Chapter 5 Exponential, Logarithmic, and Inverse Trigonometric Functions

Home , Natural logarithm, Trigonometric functions

Exponential Functions

If f(x) = x2, then the base x varies and the exponent 2 is a constant. The function f(x) = 2x has constant base 2 and variable exponent x. This is called an exponential function. x f(x) = 2x −3 1/8 −2 1/4 −1 1/2 0 1 1 2 2 4 3 8

Now lim 2x = ∞ and lim 2x = 0 x→∞ x→−∞ To see that f(x) = 2x gets large quickly consider the following: Suppose you have one penny and for the entire month of February your money doubles each day. If I oﬀer you $100,000 or the amount you would have after February 28, which would you take?

Day Amount Dollars 1 2 $0.02 7 27 = 1,024 $10.24 14 214 = 16,384 $163.84 21 221 = 2,097,152 $20,971.52 28 228 = 2.6844 × 108 $2,684,400

Hence you can see how fast the exponential function rises.

127 ³ ´x 1 −x Now let f(x) = 2 = 2

x 2−x −2 4 −1 2 0 1 1 1/2 2 1/4

The most important exponential function is f(x) = ex. We base our development on the following deﬁnition:

lim(1 + x)1/x = e x→0 x (1 + x)1/x 1 (1.1)10 = 10 1 (1.01)100 = 100 1 (1.001)1,000 = 1,000 1 (1.0001)10,000 = 10,000 1 (1.00001)100,000 = 100,000

Do the above calculations on your calculator and ﬁnd that e ≈ 2.7182818.

128 Now let f(x) = ex and complete the table of values on your calculator.

x ex 0 1 10 100 1000 −1 −10 −100

Thus the graph of f(x) = ex is

lim ex = ∞ and lim ex = 0 x→∞ x→−∞ If 0 < a, a 6= 1, then f(x) = ax is the general exponential function. All of the familiar laws of exponents hold. If 1 < a, then y = ax is increasing and if 0 < a < 1, then y = ax is x1 x2 x decreasing. Thus, if a = a , then x1 = x2. Also note that a > 0.

129 Exercises on Exponential Functions

1. Graph

a. f(x) = 3x

µ1¶x b. f(x) = 3

c. f(x) = e−x

d. f(x) = −(2−x)

2. Find the limits

a. lim (1.001)x x→∞

b. lim (.999)x x→∞

c. lim e−x x→∞

d. lim e−x x→−∞

e2x − e−2x e. lim x→∞ e2x + e−2x

f. lim e1/(1−x) x→1+

g. lim e1/(1−x) x→1−

3. Suppose the graphs of f(x) = x2 and g(x) = 2x are drawn where the unit of

measurement is 1 in. If x = 24 in, then ﬁnd f(x) in feet and g(x) in miles.

130 Logarithmic Functions y Deﬁnition: loga x = y means a = x. loga x is read logarithm of x to the base a.

Example 1 Let y = log2 x

x log2 x 1 0 2 1 4 2 8 3 1/2 −1 1/4 −2

y Remarks: log2(0) = y means 2 = 0 which is impossible. y Also log2(−1) = y means 2 = −1 which is impossible. Thus the domain of y = log2 x is (0, ∞).

Common Logarithm

The common logarithm has base a = 10. We write y = log10 x = log x which means y 1 10 = x. Thus log 1 = 0, log 10 = 1, log 100 = 2, log 10 = −1, etc. Natural Logarithm y The natural logarithm has base a = e. We write y = loge x = ln x which means e = x. Complete the following table on your calculator.

x ln x 1 e e2 10 1000 1 e 1 1000

131 Note that ln(ex) = x, because if ln(ex) = y, then ey = ex, so y = x. Also eln x = x, because if ln x = y, then ey = x, so eln x = x. x This³ means´ that f³(x) =´ e and g(x) = ln x are what are called inverse functions. That is f g(x) = x and g f(x) = x.

We will see when we get to the calculus of y = loga x why ln x is the most important logarithmic function.

Note that after completing the above table for y = ln x, we get the graph.

