To evaluate the limit

$\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$

we can use the technique of rationalizing the numerator. This involves multiplying the numerator and the denominator by the conjugate of the numerator.

Steps:

1. Multiply by the conjugate: $\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2} = \lim_{x \to 0} \frac{(x+4) - 4}{x(\sqrt{x+4} + 2)}$

2. Simplify the expression: $= \lim_{x \to 0} \frac{x}{x(\sqrt{x+4} + 2)}$

3. Cancel $x$ in the numerator and denominator: $= \lim_{x \to 0} \frac{1}{\sqrt{x+4} + 2}$

4. Substitute $x = 0$ in the simplified expression: $= \frac{1}{\sqrt{0+4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$

Final Answer:

$\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} = \frac{1}{4}$

Would you like further clarification on any step?

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1. What is the conjugate, and why is it useful in solving limits?
2. How would this limit change if $x$ approached a different value?
3. Can L'Hôpital's Rule be used here, and would it yield the same answer?
4. How would this problem be different if the expression were $\frac{\sqrt{x+4} - 2}{x+1}$?
5. What are common pitfalls in evaluating limits involving square roots?

Tip: Rationalizing the numerator is a useful technique in handling indeterminate forms involving square roots.