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CALCULUS I – REVIEW Things to Know Precalculus (Prerequisite Material). Sections 1.1 – 1.3 • Functions, Graphs, Function C

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CALCULUS I – REVIEW Things to Know Precalculus (Prerequisite Material). Sections 1.1 – 1.3 • Functions, Graphs, Function C CALCULUS I { REVIEW ALLAN YASHINSKI Abstract. The following document evolved from several \study guides" which were written to help Math 241 (Calculus I) students at the University of Hawaii at Manoa prepare for exams. The section references are for University Calcu- lus, Alternate Edition by Hass, Weir, and Thomas. Things to know Precalculus (Prerequisite Material). Sections 1.1 { 1.3 • Functions, graphs, function composition f ◦ g. • Trigonometry, the six basic trigonometric functions. You should know the values of the trig functions on angles like 0; π=6; π=4; π=3; π=2; and the corresponding angles in other quadrants. Limits and Continuous Functions. Sections 2.2, 2.4, 2.5, 2.6 • To say lim f(x) = L means that the values (the output) of the function x!c f(x) get closer and closer to the number L when we consider x (the input) closer and closer to c. The function f does not even have to be defined at x = c, and even if it is, the value f(c) does not influence what the limit is. In other words, it doesn't matter what is happening at x = c; limits are all about what is happening near x = c. • The Limit Laws Suppose that lim f(x) and lim g(x) exist. Then x!c x!c lim (f(x) + g(x)) = lim f(x) + lim g(x) , x!c x!c x!c lim (f(x) − g(x)) = lim f(x) − lim g(x) , x!c x!c x!c lim (f(x) · g(x)) = lim f(x) lim g(x) , x!c x!c x!c lim (k · f(x)) = k lim f(x) for any constant k, x!c x!c f(x) lim f(x) lim = x!c , provided lim g(x) 6= 0, x!c g(x) limx!c g(x) x!c r lim (f(x))r = lim f(x) , for any exponent r, as long as the expres- x!c x!c sion makes sense. (Notice it wouldn't make sense if limx!c f(x) < 0 and r = 1=2.) • Limits of polynomial functions are evaluated by substitution. That is, if p(x) is a polynomial function, then lim p(x) = p(c): x!c 1 2 ALLAN YASHINSKI This is because polynomial functions are continuous. f(x) • For a limit of a rational function g(x) , if g(c) 6= 0 then f(x) f(c) lim = : x!c g(x) g(c) (This is the easy case.) However, if g(c) = 0, something else needs to be done. Usually one can try to factor and cancel factors, e.g. x2 − 9 (x − 3)(x + 3) x + 3 3 + 3 lim = lim = lim = = 6: x!3 x2 − 5x + 6 x!3 (x − 3)(x − 2) x!3 x − 2 3 − 2 • One can usually rationalize to compute tricky limits with square roots, e.g. p p p 4 + x − 2 4 + x − 2 4 + x + 2 lim = lim · p x!0 x x!0 x 4 + x + 2 (4 + x) − 4 = lim p x!0 x( 4 + x + 2) x = lim p x!0 x( 4 + x + 2) 1 = lim p x!0 4 + x + 2 1 1 = p = : 4 + 0 + 2 4 • The Sandwich Theorem states that if there are three functions f; g; h such that g(x) ≤ f(x) ≤ h(x) for all x near c; and if lim g(x) = L and lim h(x) = L; x!c x!c then we can conclude that lim f(x) = L also. A picture of the graphs of x!c three such functions makes it easier to understand why this is true (see pictures in the text). • Know, and have respect for (cause it's not easy), the limit sin x lim = 1: x!0 x We also have the generalization that sin(ax) lim = 1 for any constant a: x!0 ax CALCULUS I { REVIEW 3 We can use this result to compute other limits through algebraic manipu- lation, e.g. tan(2x) sin(2x) lim = lim x!0 x x!0 x cos(2x) sin(2x) 1 = lim · x!0 x cos(2x) sin(2x) 1 = lim 2 · · x!0 2x cos(2x) sin(2x) 1 = 2 lim lim x!0 2x x!0 cos(2x) 1 = 2 · 1 · = 2: cos(0) • For the one-sided limit lim f(x), we only consider x which are less than x!c− c, whereas for the one-sided limit lim f(x), we only consider x which are x!c+ greater than c. • The two-sided limit lim f(x) exists if and only if both one-sided limits x!c lim f(x) and lim f(x) exist and are the same value. x!c− x!c+ • We say lim f(x) = L if whenever we consider larger and larger