Question

Which one of the following is the value of \( \displaystyle\lim_{x \to 0} \frac{e^{x} - x - 1}{\cos x - 1} \)?

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

Hint:

Both numerator and denominator vanish at x=0; apply L'Hopital's rule twice (or expand as Taylor series) - the leading terms are x^2/2 over -x^2/2.

Answer 1

The Correct Option is A

Expand as Maclaurin series: numerator = x^2/2 + x^3/6 + ..., denominator = -x^2/2 + x^4/24 - .... Dividing both by x^2 and taking x to 0 leaves only the leading constant terms, giving ratio = (1/2)/(-1/2) = -1, matching L'Hopital's result exactly.

\[ -1 \ \Rightarrow \ \text{Option (A)} \]