Note that lim (ln x) = ∞ and lim (ln x) = −∞. x→∞ x→0+ Also remember that the domain of ln x is (0, ∞). Laws of LOGARITHMS

I. loga(xy) = loga x + loga y Ã ! x II. log = log x − log y a y a a

r III. loga(x ) = r loga x

u v Proofs: Let loga x = u and loga y = v. Then a = x and a = y. I. xy = auav = au+v, so

loga(xy) = u + v = loga x + loga y

132 x au II. = = au−v, so y av Ã ! x log = u − v = log x − log y a y a a

III. xr = (au)r = aru, so

r loga(x ) = ru = r loga x ³ ´ 1 Note that loga x = loga 1 − loga x = 0 − loga x = − loga x

The Laws of Logarithms for the Natural Logarithm are

I. ln(xy) = ln x + ln y Ã ! x II. ln = ln x − ln y y

III. ln(xr) = r ln x

Also do not forget that ln(ex) = x and eln x = x.

Example 2 Write ln a + 2 ln b − ln c as a single logarithm.

ab2 ln a + 2 ln b − ln c = ln a + ln(b2) − ln c = ln(ab2) − ln c = ln c r xy Example 3 Expand ln z r µ ¶ 1 xy xy 2 ln = ln z z 1 µxy ¶ = ln 2 z 1 = [ln xy − ln z] 2 1 = [ln x + ln y − ln z] 2 133 Example 4 Solve for x.

e5x+1 − 2 = 0

e5x+1 = 2

ln(e5x+1) = ln 2

5x + 1 = ln 2 ln 2 − 1 x = ≈ −.0613706 5 Example 5 Solve for x in terms of y.

y = ln(x + 10)

ey = x + 10

x = ey − 10 which is the inverse function

CHANGE OF BASE FORMULA

ln x log x = a ln a

To see this let loga x = y Then ay = x ln(ay) = ln x y ln a = ln x

ln x Hence y = and the Change of Base Formula is proved. ln a Example 6 Solve 32x = 5 for x

Soln 1 ln(32x) = ln 5

2x ln 3 = ln 5 ln 5 x = ≈ .7324868 2 ln 3 134 Soln 2 32x = 5

2x log3(3 ) = log3 5

2x = log3 5 log 5 x = 3 2 Using the Change of Base Formula, we see these answers

ln 5 are the same because log 5 = 3 ln 3

135 Exercises on Logarithms

2x3 1. Expand ln (x + 4)6

2. Write as a single logarithm

1 a. 2 ln x − 6 ln(2x − 1)

1 b. ln 3 + 3 ln 27

3. Find

a. log 100,000

b. ln(e15)

c. eln 15

d. log2(5) − log2(90) + 2 log2 3

4. Solve for x

a. 3 ln x = 1

b. e−x = 4

c. ex+1 − 5 = 0

d. ln(5 − 2x) = −3

e. 5x+1 = 10

f. ln(ln x) = 1

g. eax = cebx, where a 6= b

h. ln x2 = ln x + ln 6

i. 24x−1 = 20

136 5. Solve for x in terms of y. This means that we are ﬁnding the inverse function.

a. y = ln(x + 1)

b. y = 310x

c. y = ex2

d. y = (ln x)2, 1 ≤ x

ex + 1 e. y = ex − 1 6. Compute on your calculator

a. log2 6

b. log6 2

7. Let m = 24 · 2−t/25 is the mass of 90Sr that remains from a 24 mg sample after t

years. Find the time for 12 mg to remain.

8. Suppose the graph of y = log2 x is drawn where the unit of measurement is 1 in.

How many miles to the right of the origin do we have to move before the height

of the curve reaches 3 ft.?

137 The Calculus of Exponential Functions and Logarithmic Functions

x x We now find formulas for the derivatives of y = ln x, y = loga x, y = e , and y = a . Each time we get a new formula, we also find a new antiderivative. We will use Defn I to derive the formula for the derivative of f(x) = ln x. All other formulas will be found by taking the logarithm of both sides. This is called logarithmic differentiation.

