Question

lim x->1 (x^n-1)/(x^m-1) =?

Answer 1

You can apply L'Hopital's Rule, since it's of the \(\frac{0}{0}\) form:
\[\lim_{x \rightarrow 1}{nx^{n-1} \over mx^m-1}={n \over m}\]

Answer 2

\[\lim_{x \rightarrow 1}{nx^{n-1} \over mx^{m-1}}={n \over m} \]

Answer 3

\[\lim_{x \rightarrow 1}\frac{x^{n}-1}{x^{m}-1}=\lim_{x \rightarrow 1}\frac{(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)}{(x-1)(x^{m-1}+x^{m-2}+\cdots+x+1)}\]

\[\lim_{x \rightarrow 1}\frac{x^{n-1}+x^{n-2}+\cdots+x+1}{x^{m-1}+x^{m-2}+\cdots+x+1}\]

\[=\begin{array}{cc}n \text{ terms}\\ \frac{\overbrace{1+1+\cdots+1}}{\underbrace{1+1+\cdots+1}}\\m \text{ terms}\end{array}={n \over m}\]

Answer 4

Nice Zarkon!