Calculus I, Fall 2021

Brief solutions to Quiz 3

Oct 26, 2021

1. (a) (16 pts) State precise definition of lim_{x→∞}f(x) = L.

(b) (20 pts) Use it to prove that lim_{x→∞} \frac{1}{x} = 0. Ans: See page 119 of the textbook.

Figure 1: Common mistakes to problem 1 (b)

2. For (a) and (b), just answer whether the statement is true or false. Need not explain.

(a) (8 pts) True or False? If f(x) is continuous at x = c, then it is differentiable at x = c.

(b) (8 pts) True or False? If f(x) is differentiable at x = c, then it is continuous at x = c.

(c) (16 pts) Choose either (a) or (b) to elaborate: prove it if true, find a counter example if false.

Ans: (a): False. (b): True. (either 8 pts or 0 pts, no partial credits) (c): If answer (a): counter example: see page 5 of Lecture 05 note. 8 pts for correct example. 8 pts more for correct explanation. If answer (b): see Theorem 1 in section 3.2 of the textbook.

Figure 2: A common mistake to problem 2(c)

Remark on part (c): Many people chose to answer (a) for part (c) with the counter example f(x) = |x|. The true reason for this example to be ’not differentiable at x = 0’ is

lim_{h→0^{+}} \frac{f(h) − f(0)}{h} ≠ lim_{h→0^{−}} \frac{f(h) − f(0)}{h},

and not

lim_{x→0^{+}} f′(x) ≠ lim_{x→0^{−}} f′(x),

since lim_{h→0^{+}} \frac{f(h) − f(0)}{h} and lim_{x→0^{+}} f′(x) are two different quantities in general. They are the same for the function f(x) = |x| but not always the same. The following is an example with lim_{h→0^{+}} \frac{f(h) − f(0)}{h} = 0, while lim_{x→0^{+}} f′(x) does not exist:

f(x) = \{ \begin{matrix}x^{2} sin(\frac{1}{x}), & x > 0, \\ 0, & x = 0. \end{matrix}

3. Use the derivative product rule to evaluate

(a)(16 pts) \frac{d^{4}}{dx^{4}} (f(x)g(x)) (b)(16 pts) \frac{d}{dx}(f_{1}(x)f_{2}(x)f_{3}(x)f_{4}(x))

Correct answer without derivation receives full credits. If you are sure, just write down the answer directly (and check twice). Incorrect answer with detailed derivation (starting from n = 2, then n = 3, then n = 4, etc.) will probably receive partial credits if the mistakes are only minor. If you are not sure, try to derive it step by step. Ans: (a):

f′′′′(x)g(x) + 4f′′′(x)g(x)′ + 6f′′(x)g(x)′′ + 4f′(x)g(x)′′′ + f(x)g(x)′′′′

(b):

f′_{1}(x)f_{2}(x)f_{3}(x)f_{4}(x)+f_{1}(x)f′_{2}(x)f_{3}(x)f_{4}(x)+f_{1}(x)f_{2}(x)f′_{3}(x)f_{4}(x)+f_{1}(x)f_{2}(x)f_{3}(x)f′_{4}(x)