). (f) f(, ) = +. (a) [, ). (b) [, ]. (c) [, ]. (d) (, ). (e) [0, ). (f) [0, ). -5. Draw the level curves of the following functions. (a) f(, ) =, c =,, 3. (b) f(, ) = e, c =,, 3. (c) f(, ) = ln(), c = 0,,. (d) f(, ) = ( + )/( ), c = 0,,. (e) f(, ) =, c = 0,,. (f) f(, ) = e, c = 0,,. (a) The level curves are determined b the equation = c. For c 0, this equation is equivalent = c so the level curves are
9 4 c = 3 0 c = - c = - -4-4 - 0 4 (b) The level curves are determined b the equation e = c. Therefore, the level curve corresponding to c 0 is the empt set. In particular, there is no level curve corresponding to c =. For c > 0, the level curve satisfies the equation e = c, so = ln c. For c =, the level curve consists of the points (, ) R such that = 0. For c = 3, the level curve consists of the points (, ) R such that = ln 3 Graphicall, 4 c = 3 0 c = - -4-4 - 0 4 (c) The level curves are determined b the equation log() = c which is the same as = e c. Graphicall, 5 c = -0-5 5 0 c = - -5 c = 3 (d) The level curves are determined b the equation + = c( ), which is the same as (+c) = (c ). Graphicall,
0 4 c = c = - 0 - c = 0-4 -4-0 4 (e) The level curves satisf the equation = c. Graphicall, 0 8 c = - 6 4 0 - c = 0 c = -4-0 4 (f) The level curves satisf the equation = ce. Graphicall, 4 c = 0 c = 0 - -4 c = - -3 - - 0 3-6. Let f(, ) = C α α, with 0 < α < and C > 0 be the Cobb-Douglas production function, where (resp. ) represents units of labor (resp. capital) and f are the units produced. (a) Represent the level curves of f. (b) Show that if one duplicates labor and capital then, production is doubled, as well. (a) The level curves are,
0 0 0 8 8 8 6 6 6 4 4 4 0 0 4 6 8 0 α < / 0 0 4 6 8 0 α = / 0 0 4 6 8 0 α > / (b) f(, ) = C α α, f(, ) = C () α () α = C α α = f(, ). -7. Stud the eistence and the value of the following limits. (a) lim (,) (0,0) +. (b) lim (,) (0,0) +. (c) lim (,) (0,0) 3 4 +. (d) lim (,) (0,0) (e) lim (,) (0,0) +. +. +. (f) lim (,) (0,0) (g) lim 3 (,) (0,0) +. [ ] (a) ( + ) =k (b) We show that = +k = (+k ) and lim 0 ( (+k ) lim f(, ) = 0. (,) (0,0) ) does not eist. The limit does not eist. Let ε > 0. Take δ = ε and suppose that 0 < (, ) (0, 0) = + < δ. Then, f(, ) 0 = + ( + ) + = = + = δ = ε. (c) On the one hand, (d) (e) so [ ] ( 3 4 + ) = 3 3 k =k 4 + k = 3 ] lim 0 [ ( 3 4 + ) =k = 0 k + k On the other hand, [ ] ( 3 4 + ) = 3 = Therefore, ] the limit does not eist. [( + ) = k +k = k +k, depends on k. Therefore, the limit does not eist. [ ( ] =k + ) =k (f) We show that = k +k = k +k, depends on k lim f(, ) = 0. (,) (0,0) Let ε > 0. Take δ = ε and suppose that 0 < (, ) (0, 0) = + < δ. Then, f(, ) 0 = + ( + ) + = = + = δ = ε.
(g) We show that lim (,) (0,0) f(, ) = 0. Let ε > 0. Take δ = ε and suppose that 0 < (, ) (0, 0) = + < δ. Then, f(, ) 0 = 3 + = + + + = = = And the limit is 0 + + = + = δ = ε -8. Stud the continuit { of the following functions. (a) f(, ) = 3 + if (, ) (0, 0) 3. 0 if (, ) = (0, 0) { + (b) f(, ) = if 0 0 if = 0. { 4 (c) f(, ) = 6 + if 3 0 if =. { 3 (d) f(, ) = + si (, ) (0, 0) 0 si (, ) = (0, 0). (a) The function 3 + 3 is not continuous at the points {(, ) : = }. (The limit lim (,) (0,0) 3 + 3 does not eist. This can be shown b taking curves of the form = k.) (b) The function +, (i) is continuous at the points (, ) such that 0. (ii) is not continuous at the points of the form ( 0, 0) with 0 0. Since, the limit lim (f( 0, )) = lim 3 0 + 0 0 0 does not eist if 0 0. (iii) It is not continuous at (0, 0) because which depends on k. (c) The function lim f(, 0 k ) = lim 0 ( 3 + ) = k k 4 6 + 3 is continuous at the points (, ) such that. On the other hand, at the points of the form (a, a ) is not continuous because, (i) If a 0, we have that lim f(a, ) a does not eist because the numerator approaches a 6 0 whereas the denominator approaches 0. (ii) The limit (d) The function does not eist because lim f(, ) (,) (0,0) lim f(t, t 6 t 0 t ) = lim t 0 t 6 + t 6 = whereas the value of the iterated limits is 0. 3 + is a quotient of polnomials and the denominator onl vanishes at the point (, ) = (0, 0). Hence, the function is continuous at ever point (, ) (0, 0). At the point (0, 0) the function is also continuous because we have alread proved in another problem that lim (,) (0,0) 3 + = 0
