Question

\[ \lim_{x \to 0} \frac{\tan^{-1} \left( \frac{-x}{\sqrt{1 - x^2}} \right)}{\ln(1 - x)} = ? \]

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(\infty\)

Answer 1

The Correct Option is C

Let us evaluate the numerator and denominator around \(x = 0\): \[ \tan^{-1}\left( \frac{-x}{\sqrt{1 - x^2}} \right) \approx \tan^{-1}(-x) \approx -x
\ln(1 - x) \approx -x \Rightarrow \lim_{x \to 0} \frac{-x}{-x} = 1 \Rightarrow \text{But original had } -x \text{ so answer is } -1 \]