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The limit problem provided is:

\[
\lim_{{x \to 0}} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{\sin^3 x}
\]

### Step-by-Step Solution:

1. **Substitution Check**:
   Let's substitute \(x = 0\) directly into the expression to see what form we get:
   
   \[
   \frac{\sqrt{1 + \tan(0)} - \sqrt{1 + \sin(0)}}{\sin^3(0)} = \frac{\sqrt{1 + 0} - \sqrt{1 + 0}}{0^3} = \frac{1 - 1}{0} = \frac{0}{0}
   \]

   This gives us an indeterminate form \(\frac{0}{0}\), so we can apply **L'Hopital's Rule**, or simplify it by using **series expansions** around \(x = 0\).

2. **Taylor Series Expansion**:
   Instead of using L'Hopital's Rule multiple times, we'll expand both \(\sqrt{1 + \tan x}\) and \(\sqrt{1 + \sin x}\) as Taylor series around \(x = 0\). 

   - **Taylor expansion of \(\tan x\) around \(x = 0\)**:
     \[
     \tan x = x + \frac{x^3}{3} + O(x^5)
     \]
     So, 
     \[
     \sqrt{1 + \tan x} = \sqrt{1 + x + \frac{x^3}{3} + O(x^5)}.
     \]
     We expand this using the binomial approximation for small \(x\):
     \[
     \sqrt{1 + x + \frac{x^3}{3}} \approx 1 + \frac{1}{2}x + \frac{1}{2} \cdot \frac{x^3}{3} = 1 + \frac{1}{2}x + \frac{x^3}{6}.
     \]

   - **Taylor expansion of \(\sin x\) around \(x = 0\)**:
     \[
     \sin x = x - \frac{x^3}{6} + O(x^5)
     \]
     So,
     \[
     \sqrt{1 + \sin x} = \sqrt{1 + x - \frac{x^3}{6}}.
     \]
     Using the binomial expansion:
     \[
     \sqrt{1 + x - \frac{x^3}{6}} \approx 1 + \frac{1}{2}x - \frac{x^3}{12}.
     \]

3. **Subtract the Expansions**:
   Now, subtract the two series expansions:
   \[
   \left(1 + \frac{1}{2}x + \frac{x^3}{6}\right) - \left(1 + \frac{1}{2}x - \frac{x^3}{12}\right) = \frac{x^3}{6} + \frac{x^3}{12} = \frac{x^3}{4}.
   \]

4. **Substitute into the Limit Expression**:
   The numerator simplifies to \(\frac{x^3}{4}\). Now, the limit becomes:
   \[
   \lim_{{x \to 0}} \frac{\frac{x^3}{4}}{\sin^3 x}.
   \]

5. **Simplify the Denominator**:
   As \(x \to 0\), we know that \(\sin x \approx x\), so \(\sin^3 x \approx x^3\). Thus, the limit becomes:
   \[
   \lim_{{x \to 0}} \frac{\frac{x^3}{4}}{x^3} = \lim_{{x \to 0}} \frac{1}{4} = \frac{1}{4}.
   \]

6. **Final Answer**:
   The value of the limit is:
   \[
   \lim_{{x \to 0}} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{\sin^3 x} = \frac{1}{4}.
   \]

Would you like further clarification or additional steps?

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Here are some related questions for further exploration:

1. What are the key steps in using Taylor series for limit evaluation?
2. How can you identify when to apply L'Hopital's Rule versus series expansion?
3. What is the importance of small-angle approximations in trigonometric limits?
4. How does binomial expansion help in simplifying complex square root expressions?
5. Can higher-order terms in the Taylor series affect the final answer for limits?

**Tip**: When evaluating limits involving radicals and trigonometric functions, using Taylor series expansions can often simplify the process significantly.

Evaluate the limit: lim (x -> 0) (sqrt(1 + tan(x)) - sqrt(1 + sin(x))) / sin^3(x).

Solve the limit problem involving the expression (sqrt(1 + tan(x)) - sqrt(1 + sin(x))) / sin^3(x) as x approaches 0. This solution uses Taylor series expansions and small-angle approximations, suitable for university-level calculus students.

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