Question Number 46060 by samitoh last updated on 20/Oct/18
$${determine}\:{whether}\:{or}\:{not}\:\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\:\:\:\:\:\right. \\ $$$${is}\:{continuous} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18
$${a}\:{function}\:{f}\left({x}\right)\:{will}\:{be}\:{continuous}\:{at}\:{x}={a} \\ $$$${when}\:\underset{{x}\rightarrow{a}−} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow{a}+} {\mathrm{lim}}\:{f}\left({x}\right)={f}\left({a}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{4}−} {\mathrm{lim}}\:{f}\left({x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{4}−} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{16}}{{x}−\mathrm{4}} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}−} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{4}\right)\left({x}−\mathrm{4}\right)}{{x}−\mathrm{4}}=\mathrm{4}+\mathrm{4}=\mathrm{8} \\ $$$$\underset{{x}\rightarrow\mathrm{4}+} {\mathrm{lim}}\:{f}\left({x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{4}+} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{16}}{{x}−\mathrm{4}} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}+} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{4}\right)\left({x}−\mathrm{4}\right)}{{x}−\mathrm{4}}=\mathrm{4}+\mathrm{4}=\mathrm{8} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{8}\:\:{so}\:{f}\left({x}\right)\:{continuous}\:{at}\:{x}=\mathrm{4} \\ $$$$ \\ $$