Limit of $\frac{x^5-1}{x^2-1}$

Question

I need to determine if the following limit exists.

$$\lim_{x\to 1}\frac{x^5-1}{x^2-1}$$

I've already proved using L'Hospital that this limit exists and should equal to $\frac{5}{2}$, but unfortunately I'm not allowed to used anything more than basic analysis for functions, i.e. basic definitions of convergence(at most continuity).

Do you know that $x^{5}-1$ and $x^{2}-1$ have a factor in common? — Frenzy Li, Dec 14 '17 at 21:45
If the numerator and denominator both tend to zero, the implication is that there is a first-order factor of $x-1$. With the polynomials this becomes an actual factor. L'Hospital is in effect another way of cancelling the first order factor. — Mark Bennet, Dec 14 '17 at 21:53
Bonus exercise for N. Wolf: Limit of $\frac{x^4-1}{x^3-1}$ as $x\to 1$. — Frenzy Li, Dec 14 '17 at 22:05

score 4 · Answer 1 · answered Dec 14 '17 at 22:03

Without using l'Hopital (directly), but if you know derivatives, then by definition:

$$ \lim_{x \to 1} \frac{x^5-1}{x-1} = \big(x^5 - 1\big)'\bigg|_{x=1}=5x^4\bigg|_{x=1} = 5 \\[5px] \lim_{x \to 1} \frac{x^2-1}{x-1} = \big(x^2 - 1\big)'\bigg|_{x=1}=2 x\bigg|_{x=1} = 2 $$

It follows that the given limit is:

$$ \lim_{x \to 1} \cfrac{x^5-1}{x^2-1} = \lim_{x \to 1} \cfrac{\;\;\cfrac{x^5-1}{x-1}\;\; }{ \cfrac{x^2-1}{x-1} } = \cfrac{\;\;\lim_{x \to 1} \cfrac{x^5-1}{x-1}\;\; }{ \lim_{x \to 1} \cfrac{x^2-1}{x-1} } = \frac{5}{2} $$

score 3 · Accepted Answer · answered Dec 14 '17 at 21:46

3

$$\frac{x^5-1}{x^2-1}=\frac{(x-1)(1+x+x^2+x^3+x^4)}{(x-1)(x+1)}$$ and so the limit is $5/2$.

answered Dec 14 '17 at 21:46

uniquesolution

18,541

score 3 · Answer 3 · edited Dec 14 '17 at 22:01

3

\begin{align}\lim_{x\to1}\frac{x^5-1}{x^2-1}&=\lim_{x\to1}\frac{(x-1)(x^4+x^3+x^2+x+1)}{(x-1)(x+1)}\\&=\lim_{x\to1}\frac{(x^4+x^3+x^2+x+1)}{(x+1)}\\&=\frac52\end{align}

edited Dec 14 '17 at 22:01

answered Dec 14 '17 at 21:48

José Carlos Santos

427,504

Typo alert: You ommited $(x-1)$ in the numerator after you factored but before you cancelled. – fleablood Dec 14 '17 at 22:00
@fleablood Thank you. I would edit my answer, of course, but someone else did it for me. – José Carlos Santos Dec 14 '17 at 22:04

user · Answer 4 · 2017-12-14T21:54:52.543

2

$$\frac{x^5-1}{x^2-1}=\frac{(x-1)(x^4+x^3+x^2+x+1)}{(x-1)(x+1)}=\frac{(x^4+x^3+x^2+x+1)}{(x+1)}\to\frac52$$

edited Dec 14 '17 at 21:54

answered Dec 14 '17 at 21:47

user

154,566

There's a missing ")". – Botond Dec 14 '17 at 21:54
ops...fixed...thanks! – user Dec 14 '17 at 21:55

score 2 · Answer 5 · edited Dec 14 '17 at 21:56

2

using Horner's method:
$$x^5 -1 =(x-1)(x^4 +x^3 +x^2 +x+1)$$ $$x^2 -1=(x-1)(x+1)$$

edited Dec 14 '17 at 21:56

Botond

11,938

answered Dec 14 '17 at 21:48

score 0 · Answer 6 · answered Dec 15 '17 at 17:20

Note that $$ \lim_{x\to 1}\frac{x^5-1}{x-1}=f'(1)=5 $$ where $f(x)=x^5$ (so $f'(x)=5x^4$) by the definition of the derivative. Hence $$ \lim_{x\to 1}\frac{x^5-1}{x^2-1}=\lim_{x\to 1} \frac{1}{x+1}\times \lim_{x\to 1}\frac{x^5-1}{x-1}=\frac{1}{2}\times5=\frac{5}{2}. $$

Limit of $\frac{x^5-1}{x^2-1}$

6 Answers6