1 Limits and continuity

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1 1 Limits and continuity Question 1. Which of the following its can be evaluated by continuity ( plugging in )? sin(x) (a) x + 1 (d) x 3 x 2 + x 6 (b) e x sin(x) (e) x 2 + x 6 (c) x 2 x 2 + x 6 (f) n ( 1) n. Answer. (a), (d). In (b), the second term has no it (squeeze theorem is appropriate). In (c), you get 0/0 and have to factor and cancel. In (e), looking at highest-order terms does not count as plugging in. In (f), there is no it (also, a sequence can t be continuous). { e x x < 0 Question 2. Let f(x) =. Which of the following statements about f are x sin(x) x 0 false? 1. f has a jump discontinuity at x = f is continuous at f is continuous from the right at f(x) =. 5. x f(x) = The it f(x) can be computed using the squeeze theorem from the inequality x f(x) x. Answer. 2, 4, 6. For 2, there is a jump discontinuity ( f(x) = 1 and + f(x) = 0). For 4, in fact f(x) doesn t exist, though f does oscillate between increasingly large values. For 6, only the right-hand it can be computed using these inequalities; the left-hand it is different and the inequalities don t apply. Question 3. What is x x 4 x 3 + 1? Explain the sign. x 7 Answer. It s, because the numerator has a greater degree. The sign is because as x, the fraction becomes close to x 4 /x = x 3, which is negative when x is. Question 4. What is x 3 + x 3 x 3 x 3? Answer. It s 1, since the x 3 terms are less important than the x 3 terms. Question 5. What is x 2 e 1/(x 2)2? Answer. Tricky! As x 2, we have x 2 0, so (x 2) ; that is, it remains positive as it approaches 0. Therefore 1/(x 2) 2, and we know that u e u = 0. So the anwswer is 0. 1

2 2 Squeeze/sandwich theorem Question 6. Compute. x Answer. This is done in the book too. Rationalize it as 1 + cos x x 1 + cos x 1 cos 2 x x(1 + cos x) sin 2 x x(1 + cos x) x Question 7. Compute x 2. Answer. This is actually almost the same. Doing the same thing, we get x 2 Question 8. Compute sin(e x )e. x cos x = = cos x = 1 0 = 0. Answer. To avoid confusion which of the factors is the one to use for the inequalities, just see which one you can t take the it of. As x, in fact e x 0, so sin(e x ) sin(0) = 0, which is okay. But e has no it, since oscillates between 1 and 1. This is the one to estimate: 1 1 = e 1 e e 1 = e since e x is an increasing function (meaning that it doesn t change the order of two inputs). Multiplying by sin(e x ), we get e 1 sin(e x ) sin(e x )e e sin(e x ). There are no issues with multiplying by a negative number, since e x 0 + (that is, it remains positive as it goes to zero) and sin(u) 0 for small positive u. So we apply the squeeze theorem: so sin(e x )e = 0. e 1 sin(e x ) = 0 e sin(e x ), Question 9. What would have to be true about a function f(x) in order for the it f(x) sin(1/x) to exist? (For example, we know that f(x) = 1 does not work and f(x) = x does.) Answer. Since sin(1/x) doesn t exist, the only hope is that the squeeze theorem can apply. We would want to estimate 1 sin(1/x) 1, which gives inequalities f(x) f(x) sin(1/x) f(x). The outermost functions are on opposite sides of zero, so to have the same it, they must both go to zero. That is, we need f(x) = 0. 2

