Question

${lim}_{\alpha \to 0}\frac{x^{\alpha }-1}{\alpha }$ is equal to

• A. log x
• B. 0
• C. x log x
• D. $\infty$

Answer 1

The correct option is A log x

${lim}_{\alpha \to 0}\frac{x^{\alpha }-1}{\alpha }=\frac{1-1}{0}=\frac{0}{0}$

Apply L ' Hospital's rule

=${lim}_{\alpha \to 0}\frac{x^{\alpha }logx}{1}=logx$