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.
 A: $$
\begin{aligned}
\lim _{x\to 2}\left(\frac{\sqrt{x^3\:-\:3x^2\:+\:4}-x\:+2}{x^2\:-\:4}\right)
& = \lim _{t\to 0}\left(\frac{\sqrt{\left(t+2\right)^3\:-\:3\left(t+2\right)^2\:+\:4}-\left(t+2\right)\:+2}{\left(t+2\right)^2\:-\:4}\right)
\\& = \lim _{t\to 0}\left(\frac{\left|t\right|\sqrt{\left(t+3\right)}-t}{t\left(t+4\right)}\right)
\\& = \lim _{t\to 0^+}\left(\frac{\left|t\right|\sqrt{\left(t+3\right)}-t}{t\left(t+4\right)}\right) = \color{red}{\frac{1}{4}\left(\sqrt{3}-1\right)}
\\& = \lim _{t\to 0^-}\left(\frac{\left|t\right|\sqrt{\left(t+3\right)}-t}{t\left(t+4\right)}\right) = \color{blue}{\frac{1}{4}\left(-\sqrt{3}-1\right)}
\end{aligned}
$$
Solved with substitution $\color{green}{t = x-2}$
A: HINT: Take $x=t+2$ and take limit as $t\to 0$ also notice that $\sqrt{t^2}=|t|$ and than take $\lim_{t\to0^+}$ and $\lim_{t\to 0^-}$
A: No L'Hospital, too messy. You are almost there, rewrite the expression as 
$$
\frac{(x-2)x}{(x+2)(x-2)(1 +\sqrt{\frac{(x-2)^2(x+1)}{(x-2)^2}}}
$$
what do you get? 
A: $$\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}} =$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 2x^2-x^2 + 4}-(x -2)}{(x - 2)(x+2)}}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{x^2(x-2)-(x-2)(x+2)}}{(x - 2)(x+2)}}-\lim \limits_{x \to 2}{\frac{1}{x+2}}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{(x-2)(x^2-x-2)}}{(x - 2)(x+2)}}-\frac{1}{4}=$$
$$=\lim \limits_{x \to 2}{\frac{\sqrt{(x-2)(x-2)(x+1)}}{(x - 2)(x+2)}}-\frac{1}{4}=$$
$$=\lim \limits_{x \to 2}{\frac{|x-2| \cdot \sqrt{(x+1)}}{(x-2)(x+2)}}-\frac{1}{4}$$
Next , find limit when $x \to 2^{-}$ and when $x \to 2^{+}$
