3.2 Continuity, Derivatives, and Integrals 127Let us return to the problem. Because the limit is of the form 1∞, it is standard to
write it as
lim
x→ 0(
1 +
ax 1 +ax 2 +···+axn−n
n)ax n
1 +ax 2 +···+anx−n
·ax 1 +ax 2 +···+xxn−n
nx
.Using the fact that limt→ 0 ( 1 +t)^1 /t=e, we find this to be equal to
exp[
lim
x→ 0(
ax 1 +ax 2 +···+axn−n
nx)]
=exp[
1
nlim
x→ 0(
a 1 x− 1
x+
ax 2 − 1
x+···+
anx− 1
x)]
=exp[
1
n(lna 1 +lna 2 +···+lnan)]
= n√
a 1 a 2 ···an,the desired answer.
We continue with a problem of theoretical flavor that requires an-δargument.
Written by M. Becheanu it was given at a Romanian competition in 2004.
Example.Leta∈( 0 , 1 )be a real number andf:R→Ra function that satisfies the
following conditions:
(i) limx→∞f(x)=0;
(ii) limx→∞f(x)−xf(ax)=0.
Show that limx→∞f(x)x =0.
Solution.The second condition reads, for any>0, there existsδ>0 such that if
x∈(−δ, δ)then|f(x)−f(ax)|<|x|. Applying the triangle inequality, we find that
for all positive integersnand allx∈(−δ, δ),
|f(x)−f(anx)|≤|f(x)−f(ax)|+|f(ax)−f(a^2 x)|+···+|f(an−^1 x)−f(anx)|<|x|( 1 +a+a^2 +···+an−^1 )=1 −an
1 −a|x|≤1 −a|x|.Taking the limit asn→∞, we obtain|f(x)|≤1 −a|x|.Since>0 was arbitrary, this proves that limx→∞f(x)x =0.
380.Find the real parametersmandnsuch that the graph of the functionf(x)=
√ (^38) x (^3) +mx (^2) −nxhas the horizontal asymptotey=1.