The limit you provided is:

$\lim_{{x \to 0}} \frac{e^x + e^{-x} - 2}{1 - \cos(2x)}$

Step 1: Simplify the numerator

The numerator is $e^x + e^{-x} - 2$. This expression can be recognized as part of the expansion for the hyperbolic cosine function:

$e^x + e^{-x} = 2 \cosh(x)$

So, the numerator becomes:

$e^x + e^{-x} - 2 = 2 \cosh(x) - 2 = 2 (\cosh(x) - 1)$

Using the approximation $\cosh(x) \approx 1 + \frac{x^2}{2}$ for small $x$, we get:

$2 (\cosh(x) - 1) \approx 2 \left( \frac{x^2}{2} \right) = x^2$

Thus, near $x = 0$, the numerator approximates to $x^2$.

Step 2: Simplify the denominator

The denominator is $1 - \cos(2x)$. Using the small angle approximation $\cos(2x) \approx 1 - \frac{(2x)^2}{2} = 1 - 2x^2$, we have:

$1 - \cos(2x) \approx 2x^2$

Step 3: Compute the limit

Now, substitute these approximations back into the original limit:

$\lim_{{x \to 0}} \frac{x^2}{2x^2}$

This simplifies to:

$\lim_{{x \to 0}} \frac{1}{2} = \frac{1}{2}$

Final Answer:

The value of the limit is $\frac{1}{2}$.

Would you like further details on any of the steps or concepts? Here are a few follow-up questions to deepen your understanding:

1. What is the small-angle approximation for trigonometric functions like $\cos(x)$?
2. How is the hyperbolic cosine function $\cosh(x)$ related to exponential functions?
3. Why does the numerator $e^x + e^{-x} - 2$ simplify using the hyperbolic cosine approximation?
4. How can L'Hopital's Rule be applied to limits of the indeterminate form $\frac{0}{0}$?
5. What are some common techniques for solving limits involving trigonometric functions?

Tip: In limits involving trigonometric and exponential functions, small-angle approximations can significantly simplify the problem near $x = 0$.