<s>[INST] Is my $\epsilon$-$\delta$ calculation correct? I have to show that $\lim_{x \to 1} x^4-1 =0$. Here is how i have done it:
$\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < \epsilon \qquad$ and since we are close to 1, we can assume that the $\delta$-neighborhood of $c=1$ must be havea radius of max $\delta =1$ which implies that :   $\mid x+1 \mid \le 2 \quad and \mid x^2+1 \mid \le 2 \quad \forall x \in V_{\delta}(c) \quad$
We now choose $\delta=min \left \{1,\frac{\epsilon}{4}\right\} \quad$ and we can conclude that if $\mid x-1 \mid < \delta$, it follows that $\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < 4\frac{\epsilon}{4} =\epsilon$. Is this calculation correct? Do I miss something? Or some details?
 [/INST] You want
$$
|x^2+1||x+1||x-1|<\varepsilon
$$
near $x=1$.
First step is to control the quantity $|x^2+1||x+1|$ near $x=1$. So first
restrict $x$ so that $|x-1|<1=\delta_1$. This means
$0<x<2$, and hence $|x+1|<3$ while $x^2+1<5$. Hence
$$
|x-1|<1=\delta_1 \quad\Longrightarrow\quad |x^2+1||x+1|<15,
$$
and thus
$$
|x-1|<1=\delta_1 \quad\Longrightarrow\quad |x^4-1|=|x^2+1||x+1||x-1|<15|x-1|,
$$
Now, we have
$$
|x-1|<\delta=\min\Big\{1,\frac{\varepsilon}{15}\Big\}\quad\Longrightarrow\quad
|x^4-1|<15|x-1|<15\cdot\frac{\varepsilon}{15}=\varepsilon.
$$
</s>