Behaviour of little-o in equivalence The equivalent relations I know that little-o possesses are the below:
1) $f \sim g$ for $x \rightarrow a$
2) $f = g + o(g)$ for $x \rightarrow a$
3) $f = g + o(f)$ for $x \to a$
I also know about the following properties:
1) $f \cdot o(g) = o(fg)$
2) If f is bounded near $a$, then $o(fg)=o(g)$
3) $o(h) \pm o(g) = o(g)$
4) If $f \sim g$, then $o(f)=o(g)$.
And then the lecturer is making the following statement (for $x \rightarrow 0$):
$$-\frac{x^2}{2} + o\left( x^2 \right) \sim -\frac{x^2}{2}$$
From the properties that I have above I cannot make such a statement. I am trying to rigorously understand why we can write that. Perhaps using the definition of little-o should be used here? The definition that I have been introduced to, is the following:

For $f,g : E \rightarrow \mathbb{R}$, and $a$ is the interior point of
  $E$. If there exists $\overset{\circ}{U}_a$ and a function $\phi:E
> \rightarrow \mathbb{R}$, such that $f(x)=\phi(x)g(x)$ and $\lim_{x \to
> a} \phi(x) = 0$, then $f=o(g)$ for $x \rightarrow a$.

I think the clearest definition for me would be using the limits. I would start with writing that statement as:
$$\displaystyle{\lim_{x \to 0} \frac{-\frac{x^2}{2}+o\left( x^2 \right)}{-\frac{x^2}{2}}} = 1$$
But this is weird. How would I proceed next?
 A: It seems that you are mixing up different things.
Firstly we define the little-o notation as $x \to a$
$$f(x)=o(g(x)) \iff f(x)=\phi(x)\cdot g(x) \quad \phi(x) \to 0$$
which is the more general definition and it is equivalent to the following alternative definition
$$f(x)=o(g(x)) \iff \lim_{x\to a} \frac{f(x)}{g(x)}=0$$
which holds when $g(x)\neq 0$ for $x \neq a$.
Secondly we define the asymptotic equivalence
$$f(x)\sim o(g(x)) \iff f(x)=\phi(x)\cdot g(x) \quad \phi(x) \to 1$$
which is the more general definition and it is equivalent to the following alternative definition
$$f(x)\sim o(g(x)) \iff \lim_{x\to a} \frac{f(x)}{g(x)}=1$$
which holds when $g(x)\neq 0$ for $x \neq a$.
In your example we have
$$\lim_{x \to 0} \frac{-\frac{x^2}{2}+o\left( x^2 \right)}{-\frac{x^2}{2}} = \lim_{x \to 0} \frac{-\frac{1}{2}+o\left( 1 \right)}{-\frac{1}{2}} =1$$
and according to the definition of asymptotic equivalent we say that as $x \to 0$
$$-\frac{x^2}{2} + o\left( x^2 \right) \sim -\frac{x^2}{2}$$
To evaluate the limit we have simply cancel out a factor $x^2$ using the properties for little-o notation. For that point, I also suggest to refer to the related


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*Why are we allowed to cancel fractions in limits?
