Solutions to Problem Sheet for Week 6

THE UNIVERSITY OF SYDNEY SCHOOL OF MATHEMATICS AND STATISTICS Solutions to Problem Sheet for Week 6 MATH90: Differential Calculus (Advanced) Semester, 07 Web Page: sydney.edu.au/science/maths/u/ug/jm/math90/ Lecturer: Daniel Daners Material covered Limits (continued). Squeeze Law (see also last week s tutorial). Limits as, or. Continuity, left continuity, right continuity. Outcomes After completing this tutorial you should work with its; understand the definition of continuity, left and right continuity; be able to prove that certain functions are continuous, right continuous or left continuous. Summary of essential material Limits as ±. We say that f() l if for every ε > 0 there eists M > 0 such that > M f() l < ε. Improper its. We say that a f() if for every m R there eists δ > 0 such that 0 < a < δ f() > m. The latter is called an improper it or divergence to infinity. There are more such concepts (it to as a, or as etc.) We can also look at right and left hand its. Continuity. A function f() is continuous at a if f() f(a). a We can also give an ε-δ definition of it: f() is continuous at a if for every ε > 0 there eists δ > 0 such that a < δ f() f(a) < ε. Note that we don t require 0 < a < δ, because if a then f() f(a) is automatic. Left and Right Continuity We say f is right or left continuous at a if f() f(a) or f() a+ a f(a) respectively. A function is continuous at a if and only if it is left continuous and right continuous at a. Continuity on Intervals. A function f() is continuous on an open interval (a, b) if it is continuous at each point of (a, b). It is continuous on a closed interval [a, b] if it is continuous on (a, b), right continuous at a, and left continuous at b. How to show continuity of functions. As with its, we use that elmentary functions are continuous such as α, sin, cos, e, ln, sin, cos on their natural domains. From the it laws, sums, products and quotients of continuous functions are continuous (denominator non-zero as always). By the composition/substitution law, compositions of continuous functions are continuous. Copyright 07 The University of Sydney

Questions to complete during the tutorial. Let f(), the largest integer less than or equal to. Sketch the graph of f. At which points is f continuous? At which points is f right continuous, and at which points is it left continuous? Solution: The graph is shown below: y 4-4 - - - - 4 - - -4 This function is continuous at every non-integer value. It is left continuous only at non-integer values. It is right continuous everywhere.. Provide a careful step-by-step argument to eplain why f() is continuous at π, where f() ln (cos + sin + ) + e. Solution: We break the function down into pieces: The functions cos, sin, and are all continuous at π. Thus by it laws the function f () cos + sin + is continuous at π. The function f () ln is continuous at cos π + sin π + π π (note that π > 0 is in the domain of ln). Hence by the Composition Law f () f f () ln(cos + sin + ) is continuous at π. The function f 4 () e is continuous at π. Hence by the it laws the function f 5 () f ()+f 4 () ln(cos +sin +)+e is continuous at π. The function f 6 () is continuous at f 5 (π) ln(π ) + e π (note that this number is positive). Hence by the Composition Law our function f() (f 6 f 5 )() is continuous at π.. Prove that if f() is continuous at a, then the function f() is continuous at a. (Use the reversed triangle inequality from a previous tutorial.) Is the converse true? Solution: Note that for any real numbers r, s it is true that r s r s (reversed triangle inequality, see last week s tutorial). This shows that f() f(a) f() f(a).

As f is continuous at a we have that f() f(a) 0 as s a. Thus by the squeeze law also f() f(a) 0 as a. The converse assertion that if f is continuous at a, then so is f is false. For instance, let f be the function defined by {, if 0, f(), if < 0. Then f is the constant function with value, so it is continuous at 0, but f is not continuous at 0. 4. Determine whether the functions given by the following formulas are continuous the given values. (a) h() + 7, at 4. Solution: The function is continuous everywhere, and the square root function is continuous everywhere in its domain [0, ), so h() is continuous everywhere in its domain (, 7]. In particular, it is continuous at 4. (b) k(), at. + Solution: The domain of k does not include. Thus the function k() is not continuous at. sin if > 0 (c) F (), at 0. if 0 sin Solution: As 0 + F () 0 +, 0 F () 0, and F (0), we see that F is continuous at 0. if (d) K() +, at. 6 if Solution: Since K() + for, we have K(). However, K( ) 6, so K() K( ). Therefore K is discontinuous at. 5. Find a constant c so that g is continuous everywhere, where g is defined by: { c if < 4 (a) g() c + 0 if 4. Solution: The functions c and c + 0, considered on the intervals (, 4) and [4, ) respectively, are continuous for any value of c. Thus the only possible discontinuity is at 4. For g to be continuous at 4, we require 4 g() 4 + g() g(4), that is, Hence 6 c 4c + 0, giving c. { c + 4 if 4 (b) g() c if < 4. 4 ( c ) 4 +(c + 0) g(4). Solution: As in part (a), for g to be continuous at 4, we require 4 g() 4 + g() g(4), that is, 4 c 4 +( c + 4) g(4).

