Evaluate Limit: $\lim_{x \to 0} \frac{e^x - x - 1}{\cos x - 1}$

We need to find the limit: $ L = \lim_{x \to 0} \frac{e^x - x - 1}{\cos x - 1} $ Substituting $x=0$ into the expression gives $\frac{e^0 - 0 - 1}{\cos 0 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form. We can use L'Hôpital's Rule.

Apply L'Hôpital's Rule First Time

Take the derivative of the numerator and the denominator separately:

• Derivative of numerator ($e^x - x - 1$): $\frac{d}{dx}(e^x - x - 1) = e^x - 1$
• Derivative of denominator ($\cos x - 1$): $\frac{d}{dx}(\cos x - 1) = -\sin x$

The limit becomes: $ L = \lim_{x \to 0} \frac{e^x - 1}{-\sin x} $ Substituting $x=0$ gives $\frac{e^0 - 1}{-\sin 0} = \frac{1 - 1}{0} = \frac{0}{0}$. This is still an indeterminate form, so we apply L'Hôpital's Rule again.

Apply L'Hôpital's Rule Second Time

Take the derivative of the new numerator and denominator:

• Derivative of numerator ($e^x - 1$): $\frac{d}{dx}(e^x - 1) = e^x$
• Derivative of denominator ($-\sin x$): $\frac{d}{dx}(-\sin x) = -\cos x$

The limit now is: $ L = \lim_{x \to 0} \frac{e^x}{-\cos x} $

Final Limit Evaluation

Substitute $x=0$ into the final expression: $ L = \frac{e^0}{-\cos 0} = \frac{1}{-1} = -1 $ Thus, the value of the limit is -1.