Calculus: Analytic Geometry and Calculus, with Vectors

9.3 Hyperbolic functions 505

by putting s, = f(n) in Theorem 5.651, this and (5) imply existence of a number
L for which

(6) lim f (n) = L < e.
n-.

To prove that L = e, we can use (3) and (5) to obtain

1-f I 1 (1 -n/+3!\1 n/\1 2)-I-.

N (1 -n) n) ... (1 - n f(n) < e

when n > N. Letting n --+ oo gives

(7) + 1 -1- 2 i + -1- .... r L < e,

and letting N--> w shows that L = e. Supposing now that x > 2 and that x

is not necessarily an integer, we can let n be the greatest integer in x so that

n S x < n + 1. Then it is easily seen that

(8) (1 +n-1 1) < f(x) < (1 +

n)R+i

The first and last members of this inequality converge to e as n -> because

they are respectively equal to

(9)

(1+n+1)n+i(1-i

n+1)_', 1+n(1 +n1)

and the second factors converge to 1 as n -> oo. Therefore, f(x) -> e as x -> oo.

Proof of the part of (1) involving negative values ofxis provided by the calculation

( 10 ) lim 1 + I) 2 = li m(1-^1 )-x= lim( x

(

lim (1 1+- )z= lim(1 +^1 )x-1(1 -f-^1 )= rl = e.

X_. ` x-1 X_. x-1 x-1

9.3 Hyperbolic functions We begin with a peek at some formulas

that are very useful in more advanced mathematics and will appear again

in Section 12.4. The formula

(9.31) eZ = 1 + z -I- 2 + 3+ 4-I- ... ,

which has been proved to be valid when z is a real number, is used to

define e$ when z is a complex number of the form z = x + iy, where x

and y are real and i is the so-called imaginary unit for which i2 = -1.

Since i2 = -1, is = -%, i4 = 1, i5 = i, ,it can be shown that

(9.311) el-Zi-F4h-6i+...1

+ i(z-zs zs '7 ..l

31 T-1 71!