Mathematics Lecture and practice Norbert Bogya
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(2) Limit. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 2 / 31.
(3) ”Limes”. The word limes was used by Latin writers to denote a marked or fortified frontier. This term has been adapted and used by modern historians as an equivalent for the frontiers of the Roman Empire. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 3 / 31.
(4) Motivating example. How does the function f (x) =. x2 − 1 x −1. behave near x = 1? Step 1. Investigation of the domain: the domain of the function is the real numbers except x = 1.. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 4 / 31.
(5) Step 2. We substitute some values below and above 1. x. 0. 0.5 0.9 0.99 0.999 0.999999 . . .. 1. f (x) 1. 1.5 1.9 1.99 1.999 1.999999 . . .. X. 1. .... 1.000001 1.0001 1.001 1.01 1.1 1.5 2. X .... 2.000001 2.0001 2.001 2.01 2.1 2.5 3. Step 3. We can try to graph the function with computer. Wolfram Alpha. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 5 / 31.
(6) Step 4. Try to do some mathematics... f (x) =. x2 − 1 (x − 1)(x + 1) = = x + 1, x −1 x −1. 2. (x 6= 1). 2. y =x +1. y=. 1. x 2 −1 x−1. 1. -1. 1. -1. 1. We say that f (x) approaches the limit 2 as x approaches 1, and write lim f (x) = 2. x→1 Norbert Bogya (Bolyai Institute). or. x2 − 1 = 2. x→1 x − 1 lim. Mathematics. 2017. 6 / 31.
(7) Remark. The limit value does not depend on how the function is defined at x0 . ( 2 x −1 x−1. h(x) =. 1,. 2. 2. f (x) = x + 1. 1 -1. x 6= 1 x =1. 2. 2 −1 g (x) = xx−1. 1 1. ,. 1. -1. 1. -1. 1. lim f (x) = lim g (x) = lim h(x) = 2. x→1. Norbert Bogya (Bolyai Institute). x→1. Mathematics. x→1. 2017. 7 / 31.
(8) What about these? y. y. 1 1 x. x (. ( 0, x < 0 f (x) = 1, x ≥ 0. g (x) =. 1 x,. 0,. x= 6 0 x =0. (1) Both of the two functions defined over real numbers (no exclusion). (2) None of the two functions has limit at x0 = 0. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 8 / 31.
(9) Limit. Definition Let f (x) be defined on an open interval about x0 , except at x0 itself. We say that the limit of f (x) as x approaches x0 is the number L, and write lim f (x) = L, x→x0. if, for every number ε > 0, there exists a corresponding number δ > 0 such that for all x, 0 < |x − x0 | < δ. Norbert Bogya (Bolyai Institute). =⇒. Mathematics. |f (x) − L| < ε.. 2017. 9 / 31.
(10) Limit y 1.5 + ε. 1.5 1.5 − ε. x0 − δ. Norbert Bogya (Bolyai Institute). Mathematics. x0. x x0 + δ. 2017. 10 / 31.
(11) One-sided limit. To have a limit L as x approaches c, a function must be defined on both sides of c and its values f (x) must approaches c from either side. Because of this, ordinary limits are called two-sided. If f fails to have a two-sided limit c, it may still have a one-sided limit, that is, a limit if the approach is only from one side. If the approach is from the right, the limit is a right-hand limit. From the left, it is a left-hand limit.. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 11 / 31.
(12) One-sided limit y. f (2) = 3 y = f (x). 4 3. lim f (x) = 2. x→2+. 2 1 lim f (x) = 4. x→2−. x 2 Norbert Bogya (Bolyai Institute). Mathematics. 2017. 12 / 31.
(13) One-sided limit Definition If f (x) is defined on an interval (x0 , c), where x0 < c and approaches arbitrarily close to L as x approaches x0 from within that interval, then f has right-hand limit L at x0 . lim f (x) = L. x→x0+. Definition If f (x) is defined on an interval (a, x0 ), where a < x0 and approaches arbitrarily close to M as x approaches x0 from within that interval, then f has left-hand limit M at x0 . lim f (x) = M. x→x0− Norbert Bogya (Bolyai Institute). Mathematics. 2017. 13 / 31.
(14) Connection between one- and two-sided limits. Theorem A function f (x) has a limit as x approaches c if and only if it has left-hand and right-hand limits there and these one-sided limits are equal: lim f (x) = L ⇐⇒ lim+ f (x) = L and lim− f (x) = L.. x→c. Norbert Bogya (Bolyai Institute). x→c. x→c. Mathematics. 2017. 14 / 31.
(15) One-sided limit y. lim f (x) = 2. x→2+. y = f (x) 4. lim f (x) = 4. x→2−. 3. lim f (x) doesn’t exist. x→2. 2 1 x 2 Norbert Bogya (Bolyai Institute). Mathematics. 2017. 15 / 31.
(16) What about these? y. y. 1 1 x. x (. ( 0, x < 0 f (x) = 1, x ≥ 0. g (x) =. 1 x,. 0,. x= 6 0 x =0. (1) f (0) = 1, lim− f (x) = 0, lim+ f (x) = 1 x→0. x→0. (2) g (0) = 0, lim g (x) =? x→0. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 16 / 31.
(17) One-sided limit Exercise y y = h(x) 1 x -1. 1. 2. 3. -1 -2. Find the one- and two-sided limits at x0 = 0, x1 = 1, and x2 = 2.. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 17 / 31.
