Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-7\text{x}+12}{{\text{x}^2}-3\text{x}-4}$
$\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-7\text{x}+12}{{\text{x}^2}-3\text{x}-4}$
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Answer
$\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-7\text{x}+12}{{\text{x}^2}-3\text{x}-4}$
$=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-3\text{x}-4\text{x}+12}{\text{x}^2+\text{x}-4\text{x}-4}$
$=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}(\text{x}-3)-4(\text{x}-3)}{\text{x}(\text{x}+1)-1(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow4}\frac{(\text{x}-3)(\text{x}-4)}{(\text{x}-4)(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}-3}{\text{x}+1}$
$=\frac{4-3}{4+1}$
$=\frac15$
$=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}^2-3\text{x}-4\text{x}+12}{\text{x}^2+\text{x}-4\text{x}-4}$
$=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}(\text{x}-3)-4(\text{x}-3)}{\text{x}(\text{x}+1)-1(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow4}\frac{(\text{x}-3)(\text{x}-4)}{(\text{x}-4)(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow4}\frac{\text{x}-3}{\text{x}+1}$
$=\frac{4-3}{4+1}$
$=\frac15$
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