A course of pure mathematics, by G.H. Hardy.

352 THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS

[IX

as - + so or - - oo. Since the exponential function is continuous it follows that
{1 + (y/)} = e log{l+(y/t)} -. e
asf - + oo or - o --- o: i.e. that
lim {1 + (y/:)} = lim {t + (y/l)}- = e.
- -> + Xo - o
If we suppose that -- + cc or - oo through integral values
only, we obtain the result expressed by the equations (1).
(2) If n is any positive integer, however large, and x > 1, we have
Ax dt fx dt rX dt
J1 tl + (l/) <J t J1 tl-(L,/n)
or (1 - x- /) < log x < n (xl/- 1).....................(2)
from which we deduce
<< (1 -............................(3)
if y=logx>0. It is easy to prove, as in ~67, that the first of these two
functions of y is an increasing and the second a decreasing function of n, and
therefore that each tends to a limit as n — oo; and the two limits must be
equal. For if l+(/yn) =7 and 1/{1-(y/n)}=172, we have (Ex. xxxvI. 8), at
any rate for sufficiently large values of n,
2 - 't < 7 -1 (772 - '1) =y2 l"/nl
which evidently tends to 0 as n- c oo.
191. The representation of logx as a limit. We can also prove (cf.
~ 68) that
lim n (1- X- i/n) = lim n (xl/' 1) = log x.
For n (xl - 1) -2 (1 — x-1/) = n (Xl -- 1)(1 - x-l/n)
which tends to zero as n —co, since n (x1/ -1) tends to a limit (~ 68) and
x-l'/' to 1 (Ex. xxx. 10). The result follows from the inequalities (2) of ~ 190 (2).
Examples LXXXVIII. 1. Prove, by taking n1=6 in the inequalities (3)
of ~ 190, that 2.5 < e < 2.9.
2. Prove that, if t> 1, (tl/- t-l/l)/(t-t-l) < /n, and so that, if x> 1,
x dt x dt 1 X ( I\dt I
He 1 ce t,-/) '1 t+J t )<1 (- ) t_ - =L(x+1-2).
tI -1 tlj + 'ini) V t t n X
Hence deduce the results of ~ 190.
3. If e,, is a function of n such that n$n,,-.l as n-eoo, then (1+ f)n-.el.
[Writing n log (1 + en) in the form
(n) log (1 + n,)
and using Ex. LXXXIV. 4, we see that n log (1+ 4,) —.]
4. If ne —+co then (1+$&,)-n+ oo, and if -,-oo- then (1+)n-O0.