Question Number 207519 by hardmath last updated on 17/May/24
$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\:\frac{\left(\mathrm{x}\:+\:\mathrm{2}\right)\centerdot\left(\mathrm{x}\:+\:\mathrm{1}\right)}{\mid\mathrm{x}\:+\:\mathrm{2}\mid}\:\:=\:\:? \\ $$
Answered by Frix last updated on 17/May/24
$${f}\left({x}\right)=\frac{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)}{\mid{x}+\mathrm{2}\mid}=\begin{cases}{−\left({x}+\mathrm{1}\right),\:{x}<−\mathrm{2}}\\{\left({x}+\mathrm{1}\right),\:−\mathrm{2}<{x}}\end{cases} \\ $$$$\mathrm{We}\:\mathrm{approach}\:\mathrm{2}\:\mathrm{from}\:\mathrm{the}\:\mathrm{negative}\:\mathrm{side}\:\mathrm{but} \\ $$$$\mathrm{for}\:\mathrm{0}<{x}<\mathrm{2}\:\mathrm{we}\:\mathrm{get} \\ $$$$\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\left({x}+\mathrm{1}\right)\:=\mathrm{3} \\ $$$$\mathrm{The}\:\mathrm{only}\:\mathrm{discontinuity}\:\mathrm{is}\:\mathrm{at}\:{x}=−\mathrm{2} \\ $$$$\underset{{x}\rightarrow−\mathrm{2}^{−} } {\mathrm{lim}}\:{f}\left({x}\right)\:=\mathrm{1}\:\neq\:\underset{{x}\rightarrow−\mathrm{2}^{+} } {\mathrm{lim}}\:{f}\left({x}\right)\:=−\mathrm{1} \\ $$
Commented by hardmath last updated on 18/May/24
$$\mathrm{answer}:\:\mathrm{3}\:\:\mathrm{professor}? \\ $$
Commented by Frix last updated on 18/May/24
$$\mathrm{Yes}.\:{x}=\mathrm{2}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{problem},\:\mathrm{you}\:\mathrm{can}\:\mathrm{just} \\ $$$$\mathrm{insert}\:{f}\left(\mathrm{2}\right)=\frac{\left(\mathrm{2}+\mathrm{2}\right)\left(\mathrm{2}+\mathrm{1}\right)}{\mid\mathrm{2}+\mathrm{2}\mid}=\frac{\mathrm{4}×\mathrm{3}}{\mathrm{4}}=\mathrm{3} \\ $$
Commented by hardmath last updated on 18/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$