0 1 A. Let f(x) = ln x. We prove using Defn. III that f (x) = x

f(x + h) − f(x) f 0(x) = lim (Defn III) h→0 h ln(x + h) − ln x = lim h→0 ³ ´h ln x+h = lim x (II Law of Logs) h→0 ³h ´ ln 1 + h = lim x h→0 ³h ´ x ln 1 + h = lim x (Trick) h→0 xhÃ ! 1 x h = lim ln 1 + (x is ﬁxed) x h→0 h x Ã ! x 1 h h = lim ln 1 + (III Law of Logs) x h→0 x Ã ! 1 1 h (h/x) = lim ln 1 + x h→0 x 1 · ¸ = ln lim(1 + u)1/u (ln is continuous and u = h ) x u→0 x 1 1 1 = ln e = (1) = x x x

1 Note that we are using our deﬁnition of e, namely lim(1 + x) x = e. x→0

138 Z 1 B. We immediately get the antiderivative dx = ln x + C x Since the domain of ln x is (0, ∞), x must be positive, Z 1 so we write dx = ln |x| + C Z x Recall xn dx = 1 xn+1 + C, if n 6= −1 n+1 Z Now we can do x−1 dx.

dy µ 1 ¶ 1 C. Let y = log x. We prove by logarithmic diﬀerentiation that = a dx ln a x

y = loga x ay = x ln(ay) = ln x y ln a = ln³ x ´ y = 1 ln x ³ ln a ´ dy 1 1 So dx = ln a x

This shows why natural logarithm is the best to use because ln e = 1.

dy D. Let y = ex. We prove by logarithmic diﬀerentiation that = ex. dx

y = ex ln y = ln ex = x 1 dy y · dx = 1 dy x dx = y = e Thus the slope of the tangent line at any point (x, y) is y. Z Hence we obtain ex dx = ex + C

139 dy E. Let y = ax. We prove by logarithmic diﬀerentiation that = (ln a)ax. dx y = ax

ln y = ln(ax) = x ln a

so ln y = (ln a)x

1 dy y · dx = ln a

dy dx = (ln a)y

dy x dx = (ln a)a This shows why the exponential function to the base e is the best, because ln e = 1.

Example 1 f(x) = ex2 f 0(x) = ex2 · 2x = 2xex2 dy dy du We are using the Chain Rule here letting u = x2 and = · . dx du dx In general, if f(x) = eg(x), then f 0(x) = eg(x) · g0(x). Example 2 y = e2x cos 6x dy = e2x(− sin 6x · 6) + cos 6x · e2x · 2 dx = 2e2x(−3 sin 6x + cos 6x) Example 3 y = 32x2 Soln 1 By formula F and the Chain Rule

dy 2 = (ln 3)32x · 4x dx = (4 ln 3)32x2 · x

140 Soln 2 ln y = ln(32x2 )

ln y = 2x2(ln 3)

ln y = (2 ln 3)x2

1 dy · = (2 ln 3) · 2x y dx

dy = (4 ln 3) · x · y dx

dy 2 = (4 ln 3) · x · 32x dx Example 4 f(x) = ln(x2)

Soln 1 By the Chain Rule

0 1 2 f (x) = x2 · 2x = x

Soln 2 It is easier to use a Law of Logs here

f(x) = 2 ln x

0 1 2 f (x) = 2 · x = x

So always see if you can simplify before you diﬀerentiate.

x2 + 1 Example 5 y = ln (2x − 1)3

y = ln(x2 + 1) − 3 ln(2x − 1)

dy 1 1 = · 2x − 3 · 2 dx x2 + 1 2x − 1 2x 6 = − x2 + 1 2x − 1

141 ln x Example 6 f(x) = x2 x2 1 − ln x · 2x f 0(x) = x x4 x − 2x ln x = x4 x(1 − 2 ln x) = x4 1 − 2 ln x = x3 Example 7 Let y = xx. Note that both the base and the exponent vary, so we cannot use the formulas for diﬀerentiating y = xn or y = ax. We take logs.

y = xx ln y = ln(xx) ln y = x ln x 1 dy 1 y · dx = x · x + ln x dy dx = (1 + ln x)y dy x dx = (1 + ln x)x dy THEOREM: If y = xn, then = nxn−1, for all real numbers n. dx Proof: Recall that we used the Binomial Theorem when n is a positive integer and the Quotient Rule when n is a negative integer. Now let n be any real number.

