To evaluate the limit

$\lim_{x \to 0} \left(2 - \cos x\right)^{\frac{1}{\sin^2 x}}$

we can start by rewriting the expression in a form that allows us to use logarithmic properties to simplify the calculation.

Step 1: Take the Logarithm

Let $y = \left(2 - \cos x\right)^{\frac{1}{\sin^2 x}}.$ Then $\ln y = \frac{1}{\sin^2 x} \ln (2 - \cos x).$

Now, we need to find the limit of $\ln y$ as $x \to 0$: $\lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{\ln (2 - \cos x)}{\sin^2 x}.$

Step 2: Expand $2 - \cos x$ around $x = 0$

Using the Taylor expansion of $\cos x$ around $x = 0$, we have $\cos x = 1 - \frac{x^2}{2} + O(x^4).$ Therefore, $2 - \cos x = 2 - \left(1 - \frac{x^2}{2}\right) = 1 + \frac{x^2}{2} + O(x^4).$

Thus, as $x \to 0$, $\ln (2 - \cos x) \approx \ln \left(1 + \frac{x^2}{2}\right).$

Using the approximation $\ln (1 + u) \approx u$ for $u \to 0$, $\ln (2 - \cos x) \approx \frac{x^2}{2}.$

Step 3: Substitute into the Limit

Now we have $\lim_{x \to 0} \frac{\ln (2 - \cos x)}{\sin^2 x} \approx \lim_{x \to 0} \frac{\frac{x^2}{2}}{\sin^2 x}.$

Using the small-angle approximation $\sin x \approx x$ as $x \to 0$, we get $\sin^2 x \approx x^2$. Substituting this in, we find $\lim_{x \to 0} \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2}.$

Step 4: Exponentiate to Find $y$

Since $\lim_{x \to 0} \ln y = \frac{1}{2}$, we have $\lim_{x \to 0} y = e^{\frac{1}{2}} = \sqrt{e}.$

Final Answer

$\lim_{x \to 0} \left(2 - \cos x\right)^{\frac{1}{\sin^2 x}} = \sqrt{e}.$

Would you like further explanation or have questions on any part?

Here are some related questions that may interest you:

1. How do we use Taylor series to approximate limits?
2. What other functions have similar behavior near zero?
3. Why does $\sin x \approx x$ work in limits?
4. Can this approach be used for other trigonometric limits?
5. What are other common logarithmic limit techniques?

Tip: When taking limits, using approximations like Taylor series or small-angle identities often simplifies complex expressions.