Math 100 section 104 ( LAM ) Oct 18, 2004
Additional Course Material
Two important (and famous) limits established by the use of logarithm.
( I )
Lim n → ∞
( 1 + 1 n
) n = e
Proof Let A n
=
(
1 + 1 n
) n . The strategy is to investigate first the limit of ℓ n A n as n → ∞ . By the laws of logarithm
Lim n → ∞ ℓ n A n
=
= where we have set
Lim n
1 n
→ ∞
n ℓ n (1+ h )
1 n
) = Lim n → ∞ l n ( 1
1 n
+ 1 n
)
Lim h → ∞ l n ( 1 + − l n ( 1 )
= ℓ n' (1 ) = 1.
h
= h and used ℓ n (1) = 0.
The limit, by the very definition of derivative, is ℓ n' (1), value of the derivative of the ℓ n (x ) function at x = 1. Since ℓ n' (x) = ( ℓ n x)' =
1
, that value is indeed
We now return to A n x
by exponentiating ℓ n A
Lim n → ∞
A n
That the last limit equals e 1 function at t = 1.
= Lim n → ∞ e ℓ n An = e
is because as n
1 n
= e.
→
:
∞ , ℓ n A n
→ 1 and e t
1
1
= 1 .
is a continuous
NOTE
The above proof is an excellent example showing the interplay between
limit, continuity, differentiability and inverse functions.
Suggested numerical experimentation
: Get a close approximation to e by putting
n = 2,000,000 into ( 1 + 1 n
) n , with the use of a pocket calculator.
( II )
. t
Lim
→ ∞ e t t m
= ∞ for any fixed positive integer m.
Proof Note ℓ n ( t e t m
) = ℓ n (e t ) - ℓ n ( t m ) = t - m ( ℓ n t).
As in the discussion of the limit in part ( I ) , it suffices to prove that t
Lim [ t - m ( ℓ n t) ] = ∞
(III)
→ ∞
Let the real number t be expressed as a decimal with k digits occurring before the decimal point. This means that 10 k ≥ t ≥ 10 k-1 .
Taking ℓ n gives k( ℓ n10) ≥ ℓ n t ≥ (k-1) ( ℓ n10), or kC ≥ ℓ n t ≥ (k-1)C if we write C = ℓ n 10 ≈ 2.30258 …. Now
t - m ( ℓ n t) ≥ 10 k-1 - m ( ℓ n t) ≥ 10 k-1 - m kC, or t - m ( ℓ n t) ≥ k 2 - m kC if we remember that 10 k-1 ≥ k 2 for k = 0,1,2,3, …. (The proof of this is left to you ).
As t → ∞ , t has more and more integer digits, so k → ∞ as well. It follows that t
Lim [ t - m ( ℓ n t)] ≥
→ ∞ which establishes (
III ) as well as (
II
).
Lim k ( k - mC ) = ∞ , k → ∞