y = xn ln y = ln(xn) ln y = n ln x 1 dy 1 y · dx = n · x dy y dx = n · x dy xn dx = n · x dy n−1 dx = nx

142 Remarks:

dy 1. In y = xπ, then we get = πxπ−1 dx dy 2. Let y = πx, then we get = (ln π)πx dx

143 1 − x Derivative Exercises 13. f(x) = ln 1 + x Find the derivative and simplify: (2x − 1)3 2 14. f(x) = ln 1. f(x) = xex (3x + 1)2

2. y = e3x3 15. y = ln(ln x) x 3. y = ekx 16. f(x) = , ﬁnd f 0(e) ln x ex 4. y = 17. y = xsin x x + a x 5. f(x) = etan x 18. y = (ln x)

1 6. f(x) = sin(eπx) 19. y = x ln x

aex + b 20. y = ln(sec x + tan x) 7. y = cex + d s x2 + 1 x −x 21. f(x) = ln e + e 2x + 1 8. y = x −x e − e √ 22. y = ln 4 x 9. y = ex ln x √ 23. y = t3 − 3t 10. f(x) = x ln x 24. y = 2x3 11. y = (ln x)2 25. f(x) = log (2x + 1) 12. y = ln(cos x) 3

144 Integrals Involving ex and ln x Z 1 Recall dx = ln |x| + C x Z and ex dx = ex + C

Z 1 Z Example 1 e3x dx = eu du = 1 eu + C = 1 e3x + C 3 3 3 Let u = 3x du = 3 dx 1 3 du = dx Z Z 3 1 3 Example 2 x2ex dx = eu du = 1 eu + C = 1 ex + C 3 3 3 u = x3 du = 3x2 dx 1 2 3 du = x dx Z x 1 Z du Example 3 dx = = 1 ln |u| + C = 1 ln |x2 + 1| + C x2 + 1 2 u 2 2 u = x2 + 1 du = 2x dx 1 2 du = x dx Z Z sin x Z du Example 4 tan x dx = dx = − = − ln |u| + C = − ln | cos x| + C cos x u u = cos x du = − sin x dx −du = sin x dx Z In the exercises you will show that cot x dx = ln | sin x| + C. Z ¯ e ¯e ln x 1 2¯ 1 2 2 1 2 2 1 Example 5 dx = (ln x) ¯ = 2 [(ln e) − (ln 1) ] = 2 [1 − 0 ] = 2 1 x 2 1 u = ln x 1 du = x dx Z ln x Z dx = u du = 1 u2 + C = 1 (ln x)2 + C x 2 2

145 Z 2 4 + x2 Integral Exercises 12. dx 1 x3 Z 4 −2x Ã ! 1. e dx Z 9 √ 1 2 0 13. x + √ dx Z √ 4 x 2. ex 1 + ex dx Z e 2 − x2 Z 14. dx 1 1 6x − x3 3. xe−x2 dx 0 Z e x2 + x + 1 Z 15. dx 4. etan x sec2 x dx 1 x Z cos θ Z ex + 1 16. dθ 5. dx sin θ + 1 ex Z (ln x)3 Z ex 17. dx 6. dx x ex + 1 Z 6 dx Z e1/x 18. 7. dx e x ln x x2 Z √ 19. cot x dx Z e x 8. √ dx x Z µ sec x + tan x¶ 20. sec x dx Hint: Mult. by Z sec x + tan x 9. ex cos(ex) dx Z µ csc x + cot x¶ 21. csc x dx Hint: Mult. by Z 2 dt csc x + cot x 10. 1 8 − 3t Z 2 2 11. dx 1 x

