Showing $\lim_{x\to 0} \frac{x - 1}{\sqrt{x} - 1} = 1$

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?

• check again please. i just had variable mistake. – user400359 Oct 22 '18 at 16:20
• Are you sure that $\;\sqrt{x_n}\le|x_n|\;$ ...? Also, the limit must be only from the right, as from the left the function isn't defined. – DonAntonio Oct 22 '18 at 16:29
• oops. it's false until $x = 1$ @DonAntonio – user400359 Oct 22 '18 at 16:31
• @stackofhay42 I'm not sure what you mean "until"....That inequality is true iff $\;x_n\ge1\;$ ....and you want $\;x_n\to 0\;$ ! – DonAntonio Oct 22 '18 at 16:32
• so the limit actually does not exist ? edit: never mind, i believe it exists still. I don't know how to get a value for $N'$ after getting to $|\sqrt{x_{n}}|$ – user400359 Oct 22 '18 at 16:33

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 $$N = N^\prime$$.