Question Number 190998 by mathlove last updated on 16/Apr/23
Commented by mr W last updated on 16/Apr/23
$${is}\:{it}\:{normal}\:{when}\:{somebody}\:{asks} \\ $$$${you}\:{directly}\:{a}\:{question}\:{but}\:{you}\:{totally} \\ $$$${irgnore}\:{it}\:{without}\:{giving}\:{any}\:{reply}? \\ $$$${i}\:{find}\:{this}\:{behaviour}\:{at}\:{least}\:{impolite}. \\ $$$${i}\:{mean}\:{for}\:{example}\:{the}\:{question}\:{from} \\ $$$${Tinku}\:{Tara}\:{sir}\:{in}\:{Q}\mathrm{190882}.\:{i}\:{don}'{t} \\ $$$${think}\:{you}\:{must}\:{answer}\:{it},\:{but}\:{i}\:{think} \\ $$$${you}\:{should}\:{at}\:{least}\:{give}\:{a}\:{reply}.\: \\ $$$${otherwise}\:{you}\:{are}\:{showing}\:{that}\:{you} \\ $$$${are}\:{a}\:{selfish}\:{person}\:{who}\:{only}\:{wants}\: \\ $$$${but}\:{never}\:{gives},\:{what}\:{i}\:{don}'{t}\:{think}\:{is} \\ $$$${true}. \\ $$
Commented by mr W last updated on 16/Apr/23
$${thanks}\:{for}\:{reply}\:{sir}! \\ $$$${as}\:{i}\:{said},\:{you}\:{don}'{t}\:{need}\:{to}\:{answer} \\ $$$${if}\:{you}\:{don}'{t}\:{want}\:{to}\:{or}\:{if}\:{you}\:{can}'{t}\: \\ $$$${write}\:{in}\:{english}.\:{give}\:{a}\:{short}\:{reply} \\ $$$${like}\:``{sorry},\:{i}\:{can}'{t}\:{answer}.''\:{is}\:{quite}\: \\ $$$${ok}.\:{but}\:{just}\:{simply}\:{ignore}\:{is}\:{not}\:{good}. \\ $$
Commented by mustafazaheen last updated on 16/Apr/23
$$\mathrm{ok}\:\mathrm{Mr}\:\mathrm{W}\:\mathrm{your}\:\mathrm{speak}\:\mathrm{is}\:\mathrm{very}\:\mathrm{accurate}. \\ $$
Commented by mathlove last updated on 16/Apr/23
$$ \\ $$I don't know English, brother, I have had this problem before, but I apologize
Commented by Albert12 last updated on 16/Apr/23
$${right} \\ $$
Commented by mathlove last updated on 16/Apr/23
$${ok} \\ $$
Answered by mustafazaheen last updated on 16/Apr/23
$$=\mathrm{0} \\ $$
Commented by mathlove last updated on 16/Apr/23
$${solve}\:{it} \\ $$
Answered by mehdee42 last updated on 16/Apr/23
$${it}\:{is}\:{not}\:{impossible}\:{to}\:{lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\sqrt{\mathrm{1}−{x}} \\ $$$$ \\ $$
Answered by mr W last updated on 16/Apr/23
$${due}\:{to}\:\sqrt{{x}−\mathrm{1}},\:{x}>\mathrm{1}. \\ $$$${that}\:{means}\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\Box\:{doesn}'{t}\:{exist} \\ $$$${and}\:{the}\:{question}\:{is}\:{wrong}. \\ $$$$ \\ $$$${assume}\:{the}\:{question}\:{is} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}}{\:\sqrt{{x}−\mathrm{1}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{\:\sqrt{{x}−\mathrm{1}}}×\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{\:{x}^{\mathrm{3}} −\mathrm{1}}×\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\: \\ $$$$=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\sqrt{{x}−\mathrm{1}}×\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)×\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{\:{x}^{\mathrm{3}} −\mathrm{1}}×\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\: \\ $$$$\leqslant\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\sqrt{{x}−\mathrm{1}}×\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)×\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{\:{x}^{\mathrm{3}} −\mathrm{1}}=\mathrm{0}×\mathrm{1}×\mathrm{1}=\mathrm{0}\: \\ $$$$\geqslant−\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\sqrt{{x}−\mathrm{1}}×\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)×\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{\:{x}^{\mathrm{3}} −\mathrm{1}}=−\mathrm{0}×\mathrm{1}×\mathrm{1}=\mathrm{0}\: \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}}{\:\sqrt{{x}−\mathrm{1}}}=\mathrm{0} \\ $$
Commented by mehdee42 last updated on 16/Apr/23
$${thats}\:{right} \\ $$
Commented by mathlove last updated on 16/Apr/23
$${thanks} \\ $$