146 Applications x Example 1 Graph f(x) = ex 1. V H x 100 Set ex = 0 lim = 0 ≈ 0 x→∞ ex e100 None y = 0 x −100 lim = −∞ = −100e100 large negativety x→−∞ ex e−100 ex − xex ex(1 − x) 1 − x 2. f 0(x) = = = e2x e2x ex ex(−1) − (1 − x)ex ex(−1 − 1 + x) x − 2 f 00(x) = = = e2x e2x ex 3. C.P. Set 1 − x = 0 ³x =´ 1 1, 1 4. e f 0 x < 1 + inc 1 < x − dec ³ ´ 1 5. 1, e is a local max 6. P.I.P Set x − 2 = 0 ³x =´ 2 2 2, e2 7. f 00 x < 2 − down 2 < x + up ³ ´ 2 8. 2, e2 is an inf. pt

147 Application Exercises

In Exercises 1-3, ﬁnd an equation of the tangent line at the given point. 1. f(x) = e2x cos πx (0, 1) ex 2. f(x) = (1, e) x 3. f(x) = ln x (1, 0)

dy 2 4. Find if ex y = x + y dx dy 5. Find if y = ln(x2 + y2) dx 6. Find a formula for f (n)(x) if f(x) = ln(x − 1).

7. If y = Ae−x + Bxe−x show that y00 + 2y0 + y = 0.

8. Find the values of λ for which y = eλx satiﬁes the equation y00 = y0 + y. 1 9. Let p(t) = 1 + ae−kt a. Find lim p(t) t→∞ b. Find p0(t)

In the remaining exercises, do a complete 9 step analysis and graph. (Don’t forget to ﬁnd lim f(x) when ﬁnding Horizontal Asymptotes.) x→−∞ ex 10. f(x) = x 11. f(x) = xex

12. f(x) = x2ex

13. f(x) = xex2

14. f(x) = e−x2

148 Section 5.6 Inverse Trigonometric Functions Consider the graph of f(x) = sin x.

A horizontal line throughh a, −i1 ≤ a ≤ 1, will meet the graph in inﬁnitely many points. π π If we restrict the graph to − 2 , 2 this diﬃculty is eliminated.

This leads to the deﬁnition of y = arcsin x. arcsin x = y means 1. sin y = x

π π 2. − 2 ≤ y ≤ 2 and y is in radians.

149 −1 Another notation ish arcsini x = sin x. Thus arcsin x is the inverse function of the sine π π function restricted to − 2 , 2 .

Example 1 arcsin 1 = y arcsin 1 = y arcsin 0 = y arcsin(− √1 ) = y 2 2 sin y = 1 sin y = 1 sin y = 0 sin y = − √1 2 2 π π π y = 6 y = 2 y = 0 y = − 4 ³ ´ 1 Example 2 cos arcsin 4 1 Let arcsin 4 = y 1 sin y = 4

³ ´ √ 1 15 So cos arcsin 4 = cos y = 4

Example 3

a. sin(arcsin x) = sin y = x Let arcsin x = y sin y = x

³ ´ ³ √ ´ b. arcsin sin 2π = arcsin 3 = y 3 √ 2 3 sin y = 2 π y ³= 3 ´ 2π π So arcsin sin 3 = 3 We now let y = arcsin x and derive the formula for its derivative.

y = arcsin x

sin y = x dy cos y · = 1 dx dy 1 √ = So cos y = 1 − x2 dx cos y dy 1 = √ dx 1 − x2

150 The importance of this derivative is that we get a new antiderivative formula, namely Z 1 √ dx = arcsin x + C 1 − x2 Z 1 It is useful to obtain a formula for √ dx, 0 < a. a2 − x2

Z 1 Z 1 √ dx = r dx 2 2 ³ ´ a − x 2 x2 a 1 − a2 Z u = x 1 1 a = r dx du = 1 dx a ³ ´2 a x a du = dx 1 − a 1 Z 1 = · a √ du a 1 − u2 = arcsin u + C x = arcsin a + C

Memorize these two formulas: Z 1 √ dx = arcsin x + C 1 − x2 Z 1 √ dx = arcsin x + C a2 − x2 a

151 Now consider the graph of f(x) = tan x

Once again a horizontal³ line´ will meet the graph inﬁnitely many times. If we restrict the π π graph to the interval − 2 , 2 we eliminate the diﬃculty.

