The given limit is:

$\lim_{x \to 0} \frac{\sqrt{\sin x}}{\sin (\sqrt{x})}$

To solve this limit:

1. First step: Check direct substitution. When $x = 0$, $\sqrt{\sin 0} = \sqrt{0} = 0$ and $\sin (\sqrt{0}) = \sin(0) = 0$ So, the expression gives a form of $\frac{0}{0}$, which is an indeterminate form. This suggests we can apply L'Hopital's Rule.

2. Applying L'Hopital's Rule: Since the limit is in the indeterminate form $\frac{0}{0}$, we can differentiate the numerator and denominator separately with respect to $x$.

   • Differentiate the numerator: $\frac{d}{dx} \left( \sqrt{\sin x} \right) = \frac{1}{2\sqrt{\sin x}} \cdot \cos x$

   • Differentiate the denominator: $\frac{d}{dx} \left( \sin(\sqrt{x}) \right) = \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$

3. Substitute these into L'Hopital's Rule: The new limit becomes: $\lim_{x \to 0} \frac{\frac{1}{2\sqrt{\sin x}} \cdot \cos x}{\cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}}$

   Simplifying the constants: $\lim_{x \to 0} \frac{\sqrt{x} \cdot \cos x}{\sqrt{\sin x} \cdot \cos(\sqrt{x})}$

4. Evaluate the limit as $x \to 0$:

   • $\cos(0) = 1$, so the cosine terms approach 1.
   • As $x \to 0$, $\sin x \approx x$, so $\sqrt{\sin x} \approx \sqrt{x}$.

   Therefore, the limit simplifies to: $\lim_{x \to 0} \frac{\sqrt{x}}{\sqrt{x}} = 1$

Thus, the value of the limit is: $\boxed{1}$

Would you like further clarification or details on any part of the process?

Relative Questions:

1. What conditions allow the use of L'Hopital's Rule?
2. How does the approximation $\sin x \approx x$ help in limit evaluations?
3. What are other methods besides L'Hopital's Rule to solve indeterminate forms?
4. Can the limit be solved using series expansions?
5. How do we differentiate composite functions like $\sin(\sqrt{x})$?

Tip:

For limits involving trigonometric functions near 0, small-angle approximations (like $\sin x \approx x$) are often very useful for simplifying expressions.