The given expression is:

$\lim_{x \to 0} \frac{1 - \sqrt{\cos x}}{x}.$

To evaluate this limit, we follow these steps:

Step 1: Analyze the numerator

The numerator is $1 - \sqrt{\cos x}$. Using a Taylor expansion for $\cos x$ near $x = 0$, we know: $\cos x \approx 1 - \frac{x^2}{2}.$ Taking the square root: $\sqrt{\cos x} \approx \sqrt{1 - \frac{x^2}{2}} \approx 1 - \frac{x^2}{4},$ (using the binomial expansion for small values of $x$).

So, the numerator becomes: $1 - \sqrt{\cos x} \approx 1 - \left( 1 - \frac{x^2}{4} \right) = \frac{x^2}{4}.$

Step 2: Simplify the fraction

Substitute the approximation into the limit: $\frac{1 - \sqrt{\cos x}}{x} \approx \frac{\frac{x^2}{4}}{x} = \frac{x}{4}.$

Step 3: Take the limit

As $x \to 0$: $\frac{x}{4} \to 0.$

Final Answer:

$\lim_{x \to 0} \frac{1 - \sqrt{\cos x}}{x} = 0.$

Would you like further clarification or a deeper explanation?

Here are 5 related questions you might find helpful:

1. How does the Taylor expansion of trigonometric functions work?
2. Why does the binomial expansion apply to $\sqrt{1 - x}$ for small $x$?
3. Can we use L’Hôpital’s Rule to solve this limit instead?
4. How do higher-order approximations of $\cos x$ affect this solution?
5. What are some other common limits involving square roots and trigonometric functions?

Tip: When dealing with square roots and trigonometric limits, approximating via series expansion can simplify complex expressions!