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Question Number 46060 by samitoh last updated on 20/Oct/18

determine whether or not lim_(x→0) [ is continuous

$${determine}\:{whether}\:{or}\:{not}\:\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\:\:\:\:\:\right. \\ $$$${is}\:{continuous} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

a function f(x) will be continuous at x=a when lim_(x→a−) f(x)=lim_(x→a+) f(x)=f(a) lim_(x→4−) f(x) =lim_(x→4−) ((x^2 −16)/(x−4)) =lim_(x→4−) (((x+4)(x−4))/(x−4))=4+4=8 lim_(x→4+) f(x) =lim_(x→4+) ((x^2 −16)/(x−4)) =lim_(x→4+) (((x+4)(x−4))/(x−4))=4+4=8 f(4)=8 so f(x) continuous at x=4

$${a}\:{function}\:{f}\left({x}\right)\:{will}\:{be}\:{continuous}\:{at}\:{x}={a} \\ $$$${when}\:\underset{{x}\rightarrow{a}−} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow{a}+} {\mathrm{lim}}\:{f}\left({x}\right)={f}\left({a}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{4}−} {\mathrm{lim}}\:{f}\left({x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{4}−} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{16}}{{x}−\mathrm{4}} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}−} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{4}\right)\left({x}−\mathrm{4}\right)}{{x}−\mathrm{4}}=\mathrm{4}+\mathrm{4}=\mathrm{8} \\ $$$$\underset{{x}\rightarrow\mathrm{4}+} {\mathrm{lim}}\:{f}\left({x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{4}+} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{16}}{{x}−\mathrm{4}} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}+} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{4}\right)\left({x}−\mathrm{4}\right)}{{x}−\mathrm{4}}=\mathrm{4}+\mathrm{4}=\mathrm{8} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{8}\:\:{so}\:{f}\left({x}\right)\:{continuous}\:{at}\:{x}=\mathrm{4} \\ $$$$ \\ $$

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