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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?

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  • $\begingroup$ check again please. i just had variable mistake. $\endgroup$ – user400359 Oct 22 '18 at 16:20
  • $\begingroup$ 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. $\endgroup$ – DonAntonio Oct 22 '18 at 16:29
  • $\begingroup$ oops. it's false until $x = 1$ @DonAntonio $\endgroup$ – user400359 Oct 22 '18 at 16:31
  • $\begingroup$ @stackofhay42 I'm not sure what you mean "until"....That inequality is true iff $\;x_n\ge1\;$ ....and you want $\;x_n\to 0\;$ ! $\endgroup$ – DonAntonio Oct 22 '18 at 16:32
  • $\begingroup$ 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 $|\sqrt{x_{n}}|$ $\endgroup$ – user400359 Oct 22 '18 at 16:33
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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

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