I'd like to use the sequential definition of a limit to show $\lim_{x\to 0} \frac{x – 1}{\sqrt{x} – 1} = 1$.
Here's the definition I'm using:
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 $x_{0}$, we have $$\lim_{n\to\infty} f(x_{n}) = \ell$$
Note that this is not the standard $\epsilon-\delta$ definition of a limit.
Here's my attempt at proving this claim:
Let $\{x_{n}\}$ be a sequence in $\mathbb{R} – \{0\}$ that converges to $0$. For all $\epsilon > 0$, $\exists N$ such that
$$|x_{n} – 0| < \epsilon $$
for all $n \geq N$. To prove the claim, we require $\forall \epsilon > 0$, $\exists N'$ such that
$$\left|\frac{x_{n} – 1}{\sqrt{x_{n}} – 1} – 1\right| < \epsilon$$
for all $n \geq N'$. However,
$$\left|\frac{x_{n} – 1}{\sqrt{x_{n}} – 1} – 1\right| = \left|\frac{(\sqrt{x_{n}} + 1) (\sqrt{x_{n}} – 1)}{\sqrt{x_{n}} – 1} – 1\right| = $$
$$|\sqrt{x_{n}} + 1 – 1| = |\sqrt{x_{n}}| \leq |x_{n}| = |x_{n} – 0|,$$
so setting $N' = N$ completes the proof.
Is this correct?
Best Answer
As @DonAntonio mentioned in the comments, the inequality $|\sqrt{x_n}| \leq |x_n|$ is not necessarily the case, particularly when $x < 1$ (which your limit goes through).
You can add a quick proof that $\lim_{x \to 0} \sqrt{x} = 0$ with a sequential limit. Let $\{x_n\}$ be some sequence converging to zero. So, for all $n \geq N$, we want $|\sqrt{x_n}| < \epsilon \implies |x_n| < \epsilon^2 $ for some $N$. We know that there exists an $N^\prime$ such that $|x_n| < \epsilon^\prime$ for all $n \geq N^\prime$. Let $\epsilon^\prime = \epsilon^2$; choose $N^\prime$ to fulfill this inequality, and let $N = N^\prime$.
As a side observation, an ordinary limit always corresponds to a sequential limit for real numbers. But this is not necessarily the case when the space is not metrizable