The given mathematical expression is:
$\lim_{x \to 0} (\cos(\sqrt{x}))^{\frac{1}{x}}$
We need to evaluate this limit without using L'Hôpital's rule.

Solution:

Step 1: Rewrite the function for simplification

We start by taking the natural logarithm to simplify the expression: $y = (\cos(\sqrt{x}))^{\frac{1}{x}}$ Taking the natural logarithm of both sides: $\ln y = \frac{1}{x} \ln(\cos(\sqrt{x}))$

Now, the problem reduces to finding the limit of $\ln y$ as $x \to 0$: $\ln y = \frac{\ln(\cos(\sqrt{x}))}{x}$ Thus, we need to evaluate: $\lim_{x \to 0} \frac{\ln(\cos(\sqrt{x}))}{x}$

Step 2: Expand $\cos(\sqrt{x})$ using its Taylor series

For small $x$, the Taylor expansion of $\cos(\sqrt{x})$ around $x = 0$ is: $\cos(\sqrt{x}) = 1 - \frac{x}{2} + \mathcal{O}(x^2)$ So, $\ln(\cos(\sqrt{x})) = \ln\left(1 - \frac{x}{2} + \mathcal{O}(x^2)\right)$

Using the approximation $\ln(1 + u) \approx u$ for small $u$, we get: $\ln(\cos(\sqrt{x})) \approx -\frac{x}{2}$

Step 3: Simplify the numerator

Substituting this approximation back, the numerator becomes: $\ln(\cos(\sqrt{x})) \approx -\frac{x}{2}$

Thus, the fraction becomes: $\frac{\ln(\cos(\sqrt{x}))}{x} \approx \frac{-\frac{x}{2}}{x} = -\frac{1}{2}$

Step 4: Conclude the logarithm of the limit

So, we have: $\ln y = -\frac{1}{2}$

Exponentiating both sides gives: $y = e^{-\frac{1}{2}}$

Final Answer:

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

Do you want further clarifications or step-by-step elaboration?
Here are 5 related questions to explore further:

1. How does Taylor expansion help approximate functions like $\cos(\sqrt{x})$?
2. What is the significance of taking natural logarithms when solving limits involving exponents?
3. Why is $\ln(1 + u) \approx u$ valid for small $u$?
4. What is the geometric interpretation of the result $e^{-\frac{1}{2}}$?
5. Can this limit be evaluated using a different approach, such as substitution?

Tip: Understanding Taylor series is extremely useful for solving limits involving trigonometric functions and logarithms!