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

Step 1: Check the form. Substituting x=0 gives 0/0, an indeterminate form, so L'Hopital's rule applies.

Step 2: Apply L'Hopital's rule once. \( \frac{e^x-1}{-\sin x} \), still 0/0 at x=0, apply again.

Step 3: Apply a second time. \( \frac{e^x}{-\cos x} \), now the denominator is nonzero at x=0.

Step 4: Substitute. \( \frac{e^0}{-\cos 0}=\frac{1}{-1}=-1 \).

\[ \boxed{\lim_{x\to 0}\frac{e^{x}-x-1}{\cos x-1} = -1 \ \Rightarrow \ \text{Option (A)}} \]