Evaluate the limit as x approaches 0 of (√(x+1) - √(2x+1)) / (√(3x+4) - √(2x+4)).

Steps to Solve

1. Check for Indeterminate Form

First, plug in $x=0$ into the expression to see if we get an indeterminate form:

$$ \frac{\sqrt{0+1} - \sqrt{2(0)+1}}{\sqrt{3(0)+4} - \sqrt{2(0)+4}} = \frac{\sqrt{1} - \sqrt{1}}{\sqrt{4} - \sqrt{4}} = \frac{1-1}{2-2} = \frac{0}{0} $$

Since we have the indeterminate form $\frac{0}{0}$, we can use algebraic manipulation to simplify the expression.

2. Rationalize the Numerator and Denominator

Multiply the numerator and denominator by their respective conjugates:

$$ \lim_{x \to 0} \frac{\sqrt{x+1} - \sqrt{2x+1}}{\sqrt{3x+4} - \sqrt{2x+4}} \cdot \frac{\sqrt{x+1} + \sqrt{2x+1}}{\sqrt{x+1} + \sqrt{2x+1}} \cdot \frac{\sqrt{3x+4} + \sqrt{2x+4}}{\sqrt{3x+4} + \sqrt{2x+4}} $$

3. Simplify the Expression

Simplify the numerator: $(\sqrt{x+1} - \sqrt{2x+1})(\sqrt{x+1} + \sqrt{2x+1}) = (x+1) - (2x+1) = -x$. Simplify the denominator: $(\sqrt{3x+4} - \sqrt{2x+4})(\sqrt{3x+4} + \sqrt{2x+4}) = (3x+4) - (2x+4) = x$.

So we have:

$$ \lim_{x \to 0} \frac{-x}{x} \cdot \frac{\sqrt{3x+4} + \sqrt{2x+4}}{\sqrt{x+1} + \sqrt{2x+1}} $$

4. Cancel $x$ and Evaluate the Limit

Cancel $x$ from the numerator and the denominator:

$$ \lim_{x \to 0} -1 \cdot \frac{\sqrt{3x+4} + \sqrt{2x+4}}{\sqrt{x+1} + \sqrt{2x+1}} $$

Now, plug in $x=0$:

$$ -1 \cdot \frac{\sqrt{3(0)+4} + \sqrt{2(0)+4}}{\sqrt{0+1} + \sqrt{2(0)+1}} = -1 \cdot \frac{\sqrt{4} + \sqrt{4}}{\sqrt{1} + \sqrt{1}} = -1 \cdot \frac{2+2}{1+1} = -1 \cdot \frac{4}{2} = -2 $$