L Hôpital s Rule was discovered by Bernoulli but written for the first time in a text by L Hôpital.

Transcription

1 7.5. Ineterminate Forms an L Hôpital s Rule L Hôpital s Rule was iscovere by Bernoulli but written for the first time in a text by L Hôpital. Ineterminate Forms 0/0 an / f(x) If f(x 0 ) = g(x 0 ) = 0, then lim x x0 g(x) can not be foun by substituting x = x 0. The substitution prouces 0/0, a meaningless expression known as an ineterminate form. Similarly, if lim f(x) = ± = lim g(x), then lim x x 0 x x0 g(x) prouces the ineterminate form. Sometimes, limits that lea to ineterminate forms may be foun by cancelation, rearrangement of terms, or other algebraic manipulations. Limits involving transcenental functions often can not be foun by these classical methos. L Hôpital s Rule is an alternative metho to fin limits that lea to ineterminate forms. Theorem 1. (L Hôpital s Rule) Suppose that x x0 f(x) (1) f(x 0 ) = g(x 0 ) = 0 or lim x x0 f(x) = ± = lim x x0 g(x), (2) f an g are ifferentiable on an open interval I containing x 0, (3) g (x) 0 for all x I\{x 0 }. Then provie the limit on the right han sie exists. f(x) lim x x 0 g(x) = lim f (x) x x 0 g (x), Remark. L Hôpital s Rule can be applie to one-sie limits an when x 0 is a real number or ±. Example 1. Use L Hôpital s rule to show that lim x 0 sin x x = 1. Example 2. Fin lim x 2 x x 2 Example 3. Fin lim x 1 x 3 1 4x 3 x 3. Page 1 of 25

2 Example 4. Fin lim x 0 cos x cos(3x) x 2. ln x Example 5. Fin lim x ln x. 3 x 1 Example 6. (Exam)Fin lim. x 0 x Example 7. Fin 2 sin x 1 lim x 0 e x 1 Example 8. Evaluate lim x 5 x 1 3 x 1. Page 2 of 25

3 x 2 Example 9. Fin lim x ln x. log Example 10. Fin lim 2 x x log 3 (x + 1). Example 11. (Exam)Fin the value of the constant a such that a > 0 an lim x 0 x ln(x + 1) 1 cos(ax) = 1 4. Ineterminate Forms.0 an We can hanle the ineterminate forms.0 an by using algebra to convert them to a 0/0 or / form an then we apply L Hôpital s rule. Example 1. Evaluate lim x cot x. x 0 + ( Example 2. Fin lim csc x 1 ). x 0 + x Page 3 of 25

4 Example 3. lim x 2 + ( x x 2 1 ln(x 1) ) Example 4. (Exam) Fin lim e x (3x + 1). x Ineterminate Powers (1, 0 0, an 0 ) 1, 0 0, 0 are ineterminate forms. The limits that lea to one of these forms can be hanle by L Hopital s rule as follows: lim x a [f(x)]g(x) lim g(x) ln f(x). x a If lim x a ln F (x) = L, then lim x a F (x) = lim x a e ln F (x) = e L Remark. a may be a real number or ±. ( Example 1. Evaluate lim 1 2 x. x x) Page 4 of 25

5 Example 2. Fin lim x 0 + (sin x)x. Example 3. Fin lim x (ln x) 1/x. Example 4. Fin lim x 0 +(ex + x 2 ) 1/x. Page 5 of 25

6 ( ) tan x. Example 5. Evaluate lim e 1 x x 0 + Example 6. Evaluate lim x 0 + (tan x)x. Example 7. (Exam) Fin lim (1 + 3x) 1/x. x Page 6 of 25

7 Example ( 8. (Exam) ) x 1 Fin lim. x 0 + x Example 9. (Exam) Fin lim x (x 3 + e) 1/ ln x. Example 10. Evaluate lim (cos x) 1 x 2. x 0 Page 7 of 25

