Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$
$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$
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Answer
$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-3\text{x}^2+9\text{x}-2\text{ by }\text{x}^3-\text{x}-6$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9}{\text{x}^3-\text{x}-6}=\lim\limits_{\text{x}\rightarrow2}1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-2\text{x}-6\text{x+4}}{\text{x}^3-\text{x}-6}$ $\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}-1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-\text{x}-6\text{ by}\text{ x}-2$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{(\text{x}-2)(\text{x}^2+2\text{x}+3)}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)}{(\text{x}^2-2\text{x}+3)}$ $=1+\frac{3\times2-2}{2^2+2\times2+3}$ $=1+\frac{4}{11}$ $=\frac{15}{11}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9}{\text{x}^3-\text{x}-6}=\lim\limits_{\text{x}\rightarrow2}1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-8\text{x}+4}{\text{x}^3-\text{x}-6}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{3\text{x}^2-2\text{x}-6\text{x+4}}{\text{x}^3-\text{x}-6}$ $\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3-3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}-1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{\text{x}^3-\text{x}-6}$ Dividing $\text{x}^3-\text{x}-6\text{ by}\text{ x}-2$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^3+3\text{x}^2-9\text{x}-2}{\text{x}^3-\text{x}-6}=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)(\text{x}-2)}{(\text{x}-2)(\text{x}^2+2\text{x}+3)}$ $=1+\lim\limits_{\text{x}\rightarrow2}\frac{(3\text{x}-2)}{(\text{x}^2-2\text{x}+3)}$ $=1+\frac{3\times2-2}{2^2+2\times2+3}$ $=1+\frac{4}{11}$ $=\frac{15}{11}$
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