3 We conclude that the function is continuous in all of R. -9. Consider the set A = {(, ) R : 0,, 0 } and the function f : A R, defined b f(, ) = ( + +, + ) + Are the hpotheses of Brouwer s Theorem satisfied? Is it possible to determine the fied point(s)? Brouwer s Theorem: Let A be a compact, non-empt and conve subset of R n and let f : A A be a continuous function. Then, f has a unique fied point. (That is, a point a A, such that f(a) = a). The set A is not empt, compact and conve. The function f is continuous if and. Therefore, f is continuous on A and Brouwer s Theorem applies. If (, ) is the fied point of f, then that is, = + + = + + = = Therefore = satisfies the equation + = 0 whose solutions are = ± 5 The onl solution in the set A is ( + 5, + 5 ). -0. Consider the function f(, ) = 3 defined on the set D = {(, ) R : +, 0 < /, 0}. Draw the set D and the level curves of f. Does f have a maimum and a minimum on D? The set D is the following + = (0,) = / Note that D is not compact (since it is not closed). It does not contain the point (/, 0). On the other hand, the level curves of f are of the form = C + 3 Graphicall, (the red arrow points in the direction of growth)
4 = C + / 3 (0,) ma = / We see that f attains a maimum at the point (0, ), but attains no minimum on A. -. Consider the sets A = {(, ) R 0, 0 } and B = {(, ) R, } and the function f(, ) = ( + ) ( ) + 5 + What can ou sa about the etreme points of f on A and B? The function f(, ) = ( + ) ( + 5 + ) is continuous if / and so, is continuous in the set A, which is compact. B Weierstrass Theorem, f attains a maimum and a minimum on A. But, for eample, the point (0, /) IntB and lim ( f(0, ) =, lim f(0, ) = + ) + ( ) so f does not attain neither a maimum nor a minimum on B. -. Consider the set A = {(, ) R : 0 ln, }. (a) Draw the set A, its boundar and its interior. Discuss whether the set A is open, closed, bounded, compact and/or conve. You must eplain our answer. (b) Prove that the function f(, ) = + ( ) has a maimum and a minimum on A. (c) Using the level curves of f(, ), find the maimum and the minimum of f on A. (a) The set A is L n = L n A The boundar and the interior are
5 A A º Since A A, the set A is closed. It is not open because A A. Another wa of proving this, would be to consider the sets A = {(, ) R : 0 }, A = {(, ) R : }. The set A 3 = {(, ) R : log()} is also closed since the function g(, ) = log() is continuous. Therefore, A = A A A 3 is a closed set. The set A is bounded since A B(0, r) with r > 0 large enough. Since it is closed and bounded the set A is compact. The set A is conve since is the region under the graph of f() = ln in the interval [, ] and the function ln is concave. (b) The function f is continuous in R, since it is a polnomial. In particular, the function is continuous in the set A. Furthermore, the set A is compact. B Weierstrass Theorem, the function attains a maimum and a minimum on A. (c) The equations defining the level curves of f are f(, ) = + ( ) = C These sets are circles centered at the point (, 0) and radius C, for C 0. L n A Graphicall, we see that the maimum is f(, ln ) = + (ln ) and is attained at the point (, ln ). The minimum is f(, 0) = 0 and is attained at the point (, 0). -3. Consider the set A = {(, ) R :, > 0; ln() 0}. (a) Draw the set A, its boundar and its interior. Discuss whether the set A is open, closed, bounded, compact and/or conve. You must eplain our answer. (b) Consider the function f(, ) = +. Is it possible to use Weierstrass Theorem to determine whether the function attains a maimum and a minimum on A? Draw the level curves of f, indicating the direction in which the function grows. (c) Using the level curves of f, find graphicall (i.e. without using the first order conditions) if f attains a maimum and/or a minimum on A. (a) The equation ln() 0 is equivalent to. Since, > 0, the set is A = {(, ) R : /, > 0}. Graphicall,
6 = = = A A º A The boundar is the set A = {(, ) R : = /, > 0}. he interior is the set A = {(, ) R : > /, > 0}. Since A A, the set A is not open. Furthermore, A A so the set A is closed. Graphicall, we see that A is not bounded. The set A is not compact (since is not bounded). Consider not the function g(, ) = ln() = ln + ln, defined on the conve set D = {(, ) R :, > 0}. The Hessian matri of this function is H g = ( 0 0 ), which is negative definite. From here we conclude that the function g is concave in D. Since, A = {(, ) D : g(, ) 0}, the set A is conve. (b) We ma no appl Weierstrass Theorem since the set A is not compact. The level curves of f(, ) = + are sets of the form {(, ) R : = C /} which are straigt lines. Graphicall (the vector indicates the direction of growth) (c) Looking at the level curves of f we see that the funtion does not attain a (local o global) maimum on A. The global minimum is attained at the point of tangenc of the straight line = C / with the graph of = /, This point satisfies that = that is = ±. And since > 0, the minimum is attained at the point (, / ),