504 Real Analysis
=exp(
cosu− 1
u·
u
sinu)
=e^0 ·^1 =e^0 = 1.The limit therefore exists.
382.Without loss of generality, we may assume thatm>n. Write the limit as
lim
x→ 0mn√cosnx−mn√cosmx
x^2.
Now we can multiply by the conjugate and obtain
lim
x→ 0cosnx−cosmx
x^2 (mn√
(cosnx)mn−^1 +···+mn√
(cosmx)mn−^1 )=lim
x→ 0cosnx( 1 −cosm−nx)
mnx^2
=lim
x→ 01 −cosm−nx
mnx^2=lim
x→ 0( 1 −cosx)( 1 +cosx+···+cosm−n−^1 x)
mnx^2=
m−n
mnlim
x→ 01 −cosx
x^2=
m−n
2 mn.
We are done.
383.Forx>1 define the sequence(xn)n≥ 0 byx 0 =xandxn+ 1 =x
n^2 +^1
2 ,n≥0. The
sequence is increasing because of the AM–GM inequality. Hence it has a limitL, finite
or infinite. Passing to the limit in the recurrence relation, we obtainL=L
(^2) + 1
2 ; hence
eitherL=1orL=∞. Since the sequence is increasing,L≥x 0 >1, soL=∞.We
therefore have
f(x)=f(x 0 )=f(x 1 )=f(x 2 )= ··· =nlim→∞f(xn)=xlim→∞f(x).
This implies thatfis constant, which is ruled out by the hypothesis. So the answer to
the question is negative.
384.We can assume thatm>1; otherwise, we can flip the fraction and changettom^1 t.
There is an integernsuch thatm< 2 n. Becausefis increasing,f(t)<f(mt)<f( 2 nt).
We obtain
1 <
f (mt)
f(t)
<
f( 2 nt)
f(t).
The right-hand side is equal to the telescopic product
f( 2 nt)
f( 2 n−^1 t)·
f( 2 n−^1 t)
f( 2 n−^2 t)···
f( 2 t)
f(t)