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?

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