If x→∞lim((1−ee)(e1−1+xx))x=α, then the value of 1+logeαlogeα equals :
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Official solution
α=x→∞lim[(1−ee)(e1−1+xx)]x.
Begin by rewriting the expression inside the limit. Notice that
1+xx=1−1+x1,
so
e1−1+xx=e1−(1−1+x1)=e1−1+1+x1.
Express the constant part as
1−e1=ee−1,
to obtain
e1−1+xx=−ee−1+1+x1.
Multiplying by the prefactor, we have
(1−ee)(e1−1+xx)=1−ee[−ee−1+1+x1].
Split the expression:
1−ee⋅(−ee−1)+1−ee⋅1+x1=−1−ee−1+(1−e)(1+x)e.
Since
1−e=−(e−1),
it follows that
−1−ee−1=−−(e−1)e−1=1.
Hence, the expression simplifies to
1+(1−e)(1+x)e.
Again, replacing 1−e by −(e−1),
1+−(e−1)(1+x)e=1−(e−1)(1+x)e.
Thus, the limit becomes
α=x→∞lim(1−(e−1)(1+x)e)x.
For large x, note that
1+x∼x,
so we approximate
α=x→∞lim(1−(e−1)xe)x.
Recall the limit
x→∞lim(1−xa)x=e−a,
with
a=e−1e.
Therefore,
α=e−e−1e.
Taking the natural logarithm,
lnα=−e−1e.
Now, compute
1+lnαlnα=1−e−1e−e−1e.
Express the denominator with a common denominator:
1−e−1e=e−1e−1−e−1e=−e−11.
Thus,
−e−11−e−1e=1e=e.
The final result is
1+lnαlnα=e.