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

Question

I used rationalization trick where you multiply the top and bottom with opposites of given functions and thus got $\frac{2(x^2+2)}{(x^2-2)({\sqrt {3+2x} + (\sqrt2 + 1)})}$ but by then it has become too complex. The book says answer should be $\frac{1}{2(2+\sqrt 2)}$.

Answer 1

You made a mistake in the numerator.

\begin{align} \frac{\sqrt{3+2x} - (\sqrt 2+1)}{x^2-2}\cdot \frac{\sqrt{3+2x} + (\sqrt 2+1)}{\sqrt{3+2x} + (\sqrt 2+1)} &= \frac{3+2x - (\sqrt{2}+1)^2}{(x^2-2)(\sqrt{3+2x} + (\sqrt 2+1))}\\ &= \frac{2(x-\sqrt{2})}{(x^2-2)(\sqrt{3+2x} + (\sqrt 2+1))}\\ &= \frac{2}{(x+\sqrt{2})(\sqrt{3+2x} + (\sqrt 2+1))}\\ &\xrightarrow{x\to\sqrt{2}} \frac{2}{(\sqrt{2}+\sqrt{2})(2\cdot (\sqrt{2}+1))}\\ &= \frac{1}{2(2+\sqrt2)} \end{align}

Answer 2

I would use the fact that$$\lim_{x\to\sqrt2}\frac{\sqrt{3+2x}-\sqrt2-1}{x^2-2}=\lim_{x\to\sqrt2}\frac{\sqrt{3+2x}-\sqrt2-1}{x-\sqrt2}\times\lim_{x\to\sqrt2}\frac1{x+\sqrt2}.$$The first limit is just the computation of a derivative.