Question

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

• A. $1$
• B. $\frac{m}{n}$
• C. $-\frac{m}{n}$
• D. $\frac{m^2}{n^2}$

Answer 1

Answer: (B)

$\frac{m}{n}$

${lim}_{x\to 1}\frac{x^m-1}{x^n-1}=\frac{1^m-1}{1^n-1}=\frac{0}{0}$ This is an indefinite state.

By L'Hospitals rule, if we have an indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

$\therefore {lim}_{x\to 1}\frac{x^m-1}{x^n-1}={lim}_{x\to 1}\frac{\frac{d}{dx}(x^m-1)}{\frac{d}{dx}(x^n-1)}$

$={lim}_{x\to 1}\frac{m{x}^{m-1}+(m-1)x^{m-2}+------}{n{x}^{n-1}+(n-1)x^{n-2}+------}$ -------- By Chain Rule of Differentiation

$=\frac{m{1}^{m-1}+(m-1)1^{m-2}+------}{n{1}^{n-1}+(n-1)1^{n-2}+------}$

$=\frac{m+(m-1)+(m-2)+(m-3)------+1}{n+(n-1)+(n-2)+(n-3)+------+1}$

Simplifying this we get,

$=\frac{m}{n}$

$\therefore {lim}_{x\to 1}\frac{x^m-1}{x^n-1}=\frac{m}{n}$