Question
Evaluate: $\lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}$
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Answer
We have to find the value of $\lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}$
We have,
$\lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}$
Divide $x^3+3 x^2-9 x-2$ by $x^3-x-6$\
$\begin{array}{l}\Rightarrow \lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}=\lim _{x \rightarrow 2} 1+\lim _{x \rightarrow 2} \frac{3 x^2-8 x+4}{x^3-x-6} \\ =1+\lim _{x \rightarrow 2} \frac{3 x^2-2 x-6 x+4}{x^3-x-6} \\ =1+\lim _{x \rightarrow 2} \frac{3 x^2-2 x-6 x+4}{x^3-x-6} \\ \Rightarrow \lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}=1+\lim _{x \rightarrow 2} \frac{(3 x-2)(x-2)}{x^3-x-6}\end{array}$
Divide $x^3-x-6$ by $x-2$
$\begin{array}{l}\Rightarrow \lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}=1+\lim _{x \rightarrow 2} \frac{(3 x-2)(x-2)}{(x-2)\left(x^2+2 x+3\right)} \\
=1+\lim _{x \rightarrow 2} \frac{(3 x-2)}{\left(x^2+2 x+3\right)} \\
=1+\frac{3 \times 2-2}{2^2+2 \times 2+3}\end{array}$
$\begin{array}{l}=1+\frac{4}{11} \\
=\frac{15}{11}\end{array}$
We have,
$\lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}$
Divide $x^3+3 x^2-9 x-2$ by $x^3-x-6$\
$\begin{array}{l}\Rightarrow \lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}=\lim _{x \rightarrow 2} 1+\lim _{x \rightarrow 2} \frac{3 x^2-8 x+4}{x^3-x-6} \\ =1+\lim _{x \rightarrow 2} \frac{3 x^2-2 x-6 x+4}{x^3-x-6} \\ =1+\lim _{x \rightarrow 2} \frac{3 x^2-2 x-6 x+4}{x^3-x-6} \\ \Rightarrow \lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}=1+\lim _{x \rightarrow 2} \frac{(3 x-2)(x-2)}{x^3-x-6}\end{array}$
Divide $x^3-x-6$ by $x-2$
$\begin{array}{l}\Rightarrow \lim _{x \rightarrow 2} \frac{x^3+3 x^2-9 x-2}{x^3-x-6}=1+\lim _{x \rightarrow 2} \frac{(3 x-2)(x-2)}{(x-2)\left(x^2+2 x+3\right)} \\
=1+\lim _{x \rightarrow 2} \frac{(3 x-2)}{\left(x^2+2 x+3\right)} \\
=1+\frac{3 \times 2-2}{2^2+2 \times 2+3}\end{array}$
$\begin{array}{l}=1+\frac{4}{11} \\
=\frac{15}{11}\end{array}$
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