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.
Best Answer
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 $$n\ge N\implies 1-\delta<x_n<1+\delta.$$