We are asked to evaluate the following limit:

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

Step 1: Check direct substitution

Let's first substitute $x = 0$ into the expression to see if we get an indeterminate form.

Numerator:

$e^x - x - 1 \quad \text{at} \quad x = 0 \Rightarrow e^0 - 0 - 1 = 1 - 0 - 1 = 0$

Denominator:

$\cos x - 1 \quad \text{at} \quad x = 0 \Rightarrow \cos(0) - 1 = 1 - 1 = 0$

Thus, we have the indeterminate form $\frac{0}{0}$. This suggests that we can apply L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule

L'Hopital's Rule states that if the limit of a fraction is of the form $\frac{0}{0}$, we can differentiate the numerator and the denominator and then take the limit of the resulting expression.

Differentiate the numerator:

The numerator is $e^x - x - 1$. Differentiating with respect to $x$: $\frac{d}{dx}(e^x - x - 1) = e^x - 1$

Differentiate the denominator:

The denominator is $\cos x - 1$. Differentiating with respect to $x$: $\frac{d}{dx}(\cos x - 1) = -\sin x$

Step 3: Recalculate the limit after differentiation

Now, the limit becomes:

$\lim_{x \to 0} \frac{e^x - 1}{-\sin x}$

Substitute $x = 0$ into the new expression:

$\frac{e^0 - 1}{-\sin(0)} = \frac{1 - 1}{0} = \frac{0}{0}$

Again, we have an indeterminate form, so we apply L'Hopital's Rule again.

Step 4: Apply L'Hopital's Rule again

We differentiate the numerator and denominator once more.

Differentiate the numerator:

The numerator is $e^x - 1$. Differentiating with respect to $x$: $\frac{d}{dx}(e^x - 1) = e^x$

Differentiate the denominator:

The denominator is $-\sin x$. Differentiating with respect to $x$: $\frac{d}{dx}(-\sin x) = -\cos x$

Step 5: Recalculate the limit after the second differentiation

Now, the limit becomes:

$\lim_{x \to 0} \frac{e^x}{-\cos x}$

Substitute $x = 0$:

$\frac{e^0}{-\cos(0)} = \frac{1}{-1} = -1$

Final Answer:

The value of the limit is:

$\boxed{-1}$

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Would you like more details on the steps or have any further questions? Here are 5 related questions:

1. How does L'Hopital's Rule work in general for indeterminate forms?
2. Can the limit be computed without using L'Hopital's Rule?
3. What are some common limits that lead to indeterminate forms like $\frac{0}{0}$?
4. What happens if we apply a Taylor series expansion instead of L'Hopital's Rule?
5. What is the relationship between the Taylor series for $e^x$ and $\cos x$ near $x = 0$?

Tip: Always check if you have an indeterminate form first before applying L'Hopital's Rule, as it can often simplify the process significantly.