Advanced book on Mathematics Olympiad

Real Analysis 517

is monotonic to the left, and to the right ofa, and becauseF(n−^2 )(a)=0,F(n−^2 )(x)    = 0

forx    =aandxin a neighborhood ofa. Inductively, we obtainF′(x)    =0 andf(x)  = 0

in some neighborhood ofa.

The limit from the statement can be written as

lim

x→a

eg(x)

ef(x)−g(x)− 1

f(x)−g(x)

.

We only have to compute the limit of the fraction, sinceg(x)is a continuous function.

We are in a^00 situation, and can apply L’Hôpital’s theorem:

xlim→a

ef(x)−g(x)− 1

f(x)−g(x)

=xlim→a

(f′(x)−g′(x))ef(x)−g(x)

f′(x)−g′(x)

=e^0 = 1.

Hence the limit from the statement is equal toeg(a)=eα.

(N. Georgescu-Roegen)

414.The functionh:[ 1 ,∞)→[ 1 ,∞)given byh(t)=t( 1 +lnt)is strictly increasing,

andh( 1 )=1, limt→∞h(t)=∞. Hencehis bijective, and its inverse is clearly the

functionf:[ 1 ,∞)→[ 1 ,∞),λ→f (λ). Sincehis differentiable, so isf, and

f′(λ)=

1

h′(x(λ))

=

1

2 +lnf (λ)

.

Also, sincehis strictly increasing and limt→∞h(t)=∞,f (λ)is strictly increasing, and
its limit at infinity is also infinity. Using the defining relation forf (λ), we see that

f (λ)

λ

lnλ

=lnλ·

f (λ)

λ

=

lnλ

1 +lnf (λ)

.

Now we apply L’Hôpital’s theorem and obtain

lim

λ→∞

f (λ)

λ

lnλ

= lim

λ→∞

1

f (λ)

1

λ·

1

2 +lnf (λ)

=lim

λ→∞

f (λ)

λ

( 2 +lnf (λ))= lim

λ→∞

2 +lnf (λ)

1 +lnf (λ)

= 1 ,

where the next-to-last equality follows again fromf (λ)( 1 +lnf (λ))=λ. Therefore,

the required limit is equal to 1.

(Gazeta Matematica ̆(Mathematics Gazette, Bucharest), proposed by I. Tomescu)

415.If all four zeros of the polynomialP(x)are real, then by Rolle’s theorem all three

zeros ofP′(x)are real, and consequently both zeros ofP′′(x)= 12 x^2 − 6

√

7 x+8 are

real. But this quadratic polynomial has the discriminant equal to−132, which is negative,

and so it has complex zeros. The contradiction implies that not all zeros ofP(x)are real.