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Question Number 5286 by FilupSmith last updated on 05/May/16
Can someone please explain why:  lim_(x→∞)  (x−x^2 ) = −∞
$$\mathrm{Can}\:\mathrm{someone}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{why}: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left({x}−{x}^{\mathrm{2}} \right)\:=\:−\infty \\ $$
Commented by FilupSmith last updated on 05/May/16
i normally undwrstand limits but  i dont remember how to do this
$$\mathrm{i}\:\mathrm{normally}\:\mathrm{undwrstand}\:\mathrm{limits}\:\mathrm{but} \\ $$$$\mathrm{i}\:\mathrm{dont}\:\mathrm{remember}\:\mathrm{how}\:\mathrm{to}\:\mathrm{do}\:\mathrm{this} \\ $$
Answered by 123456 last updated on 05/May/16
write it as  −x^2 +x=x^2 (−1+(1/x))  lim_(x→∞)  x−x^2   =lim_(x→∞)  x^2 (−1+(1/x))  (1/x)→0 when x→∞ so  =lim_(x→+∞) −x^2 =−∞
$$\mathrm{write}\:\mathrm{it}\:\mathrm{as} \\ $$$$−{x}^{\mathrm{2}} +{x}={x}^{\mathrm{2}} \left(−\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}−{x}^{\mathrm{2}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \left(−\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\frac{\mathrm{1}}{{x}}\rightarrow\mathrm{0}\:\mathrm{when}\:{x}\rightarrow\infty\:\mathrm{so} \\ $$$$=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}−{x}^{\mathrm{2}} =−\infty \\ $$
Commented by FilupSmith last updated on 05/May/16
Thanks!!!
$${Thanks}!!! \\ $$
Answered by enigmeyou last updated on 18/Jun/16
lim_(x→−∞) (x−x^2 )=lim_(x→−∞) x(1−x)=−∞×(+∞)=−∞  lim_(x→+∞) (x−x^2 )=lim_(x→+∞) x(1−x)=+∞×(−∞)=−∞
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left({x}−{x}^{\mathrm{2}} \right)=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{x}\left(\mathrm{1}−{x}\right)=−\infty×\left(+\infty\right)=−\infty \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left({x}−{x}^{\mathrm{2}} \right)=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{x}\left(\mathrm{1}−{x}\right)=+\infty×\left(−\infty\right)=−\infty \\ $$$$ \\ $$

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