Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-\text{x}-6}{\text{x}^3+3\text{x}^2+\text{x}-3}$
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Answer
$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-\text{x}-6}{\text{x}^3+3\text{x}^2+\text{x}-3}$Now $\text{x}^2-\text{x}-6$
$=\text{x}^2-3\text{x}+2\text{x}-6$
$=\text{x}(\text{x}-3)+2(\text{x}-3)$
$=(\text{x}+2)(\text{x}-3)\ \cdots(\text{i})$
Dividing $\text{x}^3-3\text{x}^2+\text{x}-3\text{ by }(\text{x}-3), \text{ we get}$
Thus (x - 3) is a factor of $\text{x}^3-3\text{x}^2+\text{x}-3\ \cdots(\text{ii})$
Substituting (i) and (ii) in the given expression
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}+2)(\text{x}-3)}{\big(\text{x}^2+1\big)(\text{x}-3)}$
$=\frac{\text{x}+2}{\text{x}^2+1}=\frac{3+2}{9+1}=\frac{5}{10}$
$=\frac12$
$=\text{x}^2-3\text{x}+2\text{x}-6$
$=\text{x}(\text{x}-3)+2(\text{x}-3)$
$=(\text{x}+2)(\text{x}-3)\ \cdots(\text{i})$
Dividing $\text{x}^3-3\text{x}^2+\text{x}-3\text{ by }(\text{x}-3), \text{ we get}$
Thus (x - 3) is a factor of $\text{x}^3-3\text{x}^2+\text{x}-3\ \cdots(\text{ii})$
Substituting (i) and (ii) in the given expression
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}+2)(\text{x}-3)}{\big(\text{x}^2+1\big)(\text{x}-3)}$
$=\frac{\text{x}+2}{\text{x}^2+1}=\frac{3+2}{9+1}=\frac{5}{10}$
$=\frac12$
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