The given mathematical expression is:

$\lim_{x \to 0} \frac{\sin^3(2x)}{(x + 1) - \sqrt{10 + 2x}}$

Steps to solve:

1. Evaluate the limit: The first step is to substitute $x = 0$ into the expression to check if the limit exists.

   Substituting $x = 0$: $\frac{\sin^3(2(0))}{(0 + 1) - \sqrt{10 + 2(0)}} = \frac{0^3}{1 - \sqrt{10}} = 0$ The form is $\frac{0}{0}$, which is indeterminate. Therefore, we can apply L'Hôpital's Rule.

2. L'Hôpital's Rule: Differentiate the numerator and denominator separately with respect to $x$.

   • Numerator: The numerator is $\sin^3(2x)$. Using the chain rule: $\frac{d}{dx} \left(\sin^3(2x)\right) = 3\sin^2(2x) \cdot \frac{d}{dx} (\sin(2x)) = 3\sin^2(2x) \cdot 2\cos(2x) = 6\sin^2(2x)\cos(2x)$

   • Denominator: The denominator is $(x + 1) - \sqrt{10 + 2x}$. Differentiating term by term: $\frac{d}{dx} \left((x + 1) - \sqrt{10 + 2x}\right) = 1 - \frac{1}{2\sqrt{10 + 2x}} \cdot 2 = 1 - \frac{1}{\sqrt{10 + 2x}}$

3. Recompute the limit: Now, apply the limit $x \to 0$ to the differentiated expressions:

   • For the numerator: $6\sin^2(2(0))\cos(2(0)) = 6 \cdot 0^2 \cdot 1 = 0$
   • For the denominator: $1 - \frac{1}{\sqrt{10 + 2(0)}} = 1 - \frac{1}{\sqrt{10}} = 1 - \frac{1}{\sqrt{10}}$

Thus, the limit simplifies to:

$\lim_{x \to 0} \frac{0}{1 - \frac{1}{\sqrt{10}}} = 0$

Final Answer:

The limit is $0$.

Would you like a detailed explanation of any step or further clarification?

Here are 5 related questions for practice:

1. What is L'Hôpital's Rule and when do we use it?
2. How do you differentiate a composite function like $\sin^3(2x)$?
3. What are common indeterminate forms that suggest using L'Hôpital's Rule?
4. How would this limit change if the sine function had a higher power?
5. How does the square root function affect the differentiation of the denominator?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying L'Hôpital's Rule.