Question Number 195092 by mathlove last updated on 24/Jul/23
$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{4}{x}+\mathrm{5}}{{x}−{x}^{\mathrm{2}} }=? \\ $$
Answered by tri26112004 last updated on 24/Jul/23
$${We}\:{have}: \\ $$$$\bullet\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\mathrm{4}{x}+\mathrm{5}\:=\:\mathrm{4}.\mathrm{1}+\mathrm{5}\:=\:\mathrm{9}>\mathrm{0} \\ $$$$\bullet\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:{x}−{x}^{\mathrm{2}} \:=\:\mathrm{0}\:\left(\forall{x}>\mathrm{1}\right) \\ $$$$\bullet\:{g}\left({x}\right)={x}−{x}^{\mathrm{2}} \:−>\:{a}=−\mathrm{1}<\mathrm{0} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{1}^{+} \:} {\mathrm{lim}}\:\frac{\mathrm{4}{x}+\mathrm{5}}{{x}−{x}^{\mathrm{2}} }=−\infty \\ $$
Commented by mathlove last updated on 24/Jul/23
$${way}\:{a}=−\mathrm{1} \\ $$$${ass}\:{x}\rightarrow\mathrm{1}^{+} \:\:\:\:\:{g}\left(\mathrm{0}.\mathrm{9}\right)=\mathrm{0},\mathrm{9}−\left(\mathrm{0}.\mathrm{9}\right)^{\mathrm{2}} =\mathrm{0}.\mathrm{09} \\ $$$${value}\:{is}\:{the}\:+ \\ $$$${way}\:{the}\:{limit}\:{symptom}\:{is} \\ $$$$−\infty \\ $$