# How to find the limit without L'Hospital?

$$\lim_{x\rightarrow 1} \frac{x^m-1}{x^n-1}$$

I found this in a book (recommended by the professor) without solutions or hints.

My problem is that I can't find definitions about this kind of problems and I have no idea how to start, any kind of help would be great. So far I've been dealing with "$\lim_{x\rightarrow ∞}$" only. Also without L'Hospital, because we haven't had this in lecture.

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$$\lim_{x\rightarrow 1} \frac{x^m-1}{x^n-1}$$

$$=\lim_{y\rightarrow 0} \frac{(1+y)^m-1}{(1+y)^n-1}\text{ putting } y=1+x, x\to1\implies y\to0$$

Using Generalized binomial theorem, $(1+y)^r=1+r y+\frac{r(r-1)}{2!}y^2+\frac{r(r-1)(r-2)}{3!}y^3+\cdots=1+ry+O(y^2)$

$$\implies \lim_{y\rightarrow 0} \frac{(1+y)^m-1}{(1+y)^n-1}=\lim_{y\rightarrow 0} \frac{1+my+O(y^2)-1}{1+ny+O(y^2)-1}=\frac mn\text{ as }y\ne0 \text{ as }y\to0$$

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+1 , just because this is the most generalized proof –  Fermé somme Jun 9 '13 at 14:42

Hint: $$\frac{x^m-1}{x^n-1} = \frac{(x-1)(x^{m-1} + x^{m-2} + \dots + x + 1)}{(x-1)(x^{n-1} + x^{n-2} + \dots + x + 1)}$$

(I assumed that $n, m$ stand for positive integers)

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The factorization is not applicable for $m,n \in \mathbb{R}$ –  Fermé somme Jun 9 '13 at 14:42
I assume that they are positive integers, as the notation suggests. –  Siméon Jun 9 '13 at 14:49
I can't see anywhere in the question that suggests $m$ and $n$ are naturals . And the value of the limit $\frac{m}{n}$ Is valid for every reals $m$ and $n$. So your assumption only weakens the result. –  Fermé somme Jun 9 '13 at 14:54
Why not complex numbers ? Seriously, I juste gave an elementary hint for what I thought the OP was interested in. –  Siméon Jun 9 '13 at 16:05
I wasn't interested at complex numbers just because $x^m$ is multivalued when $m$ belongs to the set of complex numbers.Well, I don't have any aversion towards your post . –  Fermé somme Jun 9 '13 at 16:37

$$\lim_{x\rightarrow 1} \frac{x^m-1}{x^n-1}$$ There is a formula : $$\lim_{x\rightarrow a}\dfrac{x^u-a^u}{x-a}=u\cdot a^{u-1}$$ proof so:

$$\lim_{x\rightarrow 1}\frac{x^m-1^m}{x-1}\times\dfrac{x-1}{x^n-1^n}$$ $$\lim_{x\rightarrow 1}\frac{x^m-1^m}{x-1}\times\dfrac{1}{\dfrac{x^n-1^n}{x-1}}$$ $$m\cdot 1^{m-1}\times \dfrac{1}{n\cdot 1^{n-1}}$$ $$\dfrac mn$$

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The "formula" you mention is just the derivative of $f(x)=x^u$ at $a$. It should be $\lim_{x\to a}$ rather than $\lim_{x\to u}$. –  egreg Jun 9 '13 at 20:28

Hint: $x^k-1=(x-1)(x^{k-1}+x^{k-2}+\ldots +x^2+x+1)$

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