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File: Calculus Pdf 170928 | 10850 Homework 5
math 10850 honors calculus 1 homework 5 solutions general and specic notes on the homework all the notes from homework 1 still apply also please read my emails from september ...

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                             Math 10850, Honors Calculus 1
                                           Homework 5
                                             Solutions
            General and specific notes on the homework
            All the notes from homework 1 still apply! Also, please read my emails from September 6, 13
            and 27 with comments on the first four homeworks.
            Reading for this homework
            Chapters 5 of Spivak.
            Assignment
            NB: I’m taking these from the 4th ed. of Spivak; the pdf easily found online is the 3th ed.,
            and there are some slight differences in the problem sets.
              1. Spivak Chapter 5, Question 1 parts (i), (ii), (iv).
                 Solution (part i): 1 is in the domain of the rational function (x2 − 1)/(x + 1), so the
                 limit is the value of the function at 1, namely 0.
                 Solution (part ii): The numerator factors as (x − 2)(x2 + 2x + 4). Since 2 is not in
                 the domain of the function, it is legitimate to cancel the factors of x − 2 above and
                 below. This leads to
                                        x3 −8
                                     lim      =lim(x2+2x+4)=12,
                                     x→2 x−2    x→2
                 the latter equality since x2 + 2x + 4 is rational, with 2 in its domain, so the limit is the
                 value at 2.
                 Solution (part iv): Viewed as a function of x, with y a constant, the domain of this
                 function is {x : x 6= y}. This means that we can divide through by x − y without
                 changing the limit (we are essentially multiplying the function by 1, with 1 written as
                (1/(x−y))/(1/(x−y)), which is valid as long as x 6= y). This leads to
                                  n   n
                             lim x −y = lim(xn−1+yxn−2+···+yn−2x+yn−1).
                             x→y x−y     x→y
                                                 1
                    This latter is a rational function (in variable x) with y in the domain, so the limit is
                    the value of the function at y, that is,
                                       yn−1 +yyn−2 +···+yn−2y+yn−1 or nyn−1.
                 2. Spivak Chapter 5, Question 3 parts (iii), (iv), (v). (Note that δ will depend on ε.)
                    Solution (part iii): f(x) = 100/x, a = 1. We claim that the limit is 100.
                    To prove this, suppose that ε > 0 is given. We want to find δ such that whenever
                    0 < |x −1| < δ, we have |100/x−100| < ε.
                    Now |100/x−100| < ε is equivalent (after a little algebra) to |x − 1|/|x| ≤ ε/100.
                    Choose δ < 1/2. Then 0 < |x − 1| < δ implies x ∈ (1/2,3/2), so |x| > 1/2 and
                    1/|x| < 2.
                    Choose also δ < ε/50. Then 0 < |x−1| < δ implies |x−1| < ε/50.
                    To get both conditions to hold, we choose δ < min{1/2,ε/50}, say for concreteness
                    δ = min{1/4,ε/100}; note that δ > 0.
                    Forthisδ, or any δ < min{1/2,ε/50}, wehavethatif0 < |x−1| < δ then|100/x−100| <
                    2(ε/50) = ε.
                    This proves that limx→1100/x = 100.
                    Solution (part iv): f(x) = x4, arbitrary a. We claim that the limit is a4.
                    To prove this, suppose that ε > 0 is given. We want to find δ such that whenever
                                               4    4
                    0 < |x −a| < δ, we have |x −a | < ε.
                           4   4                                                   3    2    2     3
                    Now|x −a |<εisequivalent (after a little algebra) to |x−a||x +ax +a x+a | ≤ ε.
                    Choose δ < 1. Then 0 < |x − a| < δ implies x ∈ (a − 1,a + 1), so x < a + 1, so
                    |x| < |a| + 1, so
                      3    2   2    3      3       2    2       3          3           2    2           3
                    |x +ax +a x+a | ≤ |x| +|a||x| +|a| |x|+|a| < (|a|+1) +|a|(|a|+1) +|a| (|a|+1)+|a| .
                                               3             2     2             3
                    Choose also δ < ε/((|a|+1) +|a|(|a|+1) +|a| (|a|+1)+|a| ). Then 0 < |x−a| < δ
                                                3             2     2             3
                    implies |x − a| < ε/(|a| + 1) + |a|(|a| + 1) + |a| (|a| + 1) + |a| .
                                                                                       3              2
                    To get both conditions to hold, we choose δ < min{1,ε/(|a| + 1) + |a|(|a| + 1) +
                      2              3                                                  3             2
                    |a| (|a| + 1) + |a| }, say for concreteness δ = min{1/2,ε/(2((|a| + 1) + |a|(|a| + 1) +
                      2              3
                    |a| (|a| + 1) + |a| ))}; note that δ > 0.
