Prove derivative at a point using sequential criterion 
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*Fixed a few errors

Suppose I have the following function:
$f(x) = x^2+3$.
I want to prove that $f'(-2)$ exists and that $f'(-2)=-4$.
I could demonstrate this and prove this in various ways, but I am working on proving this
using the Sequential Criterion.
That is,
$$f:A\longrightarrow \mathbb R \text{ is differentiable at }x_0\iff
$$
$$\text{ for every sequence } \{x_n\}_{n=1}^\infty\subset A \text{ such that } x_n\rightarrow x_0,$$
$$\dfrac {f(x_n)-f(x_0)}{x_n-x_0} \text{ converges to } f'(x_0) $$.
I am having trouble using this definition. The result is expected, which makes me think I am just using the definition wrong.
We want to show that $f'(-2)$ exists and that $f'(-2)=-4$.
Let $x_n$ be a sequence such that $x_n\rightarrow -2$, $x_n \neq -2$. Using the sequential definition,
$$\dfrac {f(x_n)-f(x_0)}{x_n-x_0} \rightarrow \dfrac {f(-2)-f(-2)}{(-2)-(-2)} = \dfrac{0}{0} \neq -4 = f'(-2)$$
I must be applying the definition wrong here. Clearly this will happen for proceeding in that fashion.
I proceeded in a fashion similar to that of a continuity or limit proof using sequential criterion.
Where am I going wrong in this derivative proof?
Thanks in advance!
 A: Looks like you are not using the definition of convergence correctly. I don't want to be mean, new contributor here! In general, in a lot of analysis, plugging in numbers at the first step is not generally a great way to go. Usually, you will want to do some algebraic manipulation along the way.
If you think that $f'(-2) = -4$, and you want to use the sequential definition then you should try to prove convergence of sequence rigorously, using the epsilon definition. To be specific, perhaps show
$$ \left| \frac{f(x_n) - f(x_0)}{x_n - x_0} - f'(x_0) \right| < \epsilon $$
given that $n$ is sufficiently large.
That is, show that $\exists N \in \mathbb{N}$ such that $\forall n \geq N$,
$$ \left| \frac{(x_n)^2 + 3 - 7}{x_n + 2} + 4 \right| < \epsilon $$.
Given that we choose large enough $N$, this yields
$$ \left| \frac{(x_n)^2 + 3 - 7}{x_n + 2} + 4 \right| = \left| (x_n - 2) + 4 \right| = \left| x_n - (-2) \right| < \epsilon $$
where the last inequality is guaranteed by the convergence of $x_n \to x_0 = -2$.
A: You need to simplify the expression before you try to evaluate:
Let $\{x_n\}_{n=1}^\infty$ be a sequence such that $\lim_{n \to \infty} x_n=-2$. Since we have
$$\lim_{n \to \infty} \frac{f(x_n)-f(-2)}{x_n-(-2)}=\lim_{n \to \infty}\frac{x_n^2 - 4}{x_n+2}=\lim_{n \to \infty}x_n-2=-2-2=-4,$$
it follows from the sequential criterion that $f'(-2)=-4$ exists.
Since $f'(-2)$ exists, $f$ is continuous at $x=-2$. But as a bonus we may notice that
$$\lim_{n \to \infty} f(x_n)=\lim_{n \to \infty} x_n^2+3 = (-2)^2+3=7=f(-2).$$