8 Example 11. Evaluate lim x 0 + ( x) x Inverse Trigonometric Functions Inverse trigonometric functions arise when we want to calculate angles from sie measurements in triangles. They also provie useful antierivatives an appear frequently in the solutions of ifferential equations. This section shows how these functions are efine, graphe, an evaluate, how their erivatives are compute, an why they appear as important antierivatives. Defining the inverses The six basic trigonometric functions are not one-to-one in their omains, but we can restrict their omains to intervals on which they are one-to-one. The arcsine function The function f(x) = sin x is one-to-one on any one of the intervals [ π 2, π 2 ], [π 2, 3π 2 choose [ π 2, π 2 ] to satisfy the relationship sin 1 (1/x) = csc 1 x. ],... But we Definition. (The arcsine function) For any x [ 1, 1], y = sin 1 x is the number y [ π 2, π ] for which sin y = x. 2 Example. Fin sin 1 (0). Remarks. (1) sin 1 x 1 sin x. (2) 1 sin x = (sin x) 1 = csc x. Page 8 of 25

9 The graph of y = sin 1 x The omain of sin 1 x is [ 1, 1]. The range of sin 1 x is [ π 2, π 2 ]. Remark. The graph of y = sin 1 x is symmetric about the origin. Therefore y = sin 1 x is an o function; that is sin 1 ( x) = sin 1 x. Example 1. Fin sin 1 ( 1 2 ). Example 2. Fin the value of ( ( )) 3 tan sin 1. 7 Remarks. (a) sin 1 (sin x) = x for x [ π 2, π 2 ]. (b) sin(sin 1 x) = x for x [ 1, 1]. Page 9 of 25

10 Example. Fin sin(sin 1 2) an sin 1 (sin 2π). The arccosine function The function f(x) = cos x is one-to-one on the interval [0, π]. Definition. (The arccossine function) For any x [ 1, 1], y = cos 1 x is the number y [0, π] for which cos y = x. ( ) 1 Example. Fin cos 1. 2 The graph of y = cos 1 x The omain of cos 1 x is [ 1, 1]. The range of cos 1 x is [0, π]. Remarks. (a) The function cos 1 x is neither even nor o. (b) cos 1 (cos x) = x for x [0, π]. (c) cos(cos 1 x) = x for x [ 1, 1]. Page 10 of 25

11 Example. Fin cos 1 (cos(π/2)) an cos 1 (cos 5π). Ientities involving arcsine an arccosine For all x [ 1, 1], we have (1) cos 1 ( x) = π cos 1 x (2) sin 1 x + cos 1 x = π 2 Proof. Example 1. Fin cos 1 ( 3/2). Example 2. Fin the value of 2 log 4 π2 cos 1 ( 1/ 2). Page 11 of 25

12 Example 3. Solve for x: ln(e 2x + 2e x + 1) = 2 sin[cos 1 ( x) sin 1 x]. Inverses of tan x, cot x, sec x, an csc x The function tan x is one-to-one on the interval ( π/2, π/2) an the function cot x is one-to-one on the interval (0, π). Thus they have inverses tan 1 x an cot 1 x. Definition. (The arctangent an arccotangent functions) (1) For any x R, y = tan 1 x is the number y ( π/2, π/2) for which tan y = x. (2) For any x R, y = cot 1 x is the number y (0, π) for which cot y = x. Remarks. (1) The omain of tan 1 x is R an omain of cot 1 x is R. (2) The range of tan 1 x is ( π/2, π/2) an the range of of cot 1 x is (0, π). (3) The graph of y = tan 1 x is symmetric about the origin. Therefore y = tan 1 x is an o function; that is tan 1 ( x) = tan 1 x. Page 12 of 25

13 Example 1. (Exam) Simplify the expression cos ( tan 1 (3/x) ). Example 2. Evaluate lim x tan 1 x. Definition. (The arcsecant an arccosecant functions) (1) For any x (, 1] [1, ), y = sec 1 x is the number y [0, π/2) (π/2, π] for which sec y = x. (2) For any x (, 1] [1, ), y = csc 1 x is the number y [ π/2, 0) (0, π/2] for which csc y = x. Useful Ientities (1) cot 1 (x) = π/2 tan 1 x. (2) sec 1 x = cos 1 (1/x). (3) csc 1 x = sin 1 (1/x). Example 1. Fin sec 1 2. Example 2. Fin ( ) sin cos 1 (1/3) + csc 1 3. Page 13 of 25