This leads to the deﬁnition of y = arctan x y = arctan x means 1. tan y = x π π 2. − 2 < y < 2 , y in radians.

152 Another notation is y = arctan x = tan−1 x and is the inverse function of tan y where π π − 2 < y < 2 . √ Example 4 arctan 1 = y arctan 0 = y arctan(− 3)√ = y tan y = 1 tan y = 0 tan y = − 3 π π y = 4 y = 0 y = − 3

Example 5 sin(arctan 2) = sin y = √2 5 arctan 2 = y tan y = 2

Example 6

a. tan(arctan x) = tan y = x arctan x = y tan y = x ³ ³ ´´ 3π b. arctan tan 4 = arctan(−1) = y tan y = −1

π y = − 4 ³ ³ ´´ 3π π so arctan tan 4 = − 4 We now let y = arctan x and derive the formula for its derivative.

y = arctan x

tan y = x dy sec2 y · = 1 dx dy 1 √ = sec y = 1 + x2 dx sec2 y dy 1 = sec2 y = 1 + x2 dx 1 + x2

153 Once again the importance of this formula is that we get a new indeﬁnite integral, Z 1 dx = arctan x + C 1 + x2 Z 1 It is useful to obtain a formula for dx , 0 < a. a2 + x2 Z 1 Z 1 dx = ³ ´ dx a2 + x2 2 x2 a 1 + a2 x Z u = a 1 1 1 = ³ ´2 dx du = dx a2 x a 1 + a a du = dx 1 Z 1 = a du a2 1 + u2 1 = arctan u + C a 1 x = arctan + C a a Memorize these two formulas: Z 1 dx = arctan x + C 1 + x2 Z 1 1 x dx = arctan + C a2 + x2 a a These are the only two inverse trig. functions that we are going to study. For example dy −1 if y = arccos x, then it can be shown that = √ . Since the main importance is for dx 1 − x2 integration, this formula is not needed. We will derive the formula for the derivative of arcsecx in the exercises. Example 7 y = arctan(x2 + 1) dy 1 = · 2x dx 1 + (x2 + 1)2 Example 8 y = arcsin(5x) dy 1 5 = q · 5 = √ dx 1 − (5x)2 1 − 25xx

154 Example 9

u = 2x

Z 1 Z 1 du = 2 dx 4 1 1 2 1 1 √ dx = √ du 2 du = dx 0 2 2 0 2 1 − 4x 1 − u x = 0 to x = 1/4 u = 0 to u = 1/2 ¯ 1 ¯ 2 1 ¯ = arcsin u¯ 2 ¯ ³ 0 ´ 1 1 = 2 arcsin 2 − arcsin 0 ³ ´ 1 π π = 2 6 − 0 = 12 Z 1 1 x Example 10 dx = arctan + C x2 + 25 5 5 a2 = 25 a = 5

Example 11

u = x2 Z x 1 Z du du = 2x dx 4 dx = 2 x + 9 2 u + 9 1 2 du = x dx 1 1 u = · arctan + C a2 = 9 a = 3 2 3 3 1 x2 = arctan + C 6 3 Example 12

u = x4 + 9 Z x3 1 Z du du = 4x3 dx 4 dx = x + 9 4 u 1 3 4 du = x dx 1 = 4 ln |u| + C 1 4 = 4 ln |x + 9| + C

155 Exercises on Inverse Trigonometric Functions

1. Find the exact value

³ √ ´ 3 a. arcsin 2 f. tan(arctan 5) ³ ´ 5π b. arctan(−1) g. arctan tan 6 √ ³ ´ 1 c. arctan 3 h. cos arcsin 2 ³ ´ ³ ´ d. arcsin − √1 i. tan arcsin 1 2 5 √ e. arcsin(−1) j. sin(2 arctan 2)

2. Simplify

a. tan(arcsin x)

b. sin(arctan x)

c. csc(arctan 2x)

3. Find the derivative and simplify. √ a. y = arctan x √ b. y = arctan x √ c. f(x) = 1 − x2 arcsin x

d. y = arcsin(x3)

e. f(x) = arctan(cos x) q x x−a f. y = arctan a + ln x+a

4. A ladder 10 feet long leans against a vertical wall. If the bottom slides away at the rate of 2 ft/sec, how fast is the angle between the ladder and the wall changing when the bottom is 6 ft. from the wall?

