Why these different forms of Triangle Removal Lemma are equivelent? Triangle removal lemma can be stated as follows.

$\forall \epsilon>0,\exists \delta>0$ such that any graph on $n$ vertices with $\leq \delta n^3$ triangles can be made triangle-free by removing at most $\epsilon n^2$ edges.

I can also find another version of this result.

Any graph on $n$ vertices with $o(n^3)$ triangles can be made triangle-free by removing $o(n^2)$ edges.

How to show these two are equivelent?
 A: The second statement is tricky to parse. You can't, looking at a graph on $n$ vertices, tell if it has $o(n^3)$ triangles or not, because $o(n^3)$ is only meaningful as $n \to \infty$. The statement really means:

Given any $f(n)$ which is $o(n^3)$, there is some $g(n)$ which is $o(n^2)$ such that any $n$-vertex graph with at most $f(n)$ triangles can be made triangle-free by removing at most $g(n)$ edges.

(The only way to know that this is a for-all, there-exists type of statement is to check that all the other possibilities either don't say anything interesting, or are false.)
Here's one way to deduce this from the first lemma. Take any $f(n) = o(n^3)$. Let $\delta(\epsilon)$ be some function realizing the dependency between $\epsilon$ and $\delta$ in the first lemma. 
For each positive $k$, let $n_k$ be the value such that for all $n > n_k$, $f(n) < \delta(\frac1k) n^3$; this must exist, because $f(n) = o(n^3)$. Define $g(n) = \frac1k n^2$ for all $n \in [n_k, n_{k+1})$. (We can set $g(n) = n^2$ for $n<n_1$, because deleting $n^2$ edges can always make any graph triangle-free.)
Then, the first lemma will tell us that for any graph with $f(n)$ triangles, we can make it triangle-free by removing $g(n)$ edges. 
