Consider the functional $$F(u)=\int_{-\infty}^{\infty}\frac{(u'^2-1)^2}{1+x^2} dx.$$ It has a global minimum $u'=\pm 1$ and a stationary point $u=0$, in fact
$$F(u+t\varphi)-F(u) = \int_{-\infty}^{\infty} \frac{4tu' \varphi'}{1+x^2}(1-u'^2).$$ If we consider the variation about the value $u=0$ we find
$$F(t\varphi)-F(0)=\int_{-\infty}^{\infty} \frac{t^2(t^2\varphi'^4-2\varphi'^2)}{1+x^2} dx$$ which is negative for all $\varphi$ as $t \to 0$. However, if we consider the sequence $u_n$ given by
$-2x+\frac{2}{n} \qquad x \in (0, \frac{1}{n})$
$2x+\frac{2}{n} \,\,\,\,\qquad x \in (-\frac{1}{n}, 0)$
$0 \qquad \qquad \,\,\,\,\,\,\,x \in \mathbb{R}- (-\frac{1}{n},\frac{1}{n})$
we have $u_n \to 0$ in $W^{1,4}$, but
$$F(u_n)-F(0)=\int_{-\frac{1}{n}}^{\frac{1}{n}} \frac{8}{1+x^2}dx > 0$$
My first thought is that this sequence cannot be written as $t\varphi$, hence the previous equation is not valid.