Question

Evaluate: $\underset{x\to 1}{lim}\frac{x^{\frac{-2}{3}}-1}{x^{\frac{-3}{4}}-1}$

• A. $\frac{5}{9}$
• B. $\frac{9}{5}$
• C. $\frac{8}{9}$
• D. $0$

Answer 1

The correct option is C $\frac{8}{9}$

Given: $\underset{x\to 1}{lim}\frac{x^{\frac{-2}{3}}-1}{x^{\frac{-3}{4}}-1}$

$={lim}_{x\to 1}\frac{\frac{-2}{3}{x}^{\frac{-5}{3}}}{\frac{-3}{4}{x}^{\frac{-7}{4}}}$.......$(Using\frac{d}{dx}{x}^n=n{x}^{n-1})$

$=\frac{8}{9}\underset{x\to 1}{lim}{x}^{\frac{1}{12}}$

$=\frac{8}{9}$

Hence, option ${}^{\prime }{C}^{\prime }$ is correct.