Proving $\lim_{x\to1}(x^3+5x^2-2)=4$ using the $\epsilon$-$\delta$ definition of a limit

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. $\blacksquare$

Note that for $0<|x-1|<1$

$$\begin{align} \left|x^3+5x^2-6\right|&=|x^2+6x+6||x-1|\\\\ &<22|x-1|\\\\ &<\epsilon \end{align}$$

whenever $|x-1|<\delta=\min\left(1,\frac{\epsilon}{22}\right)$