ANSWERS TO HOMEWORK #3 ---------------------- PROBLEM 1: Remember, a limit lim F(x) EXISTS if, as x gets closer x-->v and closer to v (no matter how it approaches it), F(x) always gets arbitrarily close to some value L (and then L is the limit). The most usual way in which limits fail to exist is if F(x) has a discontinuous JUMP in value at x>v versus x0 Does this limit exist? Why or why not? (And if it exists, then what is it?) What if we make the approach to 0 be 1-sided, such as x-->0+? ANSWER. None of these (1-sided or 2-sided) limits exist. This is because sin(1/x) oscillates an infinite number of times between -1 and +1 in any interval (0, x] no matter how small x is. So, F(x) does NOT *ALWAYS* (and "always" is the key word) get arbitrarily close to any value L as x shrinks to 0. PROOF by contradiction: suppose you think some limit L exists. Well, no matter what x>0 you pick, I can find another x, between your x and 0, with sin(1/x) = 1 or = -1, whichever is further from L. (At least one of these must be distance at least 1 from L.) I shrink your x down to my x. The result is sin(1/x) NOT arbitrarily close to L. In fact it is at least 1 away from L. Hence it is not true that ALWAYS sin(1/x) is arbitrarily close to L since I have just shown an infinite number of counterexamples exist. That contradiction proves L could not have existed. This way of avoiding having a limit is called "infinite oscillation." ANOTHER WAY TO LOOK AT IT: Substitute y = 1/x. The 1-sided limit as x-->0+ becomes lim sin(y). y-->infinity Well, this just keeps oscillationg as you make y bigger; every 2pi there is another oscillation ranging from +1 to -1. It never asymptotes to any fixed constant value. PROBLEM 2: 2 2 Is ( ( (1+x) (3-x) + 7 + x ) + 9 + x ) a polynomial? Why (or why not)? And if it is, then what is its degree? ANSWER: We may simplify 2 2 2 2 ( ( (1+x) (3-x) + 7 + x ) + 9 + x ) = (19 + 3x) = 361 + 114x + 9x which certainly is a polynomial. (1) it is a sum of monomials. (2) it is got from x and constants like 7,3,9 by using ONLY +,-, and multiplication, WITHOUT division. Either reason is enough to see it is a polynomial. It has degree 2 since the biggest exponent in a monomila is 2. [Note there was a cancelation in here, without which it would have had degree 4 and the expression would have been more complicated.] PROBLEM 3: 2 Simplify ( (1+x) (3/x - x) / (7+x) + 9 + x /(x-3) ) to rational (polynomial(x)/polynomial(x)) form. What is the degree of the top, i.e. numerator, polynomial? ANSWER: This is (to just rewrite it without changing anything) 2 (1 + x) (3/x - x) x ----------------- + 9 + ----- 7 + x x - 3 Some people did not even get this far! Why? Two reasons. ------------------------------------------------------------------ * A lot of people did not realize something VERY IMPORTANT: "precedence rules" or "priority rules" for writing formulas! Namely hi priority ^ | parentheses trump everything | B | exponentiation A | | multiplication and division | | + and - V lo priority k k Examples: A+B is NOT (A+B) because exponentiation is hi priority so you do it BEFORE and not AFTER the +. A / B + C is NOT A/(B+C) because division is hi priority so you do it BEFORE and not after the +. A B + C is NOT A(B+C) for same reason. A B / C X is AMBIGUOUS, could mean (AB)/(CX) or (AB/C)X because multiplication and division have the SAME priority. NEVER WRITE ambiguous expressions! Very very dangerous! Always use parentheses or 2-dimensional expressions like A B --- C X to get rid of any ambiguity! * Also (my fault) I had accidentally hit the "(" twice at the beginning of the formula, so I got "(formula" instead of "formula". This actually should not have confused anybody, and in fact only 1 student even noticed it (?!). He was pissed. He gets extra credit. Now if I had accidentally put a "(" in the MIDDLE of the formula, THAT really should have confused you since it really would change the meaning or make the meanign unclear. ------------------------------------------------------------------ OK, to get back to the solution... We had 2 (1 + x) (3/x - x) x ----------------- + 9 + ----- 7 + x x - 3 The first fraction is also writeable (by multiplying & dividing by x) as 2 (1 + x) (3 - x ) ----------------- x Using that fact, we have 2 2 (1 + x) (3 - x ) x ----------------- + 9 + ----- (7 + x) x x - 3 now putting everything over a common denominator yields 2 3 (1+x) (3 - x ) (x-3) 9(7+x)(x-3)x (7+x) (x-3) x ---------------------- + ------------- + --------------- (7+x) (x-3) x (7+x) (x-3) x (7+x) (x-3) x which is 2 3 (1+x) (3 - x ) (x-3) + 9(7+x)(x-3)x + (7+x) (x-3) x -------------------------------------------------------- (7+x) (x-3) x and now expanding out the polynomial on the top gives 3 2 6 x + 14 x - 65 x - 3 3 ------------------------ x (7 + x) (x - 3) where the numerator has degree=3. PROBLEM 4. We know sin(a+b) = sin(a)cos(b) + sin(b)cos(a) and we know a similar addition formula for cosine. Now use this to figure out a formula for sin(3a) in terms of sin(a) and cos(a) ONLY. Then, devise a formula for sin(4a) in terms of sin(a) and cos(a) ONLY. Hint: substitute! ANSWERS: 2 sin(3 a) = 4 sin(a) cos(a) - sin(a) 3 sin(4 a) = 8 sin(a) cos(a) - 4 sin(a) cos(a) (there are also other possible equivalent formulas which also are ok answers.) METHOD: wer start with (book p.583 bottom - "ready reference" of trig identities) sin(a+b) = sin(a)cos(b) + sin(b)cos(a) cos(a+b) = cos(a)cos(b) - sin(a)sin(b) Hence by substituting a=b in these we have the DOUBLE ANGLE FORMULAS sin(2a) = 2sin(a)cos(b) 2 2 cos(2a) = cos(a) - sin(a) now substitute b=2a inside sin(a+b) = sin(a)cos(b) + sin(b)cos(a) to get sin(3a) = sin(a+2a) = sin(a)cos(2a) + sin(2a)cos(a) and now substitute the double angle formulas into this (every time you see trig(2a) replace it with what the double angle formula says is equal to it) 2 2 2 sin(3a) = sin(a+2a) = sin(a)(cos(a) - sin(a) ) + 2sin(a)cos(a) and simplifying using 2 2 cos(a) + sin(a) = 1 yields the first answer, although, technically, it was fine just to leave this unsimplified since it is in terms of sin(a) and cos(a) only, as requested. If we substitute x=y=2a inside sin(2x) = 2sin(x)cos(y) then we get 2 2 sin(4a) = 2sin(2a)cos(2a) = 2 [2sin(a)cos(a)] [cos(a) - sin(a) ] which simplifies to the second answer.