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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.

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  • $\begingroup$ you can twice use the fact that $$(t^2 - 1) = (t - 1)(t + 1)$$ and show that the expression actually equals to $$(\sqrt{x} + 1)(x + 1)$$ $\endgroup$
    – shabunc
    Oct 21, 2018 at 21:59
  • $\begingroup$ I have used the fact, but I don't know how to proceed. I've edited my original post. $\endgroup$
    – user400359
    Oct 21, 2018 at 22:30
  • $\begingroup$ Notice that if $x_n\to 1$, then we can show $\sqrt{x_n}\to1$, for $|\sqrt{x_n}+1|>1$ if $\sqrt{x_n}>0$ (which it has to be for large n). Then if $|x_n-1|<\epsilon$, then $|x_n-1|=|\sqrt{x_n}-1||\sqrt{x_n}+1|<\epsilon\implies|\sqrt{x_n}-1|<\epsilon$. Once we have $\sqrt{x_n}\to1$, then we can use algebraic limit properties to show that $1+\sqrt{x_n}+x_n+(\sqrt{x_n])^3\to1+1+1+1=4$. $\endgroup$
    – Melody
    Oct 21, 2018 at 22:38
  • $\begingroup$ It doesn't give me the index $N_{2}$, though $\endgroup$
    – user400359
    Oct 21, 2018 at 22:43
  • $\begingroup$ You don't always need an index. From the definition you can show that for convergent sequences $a_n,b_n$ we have $\lim_{n\to\infty}a_n+b_n=\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n$. Also that $\lim_{n\to\infty}a_nb_n=\lim_{n\to\infty}a_n\lim_{n\to\infty}b_n$. Then we get take limits without necessarily needing the $\epsilon,N$ definition. $\endgroup$
    – Melody
    Oct 21, 2018 at 22:48

4 Answers 4

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

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  • $\begingroup$ I edited my original post with another attempt. Also, doesn't this way assume $x_{n} \geq 0$? $\endgroup$
    – user400359
    Oct 21, 2018 at 22:20
  • $\begingroup$ I cannot get anywhere from this hint $\endgroup$
    – user400359
    Oct 21, 2018 at 22:55
  • $\begingroup$ I see your edited post. Is there a way to do it with my sequential definition? I think yours is the $\epsilon-\delta$ definition $\endgroup$
    – user400359
    Oct 21, 2018 at 23:14
  • $\begingroup$ how can I define $N_{2}$ then, though? for example, in the other problem you looked at (here: math.stackexchange.com/questions/2965155/…), I was able to write $N_{2} = N_{1}$. can something similar be done? $\endgroup$
    – user400359
    Oct 21, 2018 at 23:17
  • $\begingroup$ Given some $\epsilon>0$ we can find some $\delta>0$ and then some $N\in \mathbb{N}.$ $\endgroup$
    – mfl
    Oct 21, 2018 at 23:26
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$\textbf{Hint}$

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?

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    $\begingroup$ Your first equality is true iff $\;x_n\ge0\;$ ... $\endgroup$
    – DonAntonio
    Oct 21, 2018 at 22:07
  • $\begingroup$ Is it? If $x_n<0$, then $\sqrt{x_n}=\sqrt{-1\cdot|x_n|}=\sqrt{-1}\sqrt{|x_n|}$, therefore $(\sqrt{x_n})^2=(\sqrt{-1}\sqrt{|x_n|})^2=(\sqrt{-1})^2(\sqrt{|x_n|})^2=-1|x_n|=x_n.$ If I'm mistaken, then a proper choice of $N$ forces $x_n>0$ for all $n>N$, so we could sidestep that issue. $\endgroup$
    – Melody
    Oct 21, 2018 at 22:13
  • $\begingroup$ I edited my original post with another attempt. $\endgroup$
    – user400359
    Oct 21, 2018 at 22:20
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    $\begingroup$ @Melody That is wrong. $\;\sqrt{ab}=\sqrt a\,\sqrt b\;$ is true only if $\;a,b\ge0\;$. Within the complex numbers the above property is false in general, and within the real numbers, of course, you can not take the square root (or any even root) of a negative number. And yes, as we're taking the limit when $\;x\to1\;$ , we can assume $\;x>0\;$ ...but imo this must be explicitly said. $\endgroup$
    – DonAntonio
    Oct 21, 2018 at 22:25
  • $\begingroup$ @DonAntonio In my edited post, I'm also assuming $x_{n} > 0$, correct? So I must explicitly state we can assume $x > 0$. $\endgroup$
    – user400359
    Oct 21, 2018 at 22:28
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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

$$\left| \frac{x_n^2-1}{\sqrt{x_n}-1} -4 \right|\leq |x_n+1||\sqrt{x_n}-1|+2|x_n-1| < 3\frac{\varepsilon}{6}+2\frac\varepsilon 4 = \varepsilon,\ n\geq N.$$

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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 $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ $$\frac{2y+y^2}{\sqrt{1+y} - 1}=\frac{2y+y^2}{\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)}=4+3 y+O\left(y^2\right)$$

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