Question

$\underset{x\to 1}{\lim }\frac{x^m-1}{x^n-1}\mathrm{is}$

​(a) 1

(b) $\frac{m}{n}$

(c) $-\frac{m}{n}$

(d) $\frac{m^2}{n^2}$

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to 1\end{array}\frac{x^m-1}{x^n-1} \\ \mathrm{since}\frac{x^n-a^n}{x-a}\stackrel{x+a}{\to }n{a}^{n-1} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to a\end{array}\frac{x^n-a^n}{x-a}=n{a}^{n-1} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 1\end{array}\frac{x^m-1}{x-1}\times \frac{x-1}{x^n-1} \\ \mathrm{i}.\mathrm{e}.\left(\begin{array}{c}\lim \\ x\to 1\end{array}\frac{x^m-1}{x-1}\right)\left(\begin{array}{c}\lim \\ x\to 1\end{array}\frac{x-1}{x^n-1}\right) \\ =m1\times \frac{1}{n} \\ =1\times \frac{m}{n}=\frac{m}{n} \end{gathered}$$

Hence, the correct answer is option B.