Question

$\underset{x\to 1}{\lim }\left(\frac{x^m-1}{x^n-1}\right)=$

• A. $\frac{n}{m}$
• B. $\frac{m}{n}$
• C. $\frac{2m}{n}$
• D. $\frac{2n}{m}$

Answer 1

The correct option is B

$\frac{m}{n}$

Explanation for the given option:

Find the value of $\underset{x\to 1}{\lim }\left(\frac{x^m-1}{x^n-1}\right)$:

The given equation is,

$$\begin{gathered} I=\underset{x\to 1}{\lim }\left(\frac{x^m-1}{x^n-1}\right) \\ I=\left(\frac{1^m-1}{1^n-1}\right) \\ I=\frac{0}{0}\left[\text{indeterminent form}\right] \end{gathered}$$

Using the L' Hospital rule

$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

$$\begin{gathered} I=\underset{x\to 1}{\lim }\left(\frac{m.x^{m-1}}{n.x^{n-1}}\right) \\ I=\frac{m}{n}\left(\underset{x\to 1}{\lim }\frac{x^{m-1}}{x^{n-1}}\right) \\ I=\frac{m}{n}\left(\frac{1^{m-1}}{1^{n-1}}\right) \\ I=\frac{m}{n}\left(1\right) \\ I=\frac{m}{n} \end{gathered}$$

Hence, the correct answer is option (B).