Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^4-81}{\text{x}^2-9}$

Answer

$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^4-81}{\text{x}^2-9}$$=\lim\limits_{\text{x}\rightarrow3}\frac{\big(\text{x}^2-9\big)\big(\text{x}^2+9\big)}{\big(\text{x}^2-9\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\text{x}^2+9=(3)^2+9=9+9=18$

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