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Then if N is the larger of N1 and N2 , we can write (1) as 1 2 1 1 jbn Bj B for all n > N < 2 1 b B jBjjbn j jBj 2 jBj n and the proof is complete. (b) From part (a) and Problem 2.12, we have an 1 1 1 A lim an A lim an lim n!1 bn n!1 n!1 n!1 bn bn B B lim This can also be proved directly (see Problem 41).

The composition of two or more functions can be accomplished with the Composition command. Note that Composition is a functional operation and as such, its arguments are functions, f, not f[x].

Netbios aliases = <Alias1 ... AliasN> Other NetBIOS Aliases that may also be used to identify the server. Interfaces = <IP-addr1 ... IP-addrN> The set of hardware interfaces used for browser communications (multi-homed).

f[x_] =

a 3n2 5n 3 5=n 3 0 3 lim n!1 5n2 2n 6 n!1 5 2=n 6=n2 5 0 0 5 lim ( ) ( ) ( ) n n 2 n3 n3 n2 2n 1 1=n 2=n2 2 lim lim n!1 n!1 n 1 n2 1 n!1 1 1=n 1 1=n2 n 1 n 1 lim 1 0 0 1 1 0 1 0

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g[x_] = x2 + 2 x + 3; h1 = Composition[f, g]; h1[x] 3 + 2 x + x2 h2 = Composition[g, f]; h2[x] 3+ 2 x + x

If we wish to compute the composition of a function with itself we could, of course, use Composition[f, f], Composition[f, f, f], and so forth. A more convenient tool is Nest or NestList.

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Nest[f, expression, n] applies f to expression successively n times. NestList[f, expression, n] applies f to expression successively n times and returns a list of all the intermediate calculations from 0 to n. (Lists are discussed in detail in 3.)

p p p p p p n 1 n 1 lim n 1 n lim n 1 n p p lim p p 0 n!1 n!1 n 1 n n!1 n 1 n

f[x_]= x2; Nest[f, x, 5] x32 NestList[f, x, 5] {x, x2, x4, x8, x16, x32} Nest[f, 2 x + 3, 5] (3 + 2 x)32 NestList[f, 2 x + 3, 5] {3 + 2 x, (3 + 2 x)2, (3 + 2 x)4, (3 + 2 x)8, (3 + 2 x)16, (3 + 2 x)32}

Nest[Framed,x,10]

NestList[Framed,x,10]

Since the limits of the numerator and denominator are 3 and 0, respectively, the limit does not exist. 3n2 4n 3n2 3n > can be made larger than any positive number M by choosing n > N, we Since 2n 1 2 2n 2 3n 4n can write, if desired, lim 1. n!1 2n 1 e

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2.72 If f(x) = sin x + 2 cos x and g(x) = 2 sin x 3 cos x, construct ( f + g) (x), ( f g) (x), ( f g) (x), and ( f /g) (x) and evaluate them at /2.

f[x_]= Sin[x]+ 2 Cos[x]; g[x_]= 2 Sin[x] 3 Cos[x]; h1[x_]= f[x]+ g[x] Cos[x]+ 3 Sin[x] h2[x_]= f[x] g[x] 5 Cos[x] Sin[x] h3[x_]= f[x] g[x] (2 Cos[x]+ Sin[x])( 3 Cos[x]+ 2 Sin[x]) h4[x_]= f[x]/g[x] 2 Cos[x] + Sin[x] 3 Cos[x] + 2 Sin[x] h1[ /2] 3 h2[ /2] 1 h3[ /2] 2 h4[ /2] 1 2

2.73 Let f ( x ) = 1 + x . Compute ( f o f o f o f o f )(x).

4 2n 3 4 2 3=n 4 2 16 lim n!1 3 7=n 2n 7 3 81 2n5 4n2 2=n2 4=n5 0 0 lim 3 10 n!1 3 1=n4 10=n7 3 n (Compare with Problem 2.5.)

2.74 Let f ( x ) = 1 . Let f n ( x ) = (14 o ...o f )( x ) . Evaluate f ( x ), f 2 ( x ), f 3 ( x ), f 4 ( x ), and f 5 ( x ) . Then evaluate fof 24 3 1+ x

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