Proving $\lim_{x\to1}(x^3+5x^2-2)=4$ using the $\epsilon$-$\delta$ definition of a limit I want to prove that the limit of $f(x)=x^3+5x^2-2$ when $x\to 1$ is $4$. 
So, I want to show that for any $\epsilon >0$ $\exists \delta_{\epsilon}$ such that for all $x$ that satisfies $|x-1|<\delta$ then $|f(x)-4|< \epsilon$. 
Thus, $|x^3+5x^2 - 2 - 4|<|x^3+5x^2|<|x^2(x+5)|=x^2|x+5|<x^2|x-1|<\epsilon.$
And we know that $x^2|x-1|<x^2\delta<\epsilon $. As such, for every $\epsilon >0$ the corresponding $\delta$ is $\min\{\epsilon/x^2, \epsilon\}, x\neq0$.
Is it right?

I concur with Arthur that your work looks OK minus a few technicalities, but those are rather a big deal in real analysis aren't they? Thus, I would write up your proof as follows:
Given $\epsilon>0$, we need $\delta>0$ such that if $|x-1|<\delta$ then $|(x^3+5x^2-2)-4)|<\epsilon$. Now,
$$
|x^3+5x^2-6|=|(x-1)(x^2+6x+6)|=|x-1||x^2+6x+6|.
$$
If $|x-1|<1$, that is, $0<x<2$, then $x^2+6x+6<(2)^2+6(2)+6=22$ and so
$$
|x^3+5x^2-6|=|x-1||x^2+6x+6|<22|x-1|.
$$
So if we take $\delta=\min\left\{1,\frac{\epsilon}{22}\right\}$, then $|x-1|<\delta$ implies that
$$
|x^3+5x^2-6|=|x-1||x^2+6x+6|<\frac{\epsilon}{22}\cdot22=\epsilon,
$$
as desired. $\blacksquare$
