Is my use of Fatou's Lemma correct in this case?

Question

I am asked to determine:

$$\lim_{n\to \infty}\int_{-\infty}^{\infty}\frac{1}{1+e^{nx}}e^{\frac{-x^2}{n}}d\lambda{x}$$

My idea:

$$\lim_{n\to \infty}\int_{-\infty}^{\infty}\frac{1}{1+e^{nx}}e^{\frac{-x^2}{n}}d\lambda{x}=\limsup_{n\to \infty}\int_{-\infty}^{\infty}\frac{1}{1+e^{nx}}e^{\frac{-x^2}{n}}d\lambda{x}\leq\int_{-\infty}^{\infty}\limsup_{n\to \infty}\frac{1}{1+e^{nx}}e^{\frac{-x^2}{n}}d\lambda{x}$$

and $\lim_{n\to \infty}\frac{1}{1+e^{nx}}e^{\frac{-x^2}{n}}=0$

Therefore, $\int_{-\infty}^{\infty}\limsup_{n\to \infty}\frac{1}{1+e^{nx}}e^{\frac{-x^2}{n}}d\lambda{x}=0$

Now note that $\frac{1}{1+e^{nx}}e^{\frac{-x^2}{n}}\geq0,$ for any $x \in ]-\infty,\infty[$

This then implies that: $\lim_{n\to \infty}\int_{-\infty}^{\infty}\frac{1}{1+e^{nx}}e^{\frac{-x^2}{n}}d\lambda{x}=0$

Answer 1

Something is wrong. The pointwise limit should be $$\lim_{n\to \infty}\frac{e^{\frac{-x^2}{n}}}{1+e^{nx}}= \begin{cases} 0& \text{if $x>0$,}\\ 1/2 &\text{if $x=0$,}\\ 1 &\text{if $x< 0$.} \end{cases}$$