Limits (with detailed solutions)

Size: px

Start display at page:

Download "Limits (with detailed solutions)"

Vanessa Philippa Bridges
6 years ago
Views:

1 Limits with detailed solutions. Verif using the definition a 3 5 = b = + c n + 3n n + = + d 3 + = 3.. Compute using the algebra of its a 3 + b c e 0 + π ± cos d f ± Compute the following its with indeterminate forms a 3 b ± c 0 ± d 5/ + 5/ e f Compute the following its rationalize the fractions a + b c + 0 d Compute the following its changing variable e 3 + a ± e + 3 b 4 + log c log. 6. Compute ± [ 3 ] studing the sign of the function [t] is the floor function.

2 7. Compute the following its using comparison theorems a + cos b cos c e g cos [] 5 + e esin d f h sin 3 + cos log3 + sin e sin i sin sin 0 l 8 + sin. 8. Compute the following its a c cos 0 sin tan 0 b d log tan sin e 0 f 3/ g i log h l Compute the following its a c e g i sin + π tan π + 0 tan tansin log cos 0 sin log + log + b d f h l π cos sinπ e + e 0 log + cos 0 cos + π cos3 +.

3 0. Compute the following its indeterminate forms of eponential tpe a c + + b d e sin 0 + f 0 +log sin g log sin h + e / 3 sin Compute the following its a sinlog 0 + b 0 + log c log 3 d log log log e f

4 Solutions. We use the definition of its. a Fi ε > 0 and consider the difference f l = 3 5 = 3 6 = 3. To obtain f l < ε we take such that 3 < ε that is < δ = ε 3. b Fi R > 0 and consider f = > R < < δ =. R R c Fi R > 0 and consider the inequalit 3n /n + > R. Since if n we have n + n + n = n we obtain 3n n + 3n n = 3 n. If 3 n > R that is n > K = 3 R we have 3n /n + > R. d Fi ε > 0 and consider f l = = 3. In order to obtain f l < ε we take such that 7 3 < ε that is > 7 3 ε with < 0 hence < 7 3ε = R.. In this eercise we use the Theorems on sums product and quotient of its. a Since both and 3 tend to as we have 3 + = 3 + =. b Since both and tends to + as + we have = = +. c-d Since in both cases the denominator does not tend to 0 we have 0 + = + 0 = and ± + = + ± = 0. e The denominator tends to 0; so the eistence of the it is not guaranteed. In this case since cos > 0 when π/ we have π cos = + ; while since cos < 0 when π/ + π + cos =. f 5 tends to + when + while tends to 0; hence the sum tends to In this eercise we deal with the following indeterminate forms: a-b Indeterminate form 0 0. We simplif the common factors in numerator and denominator 3 = = = 3 ; 4

5 5 + 6 ± = 3 3 ± = ± = where in the last it we mean that if + then the it is while if then the it is +. c Indeterminate form 0 0. We again simplif where 0 ± = 3 0 ± 3 + = 0 ± 3 0 ± = ± and 0 ± 0 ± 3 + =. 0 ± =. d-e-f Indeterminate form. We pick up the greatest power of both at the numerator and at the denominator. We have 5/ + 5/ = = = 5/ 5/ = = = 4. a-c-d We use a b = a ba b: = = = + + = = + = = = = b We use a 3 b 3 = a ba + ab + b. + + = = = + + = = = = = 0. =. 5. a Let = f = e ; since = e + when + and the function g = 3 + / + admits a it as + we have e 3 + e + = = +. 5

6 If then = e 0; g is continuous in 0 and tends to g0 = / as 0; hence e 3 + e + = =. b Let = ; we have = + as and = + 3 = c Let = f = log ; we have = log as 0 + ; hence log + log = = + since if t < 0 t = t = t = 3. + = 6. We stud the sign of f = 3 = + near =. f < 0 when < and f > 0 when >. Furthermore f is continuous in. { if δ < < [f] = 0 if < < + δ where δ > 0 is sufficientl small. [3 ] = +[3 ] = 0 7. a Since cos for ever R using the comparison Theorem we have = + + cos = +. b Since cos for ever R we have we obtain cos 0 cos. Since the right and the left its are 0 we obtain that also cos/ tends to 0 as +. c The it does not eists; indeed if we call f = cos n = π + nπ and n = nπ we have f n = 0 0 f n = nπ +. d Since sin cos we have sin cos + 3 for ever > /3. Since 3 + = 3 = + 3 from the comparison theorem we obtain sin 3 + cos = 3. 6

