Hw # 11 Solutions - OU Math

Hw # 11 Solutions

6:8#12 We apply L'Hospital Rule to obtain

lim

x !0

sin 4x tan 5x

L'H=os:

lim

x !0

4 cos 4x 5 sec2(5x)

=

4(1) 5(1)2

=

4 5

6:8# 22 Using L'Hospital Rule, we obtain

8t lim

5t L'H=os: lim 8t ln 8

5t ln 5 = ln 8

8 ln 5 = ln

t !0 t

t !0

1

5

6:8# 56 Let y = (tan 2x)x: Taking ln on both sides, we obtain ln y = x ln(tan 2x): Now taking limit both sides, we obtain

lim ln y = lim x ln(tan 2x)

x !0+

x !0+

ln(tan 2x) = lim

x !0+ 1=x

Limit on the right side is 1=1 form, applying L'Hospital Rule, we obtain

ln(tan 2x)

lim ln y = lim

x !0+

x !0+ 1=x

L'H=os: lim (1= tan 2x)(2 sec2 2x)

x !0+

1=x2

= lim x !0+

x2 :2 sec2 2x = lim

tan 2x

x !0+

x2

2

(sin 2x= cos 2x) : cos2 2x

x2 cos 2x 2

x2

2

= lim

:

= lim

:

x !0+ sin 2x cos2 2x x !0+ sin 2x cos 2x

2x x

= lim

:

x !0+ sin 2x cos 2x

2x

x

= lim

: lim

(by limit laws)

x !0+ sin 2x x !0+ cos 2x

But lim 2x= sin 2x = 1 and lim x= cos 2x = 0; therefore, limit on the right side is 1:0 = 0:

x !0

x !0

Therefore, we obtain

lim ln y = 0:

x !0+

But lim ln y = ln( lim y); and hence

x !0+

x !0+

ln( lim y) = 0 =) lim y = e0 = 1:

x !0+

x !0+

1

7:1# 14 We use integration by parts. Let u = s and dv = 2sds: Then du = ds and v = 1 22: Then we obtain

ln 2

Z s2s ds = 1 s2s ln 2 = 1 s2s ln 2

Z 1 2s ds lnZ2

1 2s ds ln 2

= 1 s2s ln 2

= 1 s2s ln 2

1 1 2s + C ln 2 ln 2

1 (ln 2)2

2s

+

C

Z 7:1#18 Let I = e cos 2 d : Put u = e and dv = cos 2 d : Then du = e d and

v = 1 sin 2 : Using integration by parts, we obtain 2

1

Z1

I = e sin 2

sin 2 ( e d )

2 1

1 Z2

= e sin 2 + e sin 2 d

2

2

(F)

Z

To ...nd e sin 2 d ; we need to use integration by parts again. Use integration by parts

by putting u = e

and dv = sin 2 d ; then you will get

Z e sin 2 d =

1 e cos 2

2

1Z e

2

1

1

= e cos 2 I

2

2

cos 2 d

Use this result in (F); we obtain

1

11

1

I = e sin 2 +

e cos 2 I

2

22

2

= 1e sin 2 1e cos 2 1I;

2

4

4

11

1

which implies I + I = e sin 2 e cos 2 ; and hence

42

4

5

1

1

4I = 2 e sin 2 4e cos 2

=)

2 I= e

5

sin 2

1 e cos 2 + C

5

2

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