Question Number 190138 by 073 last updated on 28/Mar/23
Answered by Pengu last updated on 29/Mar/23
$$\mathrm{Notice}\:\mathrm{that}\:\mathrm{for}\:{a}=\mathrm{4}: \\ $$$$\underset{{x}\rightarrow\mathrm{4}^{−} } {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{4}^{−} } {\mathrm{lim}}\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{4}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{4}^{−} } {\mathrm{lim}}\:\left(\frac{\mathrm{2}\left({x}+\mathrm{2}\right)\:−\mathrm{3}}{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}\right) \\ $$$$=\:\underset{{x}\rightarrow\mathrm{4}^{−} } {\mathrm{lim}}\:\left(\frac{\mathrm{2}}{{x}−\mathrm{2}}−\frac{\mathrm{3}}{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}\right) \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}}−\frac{\mathrm{3}}{\left(\mathrm{6}\right)\left(\mathrm{2}\right)} \\ $$$$=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{and} \\ $$$$\underset{{x}\rightarrow\mathrm{4}^{+} } {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{4}^{+} } {\mathrm{lim}}\:\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{5}} \\ $$$$=\:\frac{\mathrm{3}\left(\mathrm{4}\right)−\mathrm{1}}{\mathrm{5}} \\ $$$$=\:\frac{\mathrm{12}−\mathrm{1}}{\mathrm{5}} \\ $$$$=\:\frac{\mathrm{11}}{\mathrm{5}} \\ $$
Commented by Pengu last updated on 29/Mar/23
The trick is to remember that the limit of any "regular" function, limits hold. This function is not "regular", so we try the part where the function may potentially break. In this case, that is at x=4.
Commented by 073 last updated on 29/Mar/23
$$\mathrm{which}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}? \\ $$
Commented by 073 last updated on 29/Mar/23
Commented by mr W last updated on 29/Mar/23
$${f}\left({x}\right)\:{is}\:{not}\:{continuous}\:{at}\:{x}=−\mathrm{2},\:\mathrm{2},\:\mathrm{4} \\ $$$$\Rightarrow{when}\:{a}=−\mathrm{2},\mathrm{2},\mathrm{4}:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)\neq{f}\left({a}\right) \\ $$