14 ( 4 x Example 3. Fin csc tan 1 2 ). x Example 4. (Exam) ( )) Solve for x: log 5 (25) x + tan (sin 1 2x = cos ( sec 1 x ) 4x2 + 1 Example 5. Evaluate lim x sec 1 x. The erivatives of inverse trigonometric functions Inverse trigonometric functions provie antierivatives for a verity of functions. The erivatives of y=sin 1 x an y=cos 1 x f(x) = sin x is ifferentiable on the interval ( π/2, π/2) an f (x) = cos x > 0 in this interval. Therefore, f 1 (x) = sin 1 x is ifferentiable on the interval ( 1, 1) an Using cos y = x sin 1 x = 1 f (sin 1 x) = 1 cos(sin 1 x). 1 sin 2 y an sin(sin 1 x) = x, we get Page 14 of 25

15 x sin 1 x = 1 1 x 2, x < 1 an hence 1 1 x 2 x = sin 1 x + C Since cos 1 x = π/2 sin 1 x, we have x cos 1 x = x sin 1 x = 1 1 x 2, x < 1 Example 1. Fin y if y = sin 1 (e x2 + 3x). Example 2. Fin y if y = 9 sin 1 (3x) + cos 1 (x 2 ). Example 3. Fin x 25 x 2. Example 4. Fin x 6x x 2. Page 15 of 25

16 Example 5. Fin 1 1/2 x 3 + 4x 4x 2. Example 6. Fin y sin 1 y 1 y 2. The erivatives of y=sec 1 x an y=csc 1 x Since sec 1 x = cos 1 (1/x), we have Simplifying we get x sec 1 x = x cos 1 (1/x) = 1 1 x 2 ( 1 x 2 ). x sec 1 x = 1 x x 2 1, x > 1 an hence 1 x x 2 1 x = sec 1 x + C Since csc 1 x = π/2 sec 1 x, we have Page 16 of 25

17 x csc 1 x = x sec 1 x = 1 x x 2 1, x > 1 Example 1. Fin y if y = sec 1 (2x 2 x). Example 2. (Exam) Fin y if y = 3 x + sec 1 (3x). Example 3. Fin 2 1 x x 4x 2 1. Example 4. Fin x (x 4) x 2 8x + 7. sec 1 x Example 5. Fin lim x 1 + x2 1. Page 17 of 25

18 The erivatives of y=tan 1 x an y=cot 1 x To fin x tan 1 x we can use the theorem in section 7.1. Alternatively we may process as follows: Differentiating both sies we get Thus Therefore we have the following formula y = tan 1 x tan y = x. sec 2 y y x = 1. y x = x tan 1 x = 1 sec 2 y = tan 2 (tan 1 x). x tan 1 x = x, 2 an hence x 2 x = tan 1 x + C Since cot 1 x = π/2 tan 1 x, we have x cot 1 x = x tan 1 x = 1 x 2 + 1, Example 1. Fin y if y = log 5 (tan 1 (5x)). Example 2. Fin y if y = x cot 1 x + sin(tan 1 (x 2 )). Example 3. Fin 2 1 cot 1 x x x. Page 18 of 25

19 Example 4. Fin x 4x x + 7. Example 5. Fin x 2 x 1 + x 6. tan 1 (4x) Example 6. Fin lim. x 0 x Page 19 of 25

20 7.7. Hyperbolic Functions The hyperbolic functions ar efine as a combinations of the exponential functions e x an e x. They simplify many mathematical expressions an occur frequently in mathematical applications. Definition. (The six basic hyperbolic functions) (1) Hyperbolic sine of x: sinh x = ex e x. 2 Domain: R Range: R (2) Hyperbolic cosine of x: cosh x = ex + e x. 2 Domain: R Range: [1, ) (3) Hyperbolic tangent of x: tanh x = sinh x cosh x = ex e x e x + e x. (4) Hyperbolic cotangent of x: coth x = cosh x sinh x = ex + e x e x e x. (5) Hyperbolic secant of x: sech x = 1 cosh x = 2 e x + e x. (6) Hyperbolic cosecant of x: csch x = 1 sinh x = 2 e x e x. Remarks. (1) sinh x is an o function. (2) cosh x is an even function. Page 20 of 25

