Question
Find the limit: $\lim \limits_{x \rightarrow 2}\left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-4}\right]$
Therefore, we have,
$\lim _{x \rightarrow 2} \frac{x^{3}-4 x^{2}+4 x}{x^{2}-4}=\lim _{x \rightarrow 2} \frac{x(x-2)^{2}}{(x+2)(x-2)}$
$=\lim _{x \rightarrow 2} \frac{x(x-2)}{(x+2)}$ as $x\neq$ 2
$=\frac{2(2-2)}{2+2}=\frac{0}{4}=0$
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