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.

Solution 1:

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

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

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

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

Solution 2:

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)

Solution 3:

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

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