21 Ientities for hyperbolic functions (1) cosh 2 x sinh 2 x = 1. (2) sinh(2x) = 2 sinh x cosh x. (3) cosh(2x) = cosh 2 x + sinh 2 x. (4) cosh 2 x = cosh(2x) Example 1. Simplify cosh(2x) + sinh(2x). Example 2. Simplify tanh(2 ln x). Example 3. If sinh x = 4/3, then fin the value of the other five hyperbolic functions. Derivatives an Integrals (1) sinh x = cosh x x cosh(kx)x = 1 k sinh(kx) + C (2) cosh x = sinh x x sinh(kx)x = 1 k cosh(kx) + C Page 21 of 25

22 (3) x tanh x = sech2 x sech 2 xx = tanh x + C (4) x coth x = csch2 x csch 2 xx = coth x + C (5) sech x = sech x tanh x x sech x tanh xx = sechx + C (6) csch x = csch x coth x x csch x coth xx = csch x + C Example 1. (Exam) Fin y if y = x x + coth x. Example 2. Fin y if y = ln(cosh x 2 ). Example 3. Fin 9 4 sinh x x x. Example 4. Fin cosh 2 x x. Page 22 of 25

23 Inverse Hyperbolic Functions (1) Since sinh x = cosh x > 0 for all x R, sinh x is an increasing function on R. Thus it has x an inverse. We enote this inverse by sinh 1 x. Domain of sinh 1 x is R an its range is R. (2) cosh x is not one-to-one on its omain, but it becomes one-to-one if we take the restriction cosh x, x [0, ). Thus it has an inverse cosh 1 x. Domain of cosh 1 x is [1, ) an its range is [0, ). (3) tanh x is one-to-one on R, an hence it has an inverse tanh 1 x. Domain of tanh 1 x is ( 1, 1) an its range is R. ( ) 1 (4) sech 1 x = cosh 1. x (5) csch 1 x = sinh 1 ( 1 x (6) coth 1 x = tanh 1 ( 1 x ). ). Evaluating inverse hyperbolic functions Let us fin a formula to evaluate sinh 1 x. y = sinh 1 x sinh y = x But e y > 0 for all y an hence Similarly, ey e y 2 = x (e y x) 2 = x e y x = ± x e y = x + x ( y = sinh 1 x = ln x + ) x ( cosh 1 x = ln x + ) x 2 1. Example. Fin cosh 1 (7/4). Page 23 of 25

24 Derivatives an Integrals (1) x sinh 1 x = 1 x2 + 1 x x2 + 1 = sinh 1 x + C (2) x cosh 1 x = 1 x2 1, x > 1. x x2 1 = cosh 1 x + C (3) x tanh 1 x = 1, x < 1. 1 x2 x 1 x 2 = tanh 1 x + C Example 1. Fin x sinh 1 (5 x ). Example 2. Fin x x 2. Example 3. (Exam) Solve for x: cosh 1 (2x) = sec 2 ( cot 1 (2/π) ) 4 log 2 π Page 24 of 25

25 7.8. Relative Rates of Growth It is often important in mathematics, computer science, an engineering to compare the rates at which functions of x grow as x becomes large. Exponential functions are important in these comparisons because of their very fast growth, an logarithmic functions because of their very slow growth. In this section we show how to compare the rates at which functions of x grow as x. We restrict our attention to functions whose values eventually become an remain positive as x. Definition. (Rates of growth as x ) Let f(x) an g(x) be positive for x sufficiently large. (1) If lim x f(x) g(x) (2) If lim x f(x) g(x) (3) If lim x f(x) g(x) =, then we say that f(x) grows faster than g(x) as x. = 0, then we say that f(x) grows slower than g(x) as x. = L > 0, then we say that f(x) an g(x) grow at the same rate as x. Example 1. Show that e 5x grows faster than x 2 + 5x + 4 as x. Example 2. Determine whether f(x) = ln x grows faster, slower, or at the same rate as g(x) = log 10 x as x. Example 3. Let f(x) = x an g(x) = ln(x 2 + 1). Does f(x) grow faster, slower, or at the same rate as g(x) as x? Page 25 of 25