                                                           3             2     2             3
                    For this δ, or any δ < min{1,ε/((|a|+1) +|a|(|a|+1) +|a| (|a|+1)+|a| )}, we have
                    that if 0 < |x − a| < δ then
                      4   4           3    2   2    3                 3           2    2            3
                    |x −a | = |x−a||x +ax +a x+a | < |x−a|((|a|+1) +|a|(|a|+1) +|a| (|a|+1)+|a| ) < ε.
                                             4    4
                    This proves that lim    x =a .
                                        x→a
                    Solution (part v): f(x) = x4 +1/x, a = 1. We claim that the limit is 2.
                                                           2
                 To prove this, suppose that ε > 0 is given. We want to find δ such that whenever
                 0 < |x −1| < δ, we have |x4 +1/x−2| < ε.
                 Now |x4 +1/x−2| = |(x4 −1)+(1/x−1)| ≤ |x4 −1|+|1/x−1|, so it is enough to
                 find a δ > 0 such that whenever 0 < |x − 1| < δ, we have both |x4 − 1| < ε/2 and
                 |1/x−1| < ε/2.
                                                                    3           2    2
                 From part (iv) we know that if δ < min{1,(ε/2)/((|1| +1) +|1|(|1|+1) +|1| (|1|+
                       3                                          4
                 1) +|1| )} = min{1,ε/30} then 0 < |x−a| < δ implies |x −1| < ε/2.
                 From part (iii) we know that for δ < min{1/2,ε/50}, we have 0 < |x−1| < δ implies
                 |100/x−100| < ε so |1/x−1| < ε/100, from which it follows that if δ < min{1/2,ε},
                 we have 0 < |x−1| < δ implies |1/x−1| < ε/2.
                 Sotoshowlim    (x4+1/x) = 2, givenε > 0weshouldtakeδ < min{1,ε/30,1/2,ε/50} =
                             x→1
                 min{1/2,ε/50}.
               3. Spivak Chapter 5, Question 7.
                 Solution: The graph below gives an example of such a function: (for the record, it is
                 a portion of the graph of f(x) = p|x|).
                 Take ℓ = a = 0, and ε = δ = 1. It is certainly the case that
                                          0 < |x| < 1 =⇒ |f(x)| < 1.
                 But it is not the case that
                                        0 < |x| < 1/2 =⇒ |f(x)| < 1/2;
                 the best that we can ensure from the imposition 0 < |x| < 1/2 is |f(x)| < .7; to force
                 |f(x)| < 1/2 we need to demand that 0 < |x| < 1/4.
                                                   3
                 4. Spivak Chapter 5, Question 8.
                    Solution (part a): Yes. Consider, for example,
                                                   f(x) =  1     if x > 0
                                                             −1 if x < 0
                    and                                   
                                                   g(x) =    −1 if x > 0
                                                              1   if x < 0.
                    Certainly lim    f(x) and lim     g(x) do not exist. But (f + g)(x) = 0 unless x = 0
                                  x→0             x→0
                    (at which point the sum is undefined), so limx→0(f +g)(x) = 0.
                    Solution (part b): Yes. If lim    f(x) exists, and lim   (f(x)+g(x)) exists, then by
                                                  x→a                     x→a
                    the sum-product-reciprocal theorem,
                                             lim((f(x)+g(x))−f(x)) = limg(x)
                                            x→a                          x→a
                    exists.
                    Solution (part c): No. If lim     (f(x) + g(x)) existed then (by part b) lim     g(x)
                                                   x→a                                            x→a
                    would also exist, a contradiction.
                    Solution(partd): Itistemptingtosay“yes”. Iflim         f(x)exists, and lim   f(x)g(x)
                                                                       x→a                   x→a
                    exists, then by the sum-product-reciprocal theorem,
                                               lim(f(x)g(x))/f(x)) = limg(x)
                                               x→a                     x→a
                    should exist; but this assumes that lim  f(x) is not zero. So to find a counter-example,
                                                         x→a
                    we need to find functions f and g, and an a, with lim       f(x) = 0, lim     f(x)g(x)
                    existing, and lim    g(x) not existing.                x→a               x→a
                                     x→a
                    Taking f to be the constant 0 function, g to be the function g(x) = sin(1/x) and a = 0
                    works nicely.
                 5. Spivak Chapter 5, Question 9.
                    Solution: An implicit assumption here (contained in the writing down of the two
                    expressions) is that lim  f(x), lim    f(a+h)bothexist. Supposethatlim         f(x) =
                                          x→a          h→0                                     x→a
                    L. We will argue that lim     f(a+h)=L.
                                              h→0
                    Given ε > 0, we know that there is δ > 0 such that 0 < |x−a| < δ implies |f(x)−L| < ε
                    (this because lim    f(x) = L). We claim that the same δ works (for this ε) to show
                                     x→a
                    lim    f(a+h)=L. Indeed, we need to show that when
                       h→0
                                                         0 < |h| < δ
                    then
                                                     |f(a +h)−L| < ε.
                                                           4
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