(18) Finding limit algebraically By substitution. Exercise Find the following limits. (a) lim 4 = x→2. (b). lim 4 =. x→−13. (c) lim x = x→3. (d) lim (5 − 2x) = x→2. x 3 + 4x 2 − 3 = x→−1 x2 + 5 3x + 4 (f) lim = x→−2 x + 5. (e) lim. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 18 / 31.
(19) Finding limit algebraically (Creating and) canceling a common factor. Exercise Find the following limits. x2 + x − 2 = (a) lim x→1 x2 − x x 2 (x 2 − 3x + 2) (b) lim = x→0 x2 + x −2x − 4 = (c) lim 3 x→−2 x + 2x 2. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 19 / 31.
(20) What about f (x) = x1 ? y. 1 x 1. As x → 0+ the values of f grow without bound, eventually reaching and surpassing every positive real number. That is, given any positive real number B, however large, the values of f become larger still. lim+. x→0. Norbert Bogya (Bolyai Institute). Mathematics. 1 =∞ x. 2017. 20 / 31.
(21) What about f (x) = x1 ? As x → 0− the values of f become arbitrarily large and negative. Given any negative real number −B, the values of f eventually lie below −B.. y. 1 x 1. Norbert Bogya (Bolyai Institute). Mathematics. lim−. x→0. 1 = −∞ x. 2017. 21 / 31.
(22) Exercises. (x − 2)2 = x→2 x 2 − 4 x −2 (2) lim 2 = x→2 x − 4 x −3 = (3) lim+ 2 x→2 x − 4 (1) lim. x −3 = x→2 x 2 − 4 x −3 (5) lim 2 = x→2 x − 4 2−x = (6) lim x→2 (x − 2)3. (4) lim−. Conclusion Rational functions can behave in various ways near zeros of their denominators.. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 22 / 31.
(23) What about f (x) = x1 ? When x is positive and becomes increasingly large, 1/x becomes increasingly small.. y. 1 =0 x→∞ x lim. 1 x. When x is negative and its magnitude becomes increasingly large, 1/x again becomes small. lim. x→−∞. Norbert Bogya (Bolyai Institute). Mathematics. 1 =0 x 2017. 23 / 31.
(24) Finite limits as x → ±∞ The symbol for infinity (∞) does not represent a real number. We use ∞ to describe the behaviour of a function when the values in its domain or range outgrow all finite bounds. Definition We say that f (x) has the limit L as x approaches infinity (minus infinity) and write lim f (x) = L, lim f (x) = L. x→∞. x→−∞. if, for every number ε > 0, there exists a corresponding number M (N) such that for all x, x > M, (x < N) Norbert Bogya (Bolyai Institute). =⇒ Mathematics. |f (x) − L| < ε. 2017. 24 / 31.
(25) Exercise Find the limits of f (x) = Geometric solution.. 1 x−1. at x0 = 1 and at infinities.. y. lim f (x) = ∞. x→1+. lim f (x) = −∞. x→1−. 1 x -1. 1 2. lim f (x) = 0. x→∞. lim f (x) = 0. x→−∞. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 25 / 31.
(26) Exercise Find the limits of f (x) = Analytic solution.. 1 x−1. at x0 = 1 and at infinities.. Theorem 1 = 0, x→±∞ x lim. lim+. x→0. 1 = ∞, x. lim−. x→0. 1 = −∞ x. 1 →0 x −1 1 x → −∞ =⇒ x − 1 → −∞ =⇒ →0 x −1 1 x → 1+ =⇒ x − 1 → 0+ =⇒ →∞ x −1 1 x → 1− =⇒ x − 1 → 0− =⇒ → −∞ x −1 x →∞. =⇒. Norbert Bogya (Bolyai Institute). x −1→∞. =⇒. Mathematics. 2017. 26 / 31.
(27) Exercises 1 (1) lim 5 + = x→∞ x √ π 3 = (2) lim x→−∞ x 2 5x 2 + 8x − 3 = x→∞ 3x 2 + 2. (3) lim (4). 11x + 2 = x→−∞ 2x 3 − 1. (5) lim 7 − x→∞. 7x 3 = x→∞ x 3 − 3x 2 + 6x. (6) lim. 2x 3 = x→∞ 5x 2 + 6x. (7) lim. lim. Norbert Bogya (Bolyai Institute). 8 = x2. 2 − x5 = x→∞ x 3 + 3x. (8) lim. Mathematics. 2017. 27 / 31.
(28) Exercises. y +2 y →2 y 2 + 5y + 6. (1) lim. 2x 5/3 − x 1/3 + 7 √ x→∞ x 8/5 + 3x + x. (4) lim. 4x − x 2 √ x→4 2 − x. (2) lim. |x + 2| (3) lim + (x + 3) x→−2 x +2. Norbert Bogya (Bolyai Institute). √ 3. x − 5x + 3 x→∞ 2x + x 2/3 − 4. (5) lim. Mathematics. 2017. 28 / 31.
(29) Continuity y y = h(x) 1 x -1. 1. 2. 3. -1 -2. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 29 / 31.
(30) Continuity Definition A function y = f (x) is continuous at an interior point c of its domain if lim f (x) = f (c). x→c. Definition A function y = f (x) is continuous at a left endpoint a or at a right endpoint b of its domain if lim f (x) = f (a) or lim− f (x) = f (b), respectively.. x→a+. x→b. Definition A function is continuous if it is continuous at every point of its domain. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 30 / 31.
(31) Continuity. Exercise Is the function. 2 x −4x+4 x 2 +x−6 , if x < 2 f (x) = 0, if x = 2 x 2 −3x+2 , if x > 2 x 2 −4x+4. continuous at the point x = 2?. Norbert Bogya (Bolyai Institute). Mathematics. 2017. 31 / 31.
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