and larger x!1 values of x (e.g. x = 103; 106; 1010; 101478; 10234987235, and so on...) the values of the function f(x) get closer and closer to the number L. Similarly the limit lim f(x) is about what happens for larger and larger nega- x→−∞ tive values of x (e.g. x = −103; −106; −1010; −101478; −10234987235, and so on...) Graphically, if lim f(x) = L, then the horizontal line y = L will be x!1 a horizontal asymptote for the graph of the function y = f(x). Similarly, if lim f(x) = M, then the line y = M will be a horizontal asymptote. x→−∞ • For any positive number p, 1 1 lim = 0 and lim = 0: x!1 xp x→−∞ xp This can be used, along with algebraic manipulation, to calculate limits of rational functions as x ! ±∞. When faced with such a limit, divide the numerator and denominator by the largest power of x in the denominator, e.g. 2 2 1 4x + 2x − 7 (4x + 2x − 7) 2 lim = lim · x x!1 x2 x!1 x2 1 5 + 1000x ( 5 + 1000x) x2 4 + 2 − 7 = lim x x2 x!1 1 1000 5 + x 0 + 2 − 0 = 1 = 10: 5 + 0 4 ALLAN YASHINSKI • We say that lim f(x) = 1 if as the input x gets closer and closer to c, x!c the values f(x) grow larger and larger without bound. Similarly, we say lim f(x) = −∞ if the values f(x) are larger and larger negative numbers x!c (without bound). If lim = 1 or −∞, then the vertical line x = c will be a x!c vertical asymptote for the graph y = f(x). f(x) • Infinite limits will occur for a limit of the form lim with the property x!c g(x) that (1) lim g(x) = 0, and x!c (2) lim f(x) is some finite nonzero number. x!c In this case, the limit will be either 1 or −∞. One must pay careful attention to the signs (+ or −) in the expression to determine whether the answer is 1 or −∞. For example, consider the limit x − 1 lim : x!3+ 3x2 − x3 Notice the limit of the denominator is 0 and the limit of the numerator is 2 (a finite, nonzero number!). This indicates the limit will be either 1 or −∞. Let's factor the expression and consider the signs: x − 1 x − 1 lim = lim x!3+ 3x2 − x3 x!3+ x2(3 − x) We are supposed to consider x that are slightly larger than 3. For such an x, the numerator x − 1 will be positive (it's close to 2, so its definitely positive). The x2 in the denominator is certainly positive, because a square is never negative. The last factor 3 − x is where we must be careful. If x is a little bigger than 3, then 3 − x will be close to 0, but it will ever so slightly be less than 0, making it negative. So x − 1 (−) the signs of look like x2(3 − x) (+)(+) when x is a little bigger than 3. The overall sign is negative, and we conclude x − 1 lim = −∞: x!3+ x2(3 − x) In a similar way, we can determine x − 1 lim = 1: x!3− 3x2 − x3 f(x) • In contrast to the previous point, if a limit lim is such that x!c g(x) (1) lim g(x) = 0, and x!c (2) lim f(x) = 0; x!c then you can probably cancel factors or rationalize, see the previous dis- cussions. CALCULUS I { REVIEW 5 • A function f is called continuous at the point x = c if lim f(x) = f(c): x!c If f is continuous at every point, then f is called a continuous function. Intuitively, a function is continuous if you can draw its graph without ever lifting your pencil, though this is not a rigorous definition. Many functions are continuous: Polynomial functions are continuous. f(x) A rational function g(x) is continuous at any c for which g(c) 6= 0. Sums, differences, products, powers, and compositions of continuous functions are continuous. f(x) A quotient g(x) of continuous functions is continuous at any c for which g(c) 6= 0. The trig functions sin x and cos x are continuous. The other trig functions sin x cos x 1 1 tan x = ; cot x = ; sec x = ; csc x = cos x sin x cos x sin x are continuous at all points where their denominators are not zero. • The Intermediate Value Theorem states that if f is a continuous func- tion on a closed interval [a; b], then every value between the numbers f(a) and f(b) is in the range of f.
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