Question Number 125847 by zarminaawan last updated on 14/Dec/20
$${Discuss}\:{the}\:{continuity}\:{of}\:{functionf}\left({x}\right)={xtan}\left(\frac{\mathrm{1}}{{x}}\right)\:{at}\:{x}=\mathrm{0}\:{given}\:{tht}\:{x}\neq\mathrm{0}\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$
Answered by mindispower last updated on 14/Dec/20
$${tan}\left({y}\right)>{y},\forall{y}\geqslant\mathrm{0}...? \\ $$$${we}\:{have} \\ $$$$\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\geqslant\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right){dt}\geqslant\int_{\mathrm{0}} ^{{x}} \mathrm{1}.{dt},\forall{x}\geqslant\mathrm{0} \\ $$$$\Leftrightarrow{tg}\left({x}\right)\geqslant{x} \\ $$$${tg}\left(\frac{\mathrm{1}}{{x}}\right)\geqslant\frac{\mathrm{1}}{{x}},\forall{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}.{xtg}\left(\frac{\mathrm{1}}{{x}}\right)\geqslant{x}.\frac{\mathrm{1}}{{x}}=\mathrm{1}\Rightarrow\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{xtg}\left(\frac{\mathrm{1}}{{x}}\right)\geqslant\mathrm{1} \\ $$$${no}\:{continuity} \\ $$$$ \\ $$