7 e We recall that < [] for ever R. From the comparison Theorem it follows that < [] 0. [] =. f Since sin R we have Since ft = log t is an increasing function we have log sin 4 R. log log3 + sin log 4 R. log3 + sin 3 log 4 > 0 3 log3 + sin 3 = 0. g The it does not eists. Indeed let f = 5 + e esin n = π + nπ and n = nπ; we have f n 0 while n + f n = = 5 e = h Since e sin e R we have 5 + e sin 5 e > e = e sin = +. i Since sin/ 0 we have sinsin sin 0 l Since sin 9 we have sin = 0 sin sin 0 0 = sin 8 < sin =. cos cos 8. a = 0 0 / = 0 = 0. log + 4 log + 4 b = 4 = sin tan sin c 0 = 0 cos d We have use the result in c: 4 log + = 4 = 0 sin cos cos = 0. 0 tan sin tan = sin 0 sintan = sin tan 0 sin cos = 0. 7

8 e log e e = log = log = log. 0 0 log 3 f 3/ = = log 3. g We define = to obtain log h We define =. We have = log + = / log + =. + + / = = = =. i We have replacing / = = = l = a We replace / = and we have b We replace = and we have cos π = 5 + = 4/5 = 0. 3 = 3 = log sin = sin =. cos [ π + ] = + c We replace + π/ = and we have [ sin π ] sin π = + + = π = π. + π tan = π + tan + π = + + tan = +. d We replace = and we obtain tan tansin = 0 e 0 Replacing sin = the second it is sin π = sinπ + π sin π = = π. tan sin sin tansin = 0 0 sin cos 0 sin tansin = sin tansin. tan =. And we conclude that 0 tan tansin = 0. 8

9 e + e e e + e + + f = = e In order to compute the it of the first factor we replace + = ; we obtain While the second one is e + + e 0 + log cos log + cos g 0 sin = 0 cos e + e = = = = = e. e + e = e 0. cos sin. The it of the first factor is indeed replacing = cos we obtain Concerning the second factor we have the required it is /. log + cos h 0 The it of the first term is log + cos log + = =. 0 cos cos cos 0 sin = 0 sin =. log cos cos + = 0 = 0 log cos log cos log + cos = 0 0 cos whose first factor has it replace = cos we have log + cos +. cos log + cos log + = = ; 0 cos while the second factor has it 0. the it of the first term is 0. The second term is log + cos log + cos = 0 0 / cos cos. The first factor has it replace = / cos. The second factor tends to hence the it of the second term is. We conclude that the it is the sum 0 + =. i We compute recalling that logab = log a + log b if a b > 0 log + log + = log + log + = = log + log + log + = log + log +. 9

10 We now replace / = and we obtain log + log + log + = l We replace π = and we have cos + π + cos3 + 3π + = cos cos3 = cos log + = =. cos3. The first factor has it. In order to compute the it of the second factor we replace 3 = z; we obtain cos3 = z 0 The it is then the product 9 = a Replacing = + we have b We compute Since z /9 cosz = 9 z 0 z cosz = 9. + [ = + ] = = e e + [ = + ] + = e there eists R > 0 such that + [ > when > R. + ] > when > R. We now use the comparison Theorem to get + = +. c + [ = + ]. Since + = e we have that for ever ε > 0 in particular for 0 < ε < e there eists R > 0 such that e ε < + < e + ε for ever < R. e ε < [ + ] < e + ε when < R. We now use the comparison Theorem to obtain + = 0. 0

11 d = e log = since log = sin e = e sin log = e sin log = since f sin = 0 +log 0 The it of the eponent is sin sinlog = log = 0 = loglog esin. + sin loglog sin = loglog log log sin = 0 + loglog 0 + log log. 0 + The first it is the third is 0. For the second one we replace = log to obtain loglog log 0 + log = = 0. + sin loglog = 0 log sin sin loglog = e = g We proceed as in f: log sin log+log = esin = + h + e 3 sin / sin log+e = e We have log + e /3 = log e /3 + e /3 = 3 log e + log + e /3 then the it of the eponent is sin log + e / 3 sin = sin log + e /3. The its of the two terms of the sum are 0 sin sin = 0 + = + 3 sin log + e / = sin log + e/ = + and e 3 sin = +. sin. a sin log = sin log = log = 0 = b 0 + log = 0 + log = 0. log 3 c d = log log log = + log 3 = 0. /3 = + where log =. log e = e log e log = elog log = +. f 3 3 = 3 3 = +.

In this note we will evaluate the limits of some indeterminate forms using L Hôpital s Rule. Indeterminate Forms and 0 0. f(x)

L Hôpital s Rule In this note we will evaluate the its of some indeterminate forms using L Hôpital s Rule. Indeterminate Forms and 0 0 f() Suppose a f() = 0 and a g() = 0. Then a g() the indeterminate