Question Number 207943 by hardmath last updated on 31/May/24
$$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\frac{\left(\mathrm{x}\:+\:\mathrm{2}\right)\centerdot\left(\mathrm{x}\:+\:\mathrm{1}\right)}{\mid\mathrm{x}\:+\:\mathrm{2}\mid}\:\:=\:\:? \\ $$
Commented by mr W last updated on 31/May/24
$$=\frac{\left(\mathrm{2}+\mathrm{2}\right)\left(\mathrm{2}+\mathrm{1}\right)}{\mid\mathrm{2}+\mathrm{2}\mid}=\mathrm{3} \\ $$
Commented by mr W last updated on 01/Jun/24
$${f}\left(\mathrm{2}\right)\:{exists}.\:{so}\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}{f}\left({x}\right)={f}\left(\mathrm{2}\right). \\ $$
Commented by hardmath last updated on 01/Jun/24
$$ \\ $$Dear professor, could you please explain
Commented by mr W last updated on 01/Jun/24
$${maybe}\:{you}\:{wanted}\:{to}\:{ask} \\ $$$$\underset{\boldsymbol{\mathrm{x}}\rightarrow−\mathrm{2}^{−} } {\mathrm{lim}}\:\frac{\left(\mathrm{x}\:+\:\mathrm{2}\right)\centerdot\left(\mathrm{x}\:+\:\mathrm{1}\right)}{\mid\mathrm{x}\:+\:\mathrm{2}\mid}\:\:=\:\:? \\ $$
Commented by hardmath last updated on 01/Jun/24
$$\mathrm{Dear}\:\mathrm{professor}… \\ $$$$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{4}} \:−\:\mathrm{16}}{\mid\mathrm{x}−\mathrm{2}\mid}\:=\:? \\ $$
Commented by hardmath last updated on 01/Jun/24
$$ \\ $$Right limit and left limit
Commented by mr W last updated on 01/Jun/24
$${x}\rightarrow\mathrm{2}^{−} \:{mean}\:{x}<\mathrm{2},\:{i}.{e}.\:\mid{x}−\mathrm{2}\mid=\mathrm{2}−{x} \\ $$$$\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\frac{{x}^{\mathrm{4}} −\mathrm{16}}{\mid{x}−\mathrm{2}\mid} \\ $$$$=\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{\mathrm{2}−{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\frac{\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{−\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{2}+\mathrm{2}\right)\left(\mathrm{2}^{\mathrm{2}} +\mathrm{4}\right)}{−\mathrm{1}}=−\mathrm{32} \\ $$
Commented by hardmath last updated on 01/Jun/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Commented by hardmath last updated on 01/Jun/24
$$\mathrm{but}\:\mathrm{dear}\:\mathrm{pfofessor} \\ $$$$ \\ $$Why not -x-2 from the module for x<2 in the above exercise
Commented by mr W last updated on 01/Jun/24
$$\mid{x}−\mathrm{2}\mid={x}−\mathrm{2}\:{if}\:{x}>\mathrm{2} \\ $$$$\mid{x}−\mathrm{2}\mid=−\left({x}−\mathrm{2}\right)=\mathrm{2}−{x}\:{if}\:{x}<\mathrm{2} \\ $$
Commented by hardmath last updated on 02/Jun/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$