# Using the sequential definition of a limit to show $\lim_{x\to 1} \frac{x^2 - 1}{\sqrt{x} - 1} = 4.$

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.

• you can twice use the fact that $$(t^2 - 1) = (t - 1)(t + 1)$$ and show that the expression actually equals to $$(\sqrt{x} + 1)(x + 1)$$ Oct 21, 2018 at 21:59
• I have used the fact, but I don't know how to proceed. I've edited my original post.
– user400359
Oct 21, 2018 at 22:30
• 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 $1+\sqrt{x_n}+x_n+(\sqrt{x_n])^3\to1+1+1+1=4$. Oct 21, 2018 at 22:38
• It doesn't give me the index $N_{2}$, though
– user400359
Oct 21, 2018 at 22:43
• 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. Oct 21, 2018 at 22:48

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 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

• I edited my original post with another attempt. Also, doesn't this way assume $x_{n} \geq 0$?
– user400359
Oct 21, 2018 at 22:20
• I cannot get anywhere from this hint
– user400359
Oct 21, 2018 at 22:55
• I see your edited post. Is there a way to do it with my sequential definition? I think yours is the $\epsilon-\delta$ definition
– user400359
Oct 21, 2018 at 23:14
• 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?
– user400359
Oct 21, 2018 at 23:17
• Given some $\epsilon>0$ we can find some $\delta>0$ and then some $N\in \mathbb{N}.$
– mfl
Oct 21, 2018 at 23:26

$$\textbf{Hint}$$

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?

• Your first equality is true iff $\;x_n\ge0\;$ ... Oct 21, 2018 at 22:07
• 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. Oct 21, 2018 at 22:13
• I edited my original post with another attempt.
– user400359
Oct 21, 2018 at 22:20
• @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. Oct 21, 2018 at 22:25
• @DonAntonio In my edited post, I'm also assuming $x_{n} > 0$, correct? So I must explicitly state we can assume $x > 0$.
– user400359
Oct 21, 2018 at 22:28

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 $$-\varepsilon/4 and so, from $$(1)$$ it follows

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

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