Using the sequential definition of a limit to show $\lim_{x\to 1} \frac{x^2 - 1}{\sqrt{x} - 1} = 4.$ Very recently, I posted this thread: Using the sequential definition of a limit to show $\lim_{x\to 0} \frac{x^2}{x} = 0.$ My solution for that proof was correct, but now I'm having trouble showing that $\lim_{x\to 1} \frac{x^2 - 1}{\sqrt{x} - 1} = 4$. For reference, here is my definition:
I have the following definition for a limit:

Definition: Given a function $f : D \rightarrow \mathbb{R}$ and a limit point $x_{0}$ of its domain $D$, for a number $\ell$, we write
$$ \lim_{x\to x_{0}} f(x) = \ell$$
provided that whenever $\{x_{n}\}$ is a sequence in $D \ - \{x_{0}\}$ that converges to $x_{0}$,
$$\lim_{n\to\infty} f(x_{n}) = \ell. $$

Using this definition, here is my attempt:

Let $\{x_{n}\}$ be a sequence in $\mathbb{R} - \{1\}$ such that $\{x_{n}\}$ converges to $1$. This means for all $\epsilon > 0$, there exists an index $N$ so that
$$|x_{n} - 1| < \epsilon$$
for all $n \geq N$. Now, we need to show for all $\epsilon > 0$, there exists an index $N_{2}$ so that
$$\left|\frac{x_{n}^2 - 1}{\sqrt{x_{n}} - 1} - 4\right| < \epsilon$$
for $n\geq N_{2}$.

So, I'm having trouble finding such an index $N_{2}$. I tried writing the expression as follows:
$$\left|\frac{x_{n}^{2} - 1}{\sqrt{x_{n}} - 1} - 4\right| \\$$
$$= \left|\frac{x_{n}^{2} - 1 - 4\sqrt{x_{n}} + 4}{\sqrt{x} - 1} \right| $$
$$\leq \left|\frac{x_{n}^{2} + 3}{\sqrt{x_{n}} - 1} \right|,$$
but I couldn't get anywhere after this. Can someone please help me finish this proof?
EDIT: An attempt based on current answers:
$$\begin{align*}
\left|\frac{x_{n}^{2} - 1}{\sqrt{x_{n}} - 1} - 4\right|  = \left| \frac{(\sqrt{x_{n}} - 1)(1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}})}{\sqrt{x_{n} - 1}} - 4\right| \\[1em]
= \left|1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}} - 4\right|
\end{align*},$$
but I get nowhere from here.
 A: Hint
Use that
$$t^4-1=(t-1)(1+t+t^2+t^3)$$
with $t=\sqrt x.$
Edit
You got
$$\left|\frac{x_{n}^{2} - 1}{\sqrt{x_{n}} - 1} - 4\right|  
= \left|1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}} - 4\right|.$$
Assume that $1-\delta<x_n\le 1.$ Then we have
$$1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}}>1+\sqrt{1-\delta}+1-\delta+\sqrt{(1-\delta)^3}>1+3\sqrt{(1-\delta)^3}.$$
Thus $$0<4-(1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}})<3(1-\sqrt{(1-\delta)^3}).$$ Now
$$3(1-\sqrt{(1-\delta)^3})<\epsilon\iff \delta <1-\sqrt[3]{\left(1-\frac{\epsilon}{3}\right)^2}.$$
Assume that $1\le x_n<1+\delta.$ Then we have
$$1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}}<1+\sqrt{1+\delta}+1+\delta+\sqrt{(1+\delta)^3}<1+3\sqrt{(1+\delta)^3}.$$
Thus $$0<1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}}-4<3(\sqrt{(1+\delta)^3}-1).$$ Now
$$3(\sqrt{(1+\delta)^3}-1)<\epsilon\iff \delta <\sqrt[3]{\left(1+\frac{\epsilon}{3}\right)^2}-1.$$
Finally, we have shown that $$\delta<\min\{1-\sqrt[3]{\left(1-\frac{\epsilon}{3}\right)^2}, \sqrt[3]{\left(1+\frac{\epsilon}{3}\right)^2}-1 \}\implies \left|1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}} - 4\right|<\epsilon.$$ But since $\lim_n x_n=1$ for all $\delta>0$ there exists $N\in\mathbb{N}$ such that $$n\ge N\implies 1-\delta<x_n<1+\delta.$$ 
A: $\textbf{Hint}$
Notice that $x_n=(\sqrt{x_n})^2$. We can make a substitution $\sqrt{x_n}=y_n$. This gives us
$$\dfrac{x^2_n-1}{\sqrt{x_n}-1}=\dfrac{y_n^4-1}{y_n-1}.$$
There is a clever way to cancel out the denominator by expanding the numerator, can you find it?
A: We have
\begin{align}\left| \frac{x_n^2-1}{\sqrt{x_n}-1} -4 \right| &= \left| \frac{(\sqrt{x_n}-1)(\sqrt{x_n}+1)(x_n+1)}{\sqrt{x_n}-1} -4 \right|\\ &= \left| (\sqrt{x_n}+1)(x_n+1) -4\right|\\ &= \left| (\sqrt{x_n}+1)(x_n+1)-2(x_n+1)+2(x_n+1) -4\right|\\
&\leq |x_n+1||\sqrt{x_n}-1|+2|x_n-1|\tag{1}
\end{align}
I will leave to you to prove that if $x_n\to 1$, then $\sqrt{x_n}\to 1$ as well.
Let $0<\varepsilon\leq 1$. This is not a loss of generality, since if we can find appropriate $N$ for $\varepsilon \leq 1$, the same $N$ obviously works for bigger $\varepsilon$'s as well.
Now, choose $N$ such that $n\geq N$ implies both $|\sqrt{x_n}-1| < \varepsilon/6$ and $|x_n-1|<\varepsilon/4$ (you can choose $N$ for each of these separately and then take maximum).
We now have $$-\varepsilon/4<x_n-1<\varepsilon/4\implies 2-\varepsilon/4<x_n+1<2+\varepsilon/4 \implies |x_n+1|<3,\ n\geq N$$ and so, from $(1)$ it follows
$$\left| \frac{x_n^2-1}{\sqrt{x_n}-1} -4 \right|\leq |x_n+1||\sqrt{x_n}-1|+2|x_n-1| < 3\frac{\varepsilon}{6}+2\frac\varepsilon 4 = \varepsilon,\ n\geq N.$$
A: Not using the sequential definition of a limit.
$$\lim_{x\to 1} \frac{x^2 - 1}{\sqrt{x} - 1}=\lim_{y\to 0} \frac{(1+y)^2 - 1}{\sqrt{1+y} - 1}=\lim_{y\to 0} \frac{2y+y^2}{\sqrt{1+y} - 1}$$ Now, using binomial expansion or Taylor series
$$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$
$$\frac{2y+y^2}{\sqrt{1+y} - 1}=\frac{2y+y^2}{\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)}=4+3 y+O\left(y^2\right)$$
