Real Analysis 503Hence
1
Pn+ 1−
1
Pn=
xn+ 2
(n+ 1 )!−
xn+ 1
n!=
xn+ 2 −(n+ 1 )xn+ 1
(n+ 1 )!=
xn+^1
(n+ 1 )!, forn≥ 1.Adding up these relations for 1≤n≤N+1, and using the fact that the sum on the left
telescopes, we obtain
1
PN+ 1=
1
P 1
+
x^2
2!+
x^3
3!+···+
xN+^1
(N+ 1 )!= 1 +x
1!+
x^2
2!+···+
xN+^1
(N+ 1 )!.
Because this last expression converges toex, we obtain that limN→∞PN =e−x,as
desired.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
T. Andreescu and D. Andrica)
380.We are supposed to findmandnsuch thatxlim→∞^3√
8 x^3 +mx^2 −nx=1 or x→−∞lim^3√
8 x^3 +mx−nx= 1.We compute√ (^38) x (^3) +mx (^2) −nx= ( 8 −n (^3) )x (^3) +mx 2
√ (^3) ( 8 x (^3) +mx (^2) ) (^2) +nx√ (^38) x (^3) +mx (^2) +n (^2) x 2.
For this to have a finite limit at either+∞or−∞,8−n^3 must be equal to 0 (otherwise
the highest degree ofxin the numerator would be greater than the highest degree ofxin
the denominator). We have thus found thatn=2.
Next, factor out and cancel anx^2 to obtain
f(x)=
m
3
√(
8 +mx) 2
+ 23
√
8 +mx+ 4.
We see that limx→∞f(x)=m 12 .For this to be equal to 1,mmust be equal to 12. Hence
the answer to the problem is(m, n)=( 12 , 2 ).
381.This is a limit of the form 1∞. It can be computed as follows:lim
x→π/ 2
(sinx)
cos^1 x
= lim
x→π/ 2
( 1 +sinx− 1 )
sinx^1 − 1 ·sincosx−x^1=
(
lim
t→ 0
( 1 +t)^1 /t)limx→π/ 2 sincosx−x^1
=exp(
lim
u→ 0cosu− 1
sinu