To evaluate the limit 

\[
\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{1 - \cos x}},
\]

we'll analyze both components as \( x \) approaches \( 0 \).

1. **Limit of \(\frac{\sin x}{x}\)**:
   \[
   \lim_{x \to 0} \frac{\sin x}{x} = 1.
   \]

2. **Limit of \(1 - \cos x\)**:
   Using the Taylor series expansion, we have:
   \[
   \cos x \approx 1 - \frac{x^2}{2} \quad \text{as } x \to 0.
   \]
   Thus,
   \[
   1 - \cos x \approx \frac{x^2}{2}.
   \]
   Hence, 
   \[
   \lim_{x \to 0} (1 - \cos x) = 0.
   \]

Now, substituting these limits back into the original limit expression:

\[
\left( \frac{\sin x}{x} \right)^{\frac{1}{1 - \cos x}} \to 1^{\infty} \quad \text{as } x \to 0.
\]

This form is indeterminate, so we can apply logarithms for simplification:

Let \( L = \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{1 - \cos x}} \).

Taking the natural logarithm:
\[
\ln L = \lim_{x \to 0} \frac{1}{1 - \cos x} \ln\left( \frac{\sin x}{x} \right).
\]

Next, we analyze \( \ln\left( \frac{\sin x}{x} \right) \):
Using the fact that \(\ln a \approx a - 1\) when \( a \) is near \( 1 \):
\[
\ln\left( \frac{\sin x}{x} \right) \approx \frac{\sin x}{x} - 1 \approx 0 \quad \text{as } x \to 0.
\]

Now, we need to find the limit of the form \(\frac{0}{0}\) using L'Hôpital's rule:

Calculate:
\[
\lim_{x \to 0} \frac{\ln\left( \frac{\sin x}{x} \right)}{1 - \cos x}.
\]

Using L'Hôpital's rule:
1. Derivative of the numerator:
   \[
   \frac{d}{dx} \ln\left( \frac{\sin x}{x} \right) = \frac{\cos x}{\sin x} - \frac{1}{x} = \cot x - \frac{1}{x}.
   \]
2. Derivative of the denominator:
   \[
   \frac{d}{dx} (1 - \cos x) = \sin x.
   \]

Now applying L'Hôpital's rule gives:
\[
\lim_{x \to 0} \frac{\cot x - \frac{1}{x}}{\sin x}.
\]

As \( x \to 0 \), this limit evaluates to \(0\), and thus:
\[
\ln L = 0 \implies L = e^0 = 1.
\]

Therefore, the final limit is:

\[
\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{1 - \cos x}} = 1.
\]

If you have any questions or need further clarification, let me know! Here are some related questions:

1. How can L'Hôpital's rule be applied in other limit problems?
2. What is the significance of the Taylor series in limit evaluations?
3. Can you explain the indeterminate forms in limits?
4. What are some common limits involving trigonometric functions?
5. How does the continuity of functions play a role in limit calculations?

**Tip:** Always check if a limit leads to an indeterminate form; it may require L'Hôpital's rule or algebraic manipulation for resolution.

Explore the evaluation of the limit lim x->0 (sin x/x)^(1/(1-cos x)), detailing the steps using L'Hôpital's Rule and Taylor series expansion. This problem is relevant for advanced students studying calculus and limits.

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