156 √ Z Z 3 1 2 2 1 5. √ dt 6. 2 dt 1 2 0 t + 1 2 1 − t √ Z Z 3 1 2t 4 16 7. dt 8. 2 dx 0 t2 + 1 0 1 + 16x Z Z dt dt 9. √ 10. √ 1 − 4t2 4 − t2

Z π Z 1 2 sin x 2 arcsin x 11. √ dx 12. dx 0 1 + cos2 x 0 1 − x2 Z Z π x + 9 2 sin x 14. dx 13. dx x2 + 9 0 1 + cos x Z 5 Z arctan x t 15. dx 16. √ dt x2 + 1 1 − t6 Z Z t2 e2x 17. √ dt 18. √ dx 1 − t6 1 − e4x Z Z ex e2x 19. dx 20. dx e2x + 1 e2x + 1 21. Deﬁne y = arcsec x by

1. sec y = x and

π 2. 0 ≤ y ≤ π, y 6= 2 , y in radians.

dy Derive the formula for dx .

157 Section 5.8 Indeterminate Forms and L’Hospital’s Rule

x2 − 4 → 0 Recall lim is of indeterminate form . x→2 x − 2 → 0 x2 − 4 (x − 2)(x + 2) We can use algebra to obtain lim = lim = lim(x + 2) = 4 x→2 x − 2 x→2 (x − 2) x→2 sin x Likewise we were able to show that lim = 1 by using geometry. x→0 x ln x → 0 If we try to compute lim , we see that it is also of the type , but we have no x→1 x − 1 → 0 method to compute the limit. L’Hospital’s Rule will show us how to ﬁnd such limits of the f(x) form lim when both f(x) and g(x) are diﬀerentiable. x→a g(x) The limit can be two-sided, one-sided, or go to ±∞.

f(x) L’Hospital’s Rule: Suppose we are computing lim where f(x) and g(x) are diﬀerentiable. x→a g(x) f(x) f 0(x) 1. If lim f(x) = 0 and lim g(x) = 0, then lim = lim . x→a x→a x→a g(x) x→a g0(x) f(x) f 0(x) 2. If lim f(x) = ±∞ and lim g(x) = ±∞, then lim = lim . x→a x→a x→a g(x) x→a g0(x) → 0 ln x 1 Example 1 lim = lim x = 1 → 0 x→1 x − 1 x→1 1 → 0 sin x cos x Example 2 lim = lim = 1 → 0 x→0 x x→0 1

sin x Remark: This argument is really circular. Since we needed to know that lim = 1 x→0 x before we could prove that the derivative of sin x is cos x.

158 → ∞ ex ex → ∞ ex Example 3 lim = lim = lim = ∞ → ∞ x→∞ x2 x→∞ 2x → ∞ x→∞ 2

→ ∞ ln x ln x Example 4 lim √ = lim → ∞ x→∞ 3 x x→∞ x1/3 1 = lim x x→∞ 1 −2/3 3 x 1 = lim · 3x2/3 x→∞ x 3 = lim x→∞ x1/3 = 0

→ 0 tan x − x sec2 x − 1 → 0 Example 5 lim = lim → 0 x→0 x3 x→0 3x2 → 0 2 sec x(sec x tan x) = lim x→0 6x sec2 x tan x → 0 = lim x→0 3x → 0 sec2 x sec2 x + tan x2 sec x sec x tan x = lim x→0 3 sec4 x + 2 sec2 x tan2 x = lim x→0 3 1 + 0 = = 1 3 3 sin x cos x → −1 Example 6 lim = lim = −∞ x→π− 1 − cos x x→π− sin x → 0 This is wrong. Can you ﬁnd the mistake? Answer: We forgot to check that the hypothesis in L’Hospital’s Rule holds. → 0 sin x Correct Solution: lim = 0 → 2 x→π− 1 − cos x

159 Other Indeterminate Forms are products of the form (→ 0)(→ ∞), diﬀerences of the form (→ ∞) − (→ ∞), and powers of the form (→ 0)(→0),(→ ∞)(→0), or (→ 1)(→∞).

