Question
Solve: $\lim _{x \rightarrow 1} \frac{x^4-3 x^3+2}{x^3-5 x^2+3 x+1}$
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Answer
Dividing $x^4-3 x^3+2$ by $x^3-5 x^2+3 x+1$
$\begin{array}{l}\Rightarrow \lim _{x \rightarrow 1} \frac{x^4-3 x^3+2}{x^3-5 x^2+3 x+1}=\lim _{x \rightarrow 1}(x+2)+\lim _{x \rightarrow 1} \frac{7 x^2-7 x}{x^3-5 x^3+3 x+1} \\ =\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x(x-1)}{x^3-5 x^3+3 x+1} \\ =\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x(x-1)}{(x-1)\left(x^2-4 x-1\right)} \\ =\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x}{\left(x^2-4 x-1\right)}\end{array}$
$\begin{array}{l}=1+2+\frac{7}{(1-4-1)} \\ =3-\frac{7}{4} \\ =\frac{12-7}{4} \\ =\frac{5}{4}\end{array}$
$\begin{array}{l}\Rightarrow \lim _{x \rightarrow 1} \frac{x^4-3 x^3+2}{x^3-5 x^2+3 x+1}=\lim _{x \rightarrow 1}(x+2)+\lim _{x \rightarrow 1} \frac{7 x^2-7 x}{x^3-5 x^3+3 x+1} \\ =\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x(x-1)}{x^3-5 x^3+3 x+1} \\ =\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x(x-1)}{(x-1)\left(x^2-4 x-1\right)} \\ =\lim _{x \rightarrow 1} x+2+\lim _{x \rightarrow 1} \frac{7 x}{\left(x^2-4 x-1\right)}\end{array}$
$\begin{array}{l}=1+2+\frac{7}{(1-4-1)} \\ =3-\frac{7}{4} \\ =\frac{12-7}{4} \\ =\frac{5}{4}\end{array}$
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