Again Computing limits

Let start with the condition given $e^{a_n} + n a_n=2$

The exponential part is increasing and in the second part $n$ is increasing.But the total value is always constant So, $a_n$ should have a decreasing value.Then $a_n$ is a decreasing sequence. i.e. $a_n > a_{n+1}$ .

If $a_n <0 \implies e^{a_n}<1 \,\, \text{and}\,\, n a_n<0$ .So $e^{a_n} +n a_n<1$ that contradicts with the question.So, $a_n>0$ .

The sequence $a_n$ is decreasing and bounded below.So it is a convergent sequence.The limiting value will be the infimum value of the set .

So, $\lim_{n \to \infty} a_n=0$

Taking limits in the condition we get $e^{\lim_{n \to \infty} a_n} + \lim_{n \to \infty} n a_n=2$

So, $\lim_{n \to \infty} n a_n=1$ .

$e^{a_n} + n a_n=2 \\ \implies e^{a_n}-1=1-n a_n$

Now $\lim_{n \to \infty} n(1-n a_n) \\ =\lim_{n \to \infty} n(e^{a_n}-1) \\ =\lim_{n \to \infty} \dfrac{e^{a_n}-1}{a_n} \times \lim_{n \to \infty} n a_n \\ =1 \times 1=1$