Time complexity of Dijkstra's algorithm for sparse graph
Discrete Lizard's answer explains various aspects of the question. 

Here is an explicit calculation when $E=o(V^2/\log V)$.

$$\lim_{V\to\infty}\frac{(E+V)\log V}{V^2}=\lim_{V\to\infty}\frac{E}{\dfrac{V^2}{\log V}}+\lim_{V\to\infty}\frac{V\log V}{V^2}=0+0=0$$

That means $O((E+V)\log V)\subsetneq O(V^2)$. So $O((E+V)\log V)$ is a tighter bound than $O(V^2)$ when $E=o(V^2/\log V)$.