<s>[INST] Limit $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$ I try to calculate $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$. So, $\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4} = \frac{(x-2)x}{(x+2)(x+\sqrt{(x-2)^2(x+1)}-2)}$ but I don't know what to do next.
 [/INST] $$\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}} =$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 2x^2-x^2 + 4}-(x -2)}{(x - 2)(x+2)}}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{x^2(x-2)-(x-2)(x+2)}}{(x - 2)(x+2)}}-\lim \limits_{x \to 2}{\frac{1}{x+2}}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{(x-2)(x^2-x-2)}}{(x - 2)(x+2)}}-\frac{1}{4}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{(x-2)(x-2)(x+1)}}{(x - 2)(x+2)}}-\frac{1}{4}=$$
$$=\lim \limits_{x \to 2}{\frac{|x-2| \cdot \sqrt{(x+1)}}{(x-2)(x+2)}}-\frac{1}{4}$$
Next , find limit when $x \to 2^{-}$ and when $x \to 2^{+}$
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