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?
 A: 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
$$
|x-1|<\delta=\min\Big\{1,\frac{\varepsilon}{15}\Big\}\quad\Longrightarrow\quad
|x^4-1|<15|x-1|<15\cdot\frac{\varepsilon}{15}=\varepsilon.
$$
A: Hint
When you took $$\delta=\min(\color{red}{1},...)$$
this means that you assumed that
$$|x-1|<\color{red}{1}$$
which gives
$$-1<x-1<1 \iff 1<x+1<3$$
So, your inequality $|x+1|<2$ should be replaced by $ |x+1|<3$
and
$|x^2+1|<2 $ by $ |x^2+1|<5$.
A: You can also express $x^4-1$ in powers of $x-1$ as in
$$|x^4-1| = |((x-1)+1)^4-1| = |(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)|.$$
If we assume $|x-1| < \delta$, the triangle inequality gives
$$|x^4-1| \le |x-1|^4+4|x-1|^3+6|x-1|^2+4|x-1| < \delta^4+4\delta^3+6\delta^2+4\delta.$$
Now we want $\delta^4+4\delta^3+6\delta^2+4\delta \le \varepsilon$. Since $$\delta^4+4\delta^3+6\delta^2+4\delta =\delta(\delta^3+4\delta^2+6\delta+4)$$
we only have to bound the second factor since the first is merely $\delta$. If we set $\delta = \min\left\{1, \frac\varepsilon{15}\right\}$, we have
$$|x^4-1| \le \delta(\delta^3+4\delta^2+6\delta+4) \le \delta \cdot (1+4+6+4) = 15\delta \le 15\cdot \frac\varepsilon{15} = \varepsilon.$$