3 3 Intermediate value theorem Question 10. Let f(x) be a continuous function such that f(2) = 3 and f(4) = 1. Which of the following is true? 1. The equation f(x) = 3 has a solution x [1, 3]. 2. The equation f(x) = 2 has a solution x [1, 3]. 3. The equation f(x) = 2 has a solution x [2, 4]. 4. The equation f(x) = 3 has a solution x [2, 4]. Answer. If a = 2 and b = 4, and c = 1 and d = 3, then conditions say f(a) = d and f(b) = c, so the IVT tells us that the values of f on [a, b] = [2, 4] are the whole interval [c, d] = [1, 3]. Thus the IVT tells us about solutions to f(x) = m where m [1, 3] and x [2, 4]. In choices 1 and 2, x is in the wrong interval. Choice 3 is spot-on, since m = 2 [1, 3] is right in the middle. Choice 4 is silly, because we already know that f(2) = 3. Question 11. If f(x) is continuous and f(x) =, and if we know f(0) = 0, describe the numbers that are guaranteed to be values of f for x > 0. Give examples showing that we can t say anything about the values f(x) when x < 0. Answer. Since the it is, we know that f(x) takes large positive values for some large numbers x. For example, if we know that f takes the value 100 at b, where b > 0, then we know by the IVT that the equation f(x) = 50 is solvable for some x [0, b], since 50 is between f(0) = 0 and f(b) = 100. If we try to solve any equation f(x) = m with m > 0, we need only choose b large enough that f(b) > m (which is possible since f(b) is approaching ) and then we can find a solution x in [0, b] since m is between f(0) = 0 and f(b). This means we can get every positive number as a value of f(x) for x > 0. On the other hand, we can t necessarily get anything else when x < 0. It might be, like for f(x) = x 2, that all its values are positive. Or it might be, like for f(x) = x, that it takes only negative values when x < 0. Question 12. Prove that the equation cos(x) = x has a solution, and give a closed interval [a, b] that contains such an x. Answer. To make this an IVT problem, rewrite the equation as function equals number : cos(x) x = 0. Let f(x) = cos(x) x; to show that it hits zero, we need to find a value above zero (positive) and a value below zero (negative). It turns out that: f(0) = cos(0) 0 = 1 f(π/2) = cos(π/2) π/2 = π/2 (these are convenient values to choose because they make one of the terms diasppear, so it s easy to compute f(x)). Taking a = 0 and b = π/2, the IVT says that f(x) = 0 has a solution somewhere in [a, b] = [0, π/2]. Question 13. If f is a continuous function and f(0) = 2, f(1) = 3, and f(2) = 1, find a closed interval that must contain some zeroes of f. How many must it contain? 3

4 Answer. To find a zero of f, we need to find intervals where it starts with one sign and ends up with the other. Here, we are given one negative value at x = 0 and two positive values at x = 1, 2. So it might seem like the IVT says there is a zero in [0, 1] and also one in [0, 2]. However, there is no way to be sure that the one in [0, 2] isn t also in [0, 2] since f(1) and f(2) are both positive (so f may never cross the x-axis in this interval). So really, f may have only one zero, in [0, 1]. 4 Differentiability Question 14. Let f(x) = x Using the definition of the derivative, find f ( 1). Answer. f (( 1 + ) 2 + 1) (( 1) 2 + 1) ( 1) (2 + ) (2 + ) = 2. (1 2 + ( 2 ) Question 15. Let f(x) = x 3. Using the definition of the derivative, find f ( 1). Answer. f ( 1 + ) 3 ( 1) 3 ( 1) (3 3 + () 2 ) ( 1) 3 + 3( 1) 2 + 3( 1)() 2 + () (3 3 + ()2 ) = 3. Question 16. Let f be a function such that f (0) = 1. Which of the following are true? 1. For very small, the value f() is approximately more than f(0). 2. For very small, the value f() is approximately less than f(0). 3. For all, the value f() is smaller than f(0). 4. We have f() f(0) for close to We have f() + f(0) for close to 0. Answer. 2 and 5. The answer in 5 is the exact formula, and we see that is subtracted from f(0), so 1 is wrong. 4 is just mixed-up (perhaps because it looks a bit like the correct formula that f = f (x)). 3 is tempting, because f has a minus sign, but remember, itself may be negative, in which case f > 0, meaning that f() > f(0). (If 3 were true, then f would be what I called responsive but not sensitive, like x.) Question 17. Somehow, you measure a circle to have an area of 4π, with a measurement error of less than 0.4. Knowing that A = πr 2, you conclude that r = 2, but with what error? 4

5 Answer. The error in A is proportional to the error in r by a factor of A (r). In symbols, A = A (r) r. We know that A = πr 2, and so A (r) = 2πr (you can do this from the definition or, if you like, observe that we did the same computation for r = 2, without the π, in class, and 2 may as well be a variable called r). When r = 2, therefore, A (r) = 4π. So r = 0.4/(4π) = 0.1/π = 1/(10π). 5