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Question Number 69075 by Aditya789 last updated on 19/Sep/19

Commented by Prithwish sen last updated on 19/Sep/19

$$\mathrm{Divide}\mathrm{both}\mathrm{numerator}\mathrm{and}\mathrm{denominator}$$$$\mathrm{by}{\mathrm{x}}^3\mathrm{and}\mathrm{the}\mathrm{limit}\mathrm{will}\mathrm{be}1$$

Commented by mathmax by abdo last updated on 19/Sep/19

$$(x-2)(x+3)(x+5)=x^3+p\left(x\right)withdeg\left(p\right)⩽2$$$$(x-1)(x-3)(x-7)=x^3+q\left(x\right)withdeg\left(a\right)⩽2\Rightarrow$$$${lim}_{x\to \stackrel{-}{+}\mathrm{\infty }}\frac{(x-2)(x+3)(x+5)}{(x-1)(x-3)(x-7)}={lim}_{x\to \stackrel{-}{+}\mathrm{\infty }}\frac{x^3+p\left(x\right)}{x^3+q\left(x\right)}$$$$={lim}_{x\to \stackrel{-}{+}\mathrm{\infty }}\frac{x^3}{x^3}=1.$$

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