We are tasked with solving the following limit:

$\lim_{x \to 0} \frac{\sqrt{x+2} - \sqrt{2}}{x(1+x)}.$

Step 1: Simplify the numerator

The numerator involves a difference of square roots:
$\sqrt{x+2} - \sqrt{2}.$
To simplify, multiply and divide by the conjugate:
$\frac{\sqrt{x+2} - \sqrt{2}}{\sqrt{x+2} - \sqrt{2}} \cdot \frac{\sqrt{x+2} + \sqrt{2}}{\sqrt{x+2} + \sqrt{2}} = \frac{(\sqrt{x+2})^2 - (\sqrt{2})^2}{\sqrt{x+2} + \sqrt{2}}.$

This simplifies to:
$\frac{(x+2) - 2}{\sqrt{x+2} + \sqrt{2}} = \frac{x}{\sqrt{x+2} + \sqrt{2}}.$

Thus, the numerator becomes:
$\sqrt{x+2} - \sqrt{2} = \frac{x}{\sqrt{x+2} + \sqrt{2}}.$

Step 2: Rewrite the full expression

Substitute the simplified numerator into the original limit:
$\lim_{x \to 0} \frac{\sqrt{x+2} - \sqrt{2}}{x(1+x)} = \lim_{x \to 0} \frac{\frac{x}{\sqrt{x+2} + \sqrt{2}}}{x(1+x)}.$

Simplify the fraction by dividing $x$ out:
$\lim_{x \to 0} \frac{x}{\sqrt{x+2} + \sqrt{2}} \cdot \frac{1}{x(1+x)} = \lim_{x \to 0} \frac{1}{(\sqrt{x+2} + \sqrt{2})(1+x)}.$

Step 3: Evaluate the limit as $x \to 0$

As $x \to 0$, the terms simplify as follows:

• $\sqrt{x+2} \to \sqrt{2}$,
• $1 + x \to 1$.

Thus, the denominator becomes:
$(\sqrt{x+2} + \sqrt{2})(1+x) \to (\sqrt{2} + \sqrt{2})(1) = 2\sqrt{2}.$

Final Answer:

$\lim_{x \to 0} \frac{\sqrt{x+2} - \sqrt{2}}{x(1+x)} = \frac{1}{2\sqrt{2}}.$

Do you want further clarifications or details on any step?

Here are 5 related questions to explore:

1. How can we generalize limits involving square roots and conjugates?
2. What happens if we encounter $x \to a$ instead of $x \to 0$?
3. How does multiplying by conjugates simplify expressions?
4. Can L'Hôpital's Rule also solve this limit?
5. How do we handle more complex roots in limits?

Tip: When dealing with square roots in limits, multiplying by the conjugate is a key strategy to simplify the expression.