1549901369-Elements_of_Real_Analysis__Denlinger_

5.6 *Exponentials, Powers, and Logarithms 287

function f ( x) = ( 1 + ~ r. To make this transition we will use the "greatest

integer function,'' lx J =the greatest integer:::; x, introduced in Example 5.2.16.

Proof of Theorem 5.6.17:

( )

LxJ

(1) Claim: }2...1! 1 + L!J = e.

Proof: Let c: > 0. Since n-+(X) lim (1+.!.)n n = e, :Jn 0 EN 3 n 2: n 0 =?

I (1 + ~ r -e I < €. Then x 2: no =} l x J 2: no =} I ( 1 + L! j ) L x j - e I < €.

( )

LxJ

(2) Claim: }2...1! 1 + LxI+i = e.

Proof: ( 1 + LxI+1) LxJ = ( 1 + LxJl+l) L xJ+l ( 1 + LxI+ 1 )-1

(

1 ) LxJ+i (EW.)

1 + LxJ+l LxJ+2

Thus,

. (^1 )LxJ. (^1 )LxJ+l. (EW.)
}~1! l + LxJ+l = }~1! l + LxJ+l · }~1! LxJ+2 = e · l = e.

(3) Finally, noting that l x J + 1 > x, we have, for all x > 0,

(

1 ) LxJ ( 1) LxJ ( 1 )x ( 1 ) x ( 1 ) LxJ+l

1 + LxJ+l < 1 + x :::; 1 + x :::; 1 + LxJ < 1 + LxJ.

Applying (1), (2), and the squeeze principle to these inequalities, we h ave
lim (1 + .!.)x = e. •
x-+oo x

Corollary 5.6.18 (Function Limit fore) lim (1 + x)^1 fx = e.

x--+O

Proof. E xercise 11. •

Lemma 5.6.19 Let y "I- 0 be a fixed real number. Then lim (1 + xy)^1 fxy = e.
x--+O

Proof. Let c: > 0. By Corollary 5.6.18, :J 8 > 0 3 0 < lxl < 8 =?
1(1 +x)^1 fx -el < c:. ThenO < lx l < 8/IYI =? 0 <lxyl<8=?1(1 +xy)^1 fxy -el<
€. •

Theorem 5.6.20 (Function Limit fore"') 't/x E JR, lim(l +tx )^1 ft = ex.
t--+O

Proof. If x = 0 this equation is obviously true. Thus, suppose x is a (fixed)
nonzero real number. Notice that (l+tx)^1 / t = f(g(t)), where g(t) = (l+tx)^1 f tx
and f(u) = ux. The power function f is continuous on (0,+oo), by Theorems