\frac{x^2-x-2}{x^3-2x^2+3}=\frac{1-\frac{1}{x}-\frac{2}{x^2}}{x-2+\frac{3}{x^2}} \mathrm\[\ which \ tends\ to\ }\frac{1}{x-2)
\frac{x^2-x-2}{x^3-2x^2+3}=\frac{1-\frac{1}{x}-\frac{2}{x^2}}{x-2+\frac{3}{x^2}} \mathrm\[\ which \ tends\ to\ }\frac{1}{x-2)
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