Hence we require 6 c c. From this we see that c 0. If c 4 then we require c 6 c, that is, c 65. If 4 < c 0 then we require 6 c c, that is, c 65. Hence the given function is continuous at 4 for two values of c, namely c 65±. 6. Calculate the following its using it laws, the squeeze law, and/or the substitution law: (a) (b) 0 cos Solution: We use the Squeeze Law. Since cos and 0 ± 0, we have 0 cos 0. + 0 Solution: We can t use the it laws with the epression in its present form, so we manipulate it first. + ( + )( + + ) ( + + ) + ( + + ). + + Hence 0 +. 0 + + (c) Note that the last step used the substitution law to evaluate the it of the denominator. + sin sin + sin Solution: +. sin Note: we have used the fact that 0, which follows from an application of the Squeeze Law. Since sin, we have (for > 0) sin and ± 0. (d) 4 Solution: Divide top and bottom inside the square root sign by. We obtain 4 + 4 +. Now as, + and 4 +. By the substitution law, as the square root function is continuous, we see that 0 4. 4

(e) 4 Solution: This time we divide top and bottom by. We obtain 4 + 4 +. Now as, + 0 and 4 +. By the substitution law, as the square root function is continuous, we see that 0 0. 4 (f) ( + ) Solution: ( + ) ( +)( + +) + + + + 0. 7. (a) Suppose that f is a function such that f(). Use the definition of a it to show that a 0, where a is either finite or a. a f() Solution: Let a be finite and fi ε > 0. Then clearly As a f() there eists δ > 0 such that f() < ε f() > ε Putting the two conditions together we see that 0 < a < δ f() > ε 0 < a < δ f() 0 f() < ε. As the above argument works for any choice of ε > 0 we conclude that f() 0. We proceed similarly if a. Given m R we have f() < ε f() > ε As f() there eists m R such that a Putting the two conditions together we see that > m f() > ε > m f() 0 f() < ε. As the above argument works for any choice of ε we conclude that f() 0. (b) Hence show that e 0 as. Solution: We know that e as. Hence from part (a) we conclude that as. e e 0 5

Etra questions for further practice 8. (a) By comparing the areas of a suitable sector and triangle, show that sin θ θ, where θ R is measured in radians. Solution: Consider the diagram, where the circle is the unit circle: y sin(θ) B θ O A To begin with, suppose that 0 < θ < π. The area of OAB is less than the area of the sector OAB, which gives 0 θ sin θ π π, and so 0 sin for all 0 < θ < π. Multiplying by this gives 0 sin θ θ for all θ (0, π ), and thus since sin θ sin( θ) we have 0 sin θ θ for all θ ( π, 0). It follows that sin for all 0 < < π, and this is clearly true also for θ 0, and also for θ π, because in this case sin θ < π. Therefore sin θ θ for all θ R. y + y (b) Prove that sin sin y sin cos for all, y R. Solution: You could either use various double angle formulae, or argue as follows. Recall from class that sin and cos can be written in terms of the comple eponential function as cos ( e i + e i) and sin i ( e i e i) for all R. We have sin y cos + y ( e i( y) e i( y) ) ( e i(+y) + e i(+y) ) i ( e i + e iy e iy e i) i ( e i e i) ( e iy e iy) i i sin sin y. (c) Hence, show that sin sin y y for all, y R. Deduce that sin R R is continuous. Solution: Using the previous parts, and also the facts that sin t and cos t for all real numbers t, we have sin sin y sin y + y cos y y. (d) Using that the sine function is continuous, show that all other trigonometric functions are continuous. Use for instance that cos() sin(π ). 6

Solution: As cos() sin(π ) for all R the subsitution law implies continuity of the cosine. Alternatively we can see this using the inequality from the previous part: cos cos y sin ( π ) sin ( π y) (π ) ( π y) y y for all, y R. sin cos Net, tan, cot, sec cos sin cos quotient law. and cosec sin are continuous by the 9. Compute the following its using the it laws and the substitution law. tan t (a). t 0 t tan t sin t Solution: We have t t sin t as t 0 by using the elementary it cos t t and cos t as well as the product law. (b) (c) (d) sin(t ). t 0 t Solution: We have sin(t ) t sin(t ) 0 0 as t 0 by using the elementary it t t sin (since t 0) as well as the product law. + sin. Solution: As sin θ θ for all θ 0 we have that + sin sin + sin as, using that 0 and the elementary it [ cosh() ( cosh() sinh() )]. Solution: By definition of the hyperbolic functions cosh sinh ( (e + e ) (e e ) ) sin θ θ + substituting θ. ( (e + e ) (e e ) ) e. (e) Hence, cosh() ( cosh() sinh() ) e e + e as, using the it laws. + e +. 0 Solution: We have ( )( ) + + + + ( + + ) + 0 ( + ) ( ) ( + + ) (9 + 6 + ) (9 6 + ) ( + + ) The it of the denominator in the last epression as 0 is, so as 0. ( + + ) + + + 6 7

(f) sin() 0 sin(5). Solution: We can rewrite the epression in the form sin() sin(5) 5 sin() ( sin(5) 5 ) 5 5 as 0. 0. Show that if f() is continuous at a, and if f(a) > 0, then there is a number δ > 0 such that f() > 0 whenever a < δ. Solution: By continuity of f() at a, for each ε > 0 there eists δ > 0 such that a < δ f() f(a) < ε. In particular, taking ε f(a) > 0 there is δ > 0 such that Challenge questions (optional). Consider the function f defined by a < δ f() f(a) < f(a) f(a) f() < f(a) f() > 0. 0 if is irrational f() if 0 if p with q > 0 and with p and q integers having no factors in common. q q For eample f(6 8) 4 since 6 8 4. Prove that f is discontinuous at every rational number. Solution: Let a be any rational number, and suppose (for a contradiction) that f is continuous at a. We have f(a) > 0, so by the previous question there is δ > 0 such that f() > 0 for all satisfying a < δ. However there is an irrational number y with 0 < y < δ (see Tutorial ). Then a+y is also irrational, and a < δ. But f() 0, contradicting f() > 0. This is the desired contradiction, proving that f is discontinuous at every rational number (in particular, it has infinitely many discontinuities). Remark: Rather remarkably, it turns out that a f() 0 for all a R. Thus f() is actually continuous at every irrational number! 8