PRACTICE PROBLEM SET NOTE: On the exam, you will have to show all your work (unless told otherwise), so write down all your steps and justify them. Exercise. Solve the following inequalities: () x < 3 x < 3 3 < x < 3 2 < x < 4 So the solution set is {x R 2 < x < 4} = ( 2, 4). (2) 5x + 4 > 5x + 4 > 5x + 4 < or 5x + 4 > 5x < 5 or 5x > 3 x < or x > 3 5 So the solution set is {x R x < } {x R x > 3 5 } = (, ) ( 3 5, ). Exercise 2. Solve the equality log 2 (x) + log 2 (x 2) = 3 log 2 (x) + log 2 (x 2) = 3 log 2 [(x)(x 2)] = 3 log 2 (x 2 2x) = 3 2 log 2 (x2 2x) = 2 3 x 2 2x = 8 x 2 2x 8 = 0 (x 4)(x + 2) = 0 x = 4, 2 So the solution set is { 2, 4}. Exercise 3. Solve the equality sin x(sin x ) = 0. The equation is satisfied if sin x = 0 or sin x = 0, so for the equation to hold we need sin x = 0 or sin x =.
Note sin x = 0 for x = 0, ±π, ±2π,..., kπ,..., and sin x = for x = π 2 and x = π 2 + 2kπ for all integers k. So the solution set is {kπ k Z} { π 2 + 2kπ k Z}. Exercise 4. () Suppose {a n } is a sequence with a n+ = 2a n and initial value a 0 =. You can assume the sequence converges. Find a n We are assuming the sequence {a n } converges; let a = a n. The shifted sequence {a n+ } has the same it a n+ = a, and a needs to satisfy the recursion a = 2a. Thus a 2 = 2a, so a(a 2) = 0, so a is equal to either 0 or 2. We know that the it, when it exists, is unique, so only one of these values can be the it. By inspection, since a 0 =, a = 2, a 3 = 2 2,..., we see that the sequence is increasing and the it is not 0, so if it exists, it has to be 2. So a = 2. (2) Suppose {a n } is a sequence with a n+ = a n and initial value a 0 = 0. You can assume the sequence converges. Find a n Note that this is the constant sequence 0, 0, 0, 0,... Thus it converges and the it is 0. (3) Suppose {a n } is a sequence with a n+ = a n and initial value a 0 = 2. You can assume the sequence converges. Find a n Note that this is the constant sequence 2, 2, 2,... Thus it converges and the it is 2. Exercise 5. For each of the following diagrams, write True or False to indicate whether the given function f(x) and given values of ɛ and δ satisfy the following condition: If x is such that 0 < x c < δ, then f(x) L < ɛ. If False, mark a point on the graph at which the condition fails. 2
* FALSE (the star indicates a point where the condition fails) TRUE * 3 FALSE (the star indicates a point where the condition fails)
Exercise 6. Use it laws to compute the following its. Explain all your steps. x 2 2x 3 () x 3 x 3 Note that for any x 3, Therefore, (2) 3x 7 5x + 2 x 3 2x 2 x 2 2x 3 x 3 x 2 2x 3 x 3 x 3 = (x 3)(x + ) x 3 = x +. = (x + ) x 3 = 3 + = 4 by it law Note that for any x 0, x 0, and thus x 7 =. Therefore, 7 x 7 3x 7 5x + 2 3x 7 5x + 2 = x 3 2x 2 x 3 2x 2 3 5 + 2 x = 6 x 7 2 x 3 x 4 We compute the its of top and bottom and decide if we can apply the it law for quotients: 3 5 x + 2 = 3 5 0 + 2 0 = 3 by it laws, since we know 6 x7 x = 0 and x 2 = 0 2 0 = 0 by it laws, since we know 3 x4 x = 0 Since the it of the denominator is 0, we cannot apply the quotient it law. However, since the it of the denominator is a nonzero real number, and the it of the denominator is 0, we can conclude that the it goes to : (3) 7x 6 3x 7 5x + 2 x 3 2x 2 = Note that for any x 0, x 0, and thus x 6 =. Therefore, 6 x 6 7x 6 = 7x 6 = 4 x 5 x 6 7 x 6 x 6 x 7 x 7
We compute the its of top and bottom and decide if we can apply the it law for quotients: x 5 = 0 0 by it laws, since we know x6 x = 0 and 7 = 7. Since both its exists and the it of the denominator is not 0, we can apply the quotient law, and get that = 0 7x 6 7 = 0 NOTE: For the last two questions, you can assume we know x = 0. Exercise 7. Let f(x) = 5 + cos x () Write f as a composite of continuous elementary functions. Let h(x) = cos x, g(x) = x, k(x) =. Then f = k g h. 5+x (2) For which points x can you deduce that f is continuous? f is a composite of continuous functions, so it will be continuous at all points at which this composite is defined. Thus we need to figure out the domain of k g h. The domain of h is all of R. The domain of g is only [0, ), though, so we need to restrict to those x for which h(x) = cos x 0, and this holds for all x in the first and fourth quadrant. In other words, we need 0 x π and all 3π x 2π, and all the x obtained from points 2 2 in these intervals by adding or subtracting multiples of 2π to them. Thus we need to restrict to x in the following set D: D = {x R x = a+2kπ for a [0, π 2 ] and k Z} {x R x = b+2kπ for b [3π 2, 2π] and k Z}. Lastly, the domain of k is all x 5, but since g(h(x)) = cos(x) is always 0, it can never be 5, so there are no further restrictions. Thus the domain of the composite is just the set D from above, and thus f is continuous at all the x D. Exercise 8. Let f : R R be a function defined as x 2 for x 2; f(x) = (x 2) 2 a for x = 2. 5
For what value of a is f a continuous function? x 2 Note that for all x 2, we have (x 2) =, so this function is the same as 2 x 2 for x 2; f(x) = x 2 a for x = 2. In order for f to be continuous at x = 2, we would need f(x) = f(2) = a. x 2 But note that f(x) =, x 2 so it does not exist. Since this it does not exist, it is not possible for the value a to be equal to it, so for NO values of a can f be continuous. 6