Question Number 121576 by oustmuchiya@gmail.com last updated on 09/Nov/20 | ||
Answered by TANMAY PANACEA last updated on 09/Nov/20 | ||
$${f}\left({x}\right)=\frac{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:−\mathrm{5}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{15}−\mathrm{25}}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)}=\frac{{x}^{\mathrm{2}} −\mathrm{10}}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)} \\ $$$${at}\:{x}=\mathrm{2}\:{the}\:{function}\left[{discontinous}\right. \\ $$$${li}\underset{{x}\rightarrow\mathrm{2}} {{m}f}\left({x}\right)=\infty \\ $$$$ \\ $$ | ||
Answered by TANMAY PANACEA last updated on 09/Nov/20 | ||
$${f}\left({x}\right)=\frac{{x}\sqrt{\mathrm{1}+\frac{\mathrm{15}}{{x}^{\mathrm{2}} }}\:−\mathrm{5}}{{x}−\mathrm{2}}=\frac{\sqrt{\mathrm{1}+\frac{\mathrm{15}}{{x}^{\mathrm{2}} }}\:−\frac{\mathrm{5}}{{x}}}{\mathrm{1}−\frac{\mathrm{2}}{{x}}} \\ $$$${so}\:{li}\underset{{x}\rightarrow\infty} {{m}}\:{f}\left({x}\right) \\ $$$$=\frac{\sqrt{\mathrm{1}+\mathrm{0}}\:−\mathrm{0}}{\mathrm{1}−\mathrm{0}}=\mathrm{1} \\ $$ | ||
Commented by oustmuchiya@gmail.com last updated on 12/Nov/20 | ||
$$\mathrm{many}\:\mathrm{thanks}\:\mathrm{indeed} \\ $$ | ||