<s>[INST] Evaluating $\lim_{x\to \sqrt 2}\frac{\sqrt{3+2x} - (\sqrt 2+1)}{x^2-2}$ 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 $\frac{1}{2(2+\sqrt 2)}$.
 [/INST] 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}
</s>