Example 7 lim (sec x − tan x)(→ ∞) − (→ ∞) π − x→ 2

µ 1 sin x ¶ = lim − π − x→ 2 cos x cos x

1 − sin x → 0 = lim π − x→ 2 cos x → 0 − cos x = lim π − x→ 2 − sin x cos x → 0 = lim π − x→ 2 sin x → 1

= 0

Example 8 lim x ln x (→ 0)(→ −∞) x→0+

ln x → −∞ = lim 1 x→0+ x → ∞

1 x = lim 1 x→0+ − x2 1 = lim (−x2) = lim (−x) = 0 x→0+ x x→0+

160 Example 9 Recall that we defined lim(1 + x)1/x = e. This is of the type (→ 1)(→∞) if x→0 we come in from the right. Assuming that we know how to differentiate ln x, let’s show this limit must be e. Let y = (1 + x)1/x. Recall eln y = y. Thus if we can find the limit of ln y we can find the limit of y.

lim ln y = lim ln(1 + x)1/x x→0 x→0 1 = lim ln(1 + x) x→0 x ln(1 + x) → 0 = lim x→0 x → 0 1 = lim 1+x x→0 1 = 1

Thus, lim y = e1 or lim(1 + x)1/x = e. x→0 x→0

161 ex − 1 − x Exercises on L’Hospital’s Rule 15. lim x→0 x2 x2 − 9 ex − 1 − x − 1 x2 1. lim 16. lim 2 x→3 x − 3 x→0 x3 ex x2 − 9 17. lim 2. lim x→∞ x3 x→−3 x + 3 arcsin x 19 18. lim x − 1 x→0 x 3. lim x→1 x6 − 1 sin x − x 19. lim 3 1 − sin x x→0 x 4. lim π + cos x − 1 x→ 2 cos x 20. lim x→0 x2 x + tan x 1 − cos x 5. lim 21. lim x→0 sin x x→0 x (ln x)2 et − 1 22. lim 6. lim x→∞ x t→0 t3 x + sin x tan ax 23. lim 7. lim x→0 x + cos x x→0 tan bx x 24. lim 1 − sin θ x→∞ ln(1 + 2ex) 8. lim θ→ π k 2 csc θ x − kx + k − 1 25. lim x→1 (x − 1)2 ln x 9. lim 1 − e−2x x→∞ x 26. lim x→0 sec x ln x 10. lim 27. lim x2ex x→0+ x x→−∞ √ 28. lim cot 2x sin 6x 11. lim x ln x x→0 x→0+ 29. lim x3e−x2 12. lim x3 ln(x2) x→∞ x→0+ 30. lim xx x→0+ x e ³ ´x 13. lim 31. lim 1 + 1 x→∞ x x→∞ x ³ ´x t t k 4 − 2 32. lim 1 + x 14. lim x→∞ t→0 t

162 33. Extra Credit

L’Hospital wrote the ﬁrst calculus book in 1696. This is his ﬁrst example of his rule √ √ 2a3x − x4 − a 3 a2x (which was really proved by Bernolli). Find lim √ , 0 < a. x→a a − 4 ax3

Remarks: Now that we know how to diﬀerentiate y = ln x and we have the Chain Rule, then we are able to derive the Product Rule and the Quotient Rule. Product Rule: Let u = f(x) and v = g(x) be diﬀerentiable. If y = f(x)g(x) = uv, then dy dv du dx = u dx + v dx . Proof : Let y = uv ln y = ln(uv) = ln u + ln v 1 dy 1 du 1 dv Then y · dx = u · dx + v · dx h i dy 1 du 1 dv Thus dx = u · dx + v · dx · y h i 1 du 1 dv = u · dx + v · dx · uv du dv = v · dx + u · dx dv du = u · dx + v · dx and the Product Rule is proved.

f(x) u Quotient Rule: Let u = f(x) and v = g(x) be diﬀerentiable. If y = = , then g(x) v dy v du − u dv = dx dx . dx v2 Proof: Exercise

163