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First I tried direct substitution, which resulted in the indeterminate form. Then because of the square roots, I tried rationalizing (both numerator and denominator), but still get the indeterminate form. I can't use L'Hopital's rule because we haven't learned that yet (not sure if it would work anyway). Don't really know what else to try algebraically... Any hints?

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$\lim_{x\to0}\frac{\sqrt{x+1}-\sqrt{2x+1}}{\sqrt{3x+4}-\sqrt{2x+4}}\cdot\frac{\sqrt{3x+4}+\sqrt{2x+4}}{\sqrt{3x+4}+\sqrt{2x+4}}\cdot\frac{\sqrt{x+1}+\sqrt{2x+1}}{\sqrt{x+1}+\sqrt{2x+1}}=\lim_{x\to0}\frac{-x(\sqrt{3x+4}+\sqrt{2x+4})}{x(\sqrt{x+1}+\sqrt{2x+1})}=-\frac{4}{2}=-2$.

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I took a course on limits a few years ago. There some limits which converge, but many diverge (exercise). Limits which approach 0 are very special; there is a general theorem about these type of limits. The theorem says that $lim_{x \to 0} f{x} = f(0)$ under some special conditions. In your case, the function does not satisfy these conditions (exercise). You must look at the asymptotic behavior. By taylor expansion, near 0, $\sqrt{bx+a}=\sqrt{a}(1+\frac{bx}{2a})+O(x^2)$ Ok WE ARE GOOD TO GO NOW!!! Let's look at your limit (exercise) your limit reduces to -8/4=-2.(exercise)

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