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Question Number 197301 by uchihayahia last updated on 13/Sep/23

how do i calculate this lim_(x→-∞) ((x^4 +2x^2 +x−2)/(x^3 +2x^2 +x−1)) multiplying both numerator and denumerator by (1/x^4 ) lim_(x→-∞) ((1+(2/x^2 )+(1/x^3 )−(2/x^4 ))/((1/x)+(2/x^2 )+(1/x^3 )−(1/x^4 ))) ((1+0+0−0)/(0+0+0−0)) ∞ which is not true the answer is -∞, i tried multiplying (1/x^3 ) and got -∞ but still confused what did i do wrong using (1/x^4 )

$$ \\ $$$$\:{how}\:{do}\:{i}\:{calculate}\:{this} \\ $$$$\:\underset{{x}\rightarrow-\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{2}}{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}} \\ $$$$\:{multiplying}\:{both}\:{numerator} \\ $$$$\:{and}\:{denumerator}\:{by}\:\frac{\mathrm{1}}{{x}^{\mathrm{4}} } \\ $$$$\:\underset{{x}\rightarrow-\infty} {\mathrm{lim}}\:\frac{\mathrm{1}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{2}}{{x}^{\mathrm{4}} }}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }} \\ $$$$\:\frac{\mathrm{1}+\mathrm{0}+\mathrm{0}−\mathrm{0}}{\mathrm{0}+\mathrm{0}+\mathrm{0}−\mathrm{0}} \\ $$$$\:\infty \\ $$$$\:{which}\:{is}\:{not}\:{true}\:{the}\:{answer}\:{is}\:-\infty, \\ $$$$\:{i}\:{tried}\:{multiplying}\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:{and}\:{got}\:-\infty \\ $$$$\:{but}\:{still}\:{confused}\:{what}\:{did}\:{i}\:{do}\:{wrong} \\ $$$$\:{using}\:\frac{\mathrm{1}}{{x}^{\mathrm{4}} } \\ $$$$ \\ $$

Answered by deleteduser1 last updated on 13/Sep/23

((x(x^3 +2x^2 +x−1)−2x^3 +x^2 +2x−2)/(x^3 +2x^2 +x−1)) =x−((2x^3 +4x^2 +2x−2−5x^2 −4x+4)/(x^3 +2x^2 +x−1)) =x−2−((5x^2 +4x−4)/(x^3 +2x^2 +x−1))=x−2−(((5/x)+(4/x^2 )−(4/x^3 ))/(1+(2/x)+(1/x^2 )−(1/x^3 ))) →−∞

$$\frac{{x}\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}\right)−\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}}{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}} \\ $$$$={x}−\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}−\mathrm{5}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}} \\ $$$$={x}−\mathrm{2}−\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{4}}{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}}={x}−\mathrm{2}−\frac{\frac{\mathrm{5}}{{x}}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\frac{\mathrm{4}}{{x}^{\mathrm{3}} }}{\mathrm{1}+\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }} \\ $$$$\rightarrow−\infty \\ $$

Commented by uchihayahia last updated on 16/Sep/23

thank you, i always thought i should divide both by x^n , m is highest degree

$${thank}\:{you},\:{i}\:{always}\:{thought}\:{i}\:{should} \\ $$$$\:{divide}\:{both}\:{by}\:{x}^{{n}} ,\:{m}\:{is}\:{highest}\:{degree} \\ $$

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