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Question-190998




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 Q190882. 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.
$${is}\:{it}\:{normal}\:{when}\:{somebody}\:{asks} \\ $$$${you}\:{directly}\:{a}\:{question}\:{but}\:{you}\:{totally} \\ $$$${irgnore}\:{it}\:{without}\:{giving}\:{any}\:{reply}? \\ $$$${i}\:{find}\:{this}\:{behaviour}\:{at}\:{least}\:{impolite}. \\ $$$${i}\:{mean}\:{for}\:{example}\:{the}\:{question}\:{from} \\ $$$${Tinku}\:{Tara}\:{sir}\:{in}\:{Q}\mathrm{190882}.\:{i}\:{don}'{t} \\ $$$${think}\:{you}\:{must}\:{answer}\:{it},\:{but}\:{i}\:{think} \\ $$$${you}\:{should}\:{at}\:{least}\:{give}\:{a}\:{reply}.\: \\ $$$${otherwise}\:{you}\:{are}\:{showing}\:{that}\:{you} \\ $$$${are}\:{a}\:{selfish}\:{person}\:{who}\:{only}\:{wants}\: \\ $$$${but}\:{never}\:{gives},\:{what}\:{i}\:{don}'{t}\:{think}\:{is} \\ $$$${true}. \\ $$
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.
$${thanks}\:{for}\:{reply}\:{sir}! \\ $$$${as}\:{i}\:{said},\:{you}\:{don}'{t}\:{need}\:{to}\:{answer} \\ $$$${if}\:{you}\:{don}'{t}\:{want}\:{to}\:{or}\:{if}\:{you}\:{can}'{t}\: \\ $$$${write}\:{in}\:{english}.\:{give}\:{a}\:{short}\:{reply} \\ $$$${like}\:“{sorry},\:{i}\:{can}'{t}\:{answer}.''\:{is}\:{quite}\: \\ $$$${ok}.\:{but}\:{just}\:{simply}\:{ignore}\:{is}\:{not}\:{good}. \\ $$
Commented by mustafazaheen last updated on 16/Apr/23
ok Mr W your speak is very accurate.
$$\mathrm{ok}\:\mathrm{Mr}\:\mathrm{W}\:\mathrm{your}\:\mathrm{speak}\:\mathrm{is}\:\mathrm{very}\:\mathrm{accurate}. \\ $$
Commented by mathlove last updated on 16/Apr/23
  I don't know English, brother, I have had this problem before, but I apologize
$$ \\ $$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
$${right} \\ $$
Commented by mathlove last updated on 16/Apr/23
ok
$${ok} \\ $$
Answered by mustafazaheen last updated on 16/Apr/23
=0
$$=\mathrm{0} \\ $$
Commented by mathlove last updated on 16/Apr/23
solve it
$${solve}\:{it} \\ $$
Answered by mehdee42 last updated on 16/Apr/23
it is not impossible to lim_(x→1^− )  (√(1−x))
$${it}\:{is}\:{not}\:{impossible}\:{to}\:{lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\sqrt{\mathrm{1}−{x}} \\ $$$$ \\ $$
Answered by mr W last updated on 16/Apr/23
due to (√(x−1)), x>1.  that means lim_(x→1^− )  □ doesn′t exist  and the question is wrong.    assume the question is  lim_(x→1^+ ) ((sin (x^3 −1) cos (1/(1−x)))/( (√(x−1))))  =lim_(x→1^+ ) (((x−1)(x^2 +x+1))/( (√(x−1))))×((sin (x^3 −1))/( x^3 −1))×cos (1/(1−x))   =lim_(x→1^+ ) (√(x−1))×(x^2 +x+1)×((sin (x^3 −1))/( x^3 −1))×cos (1/(1−x))   ≤lim_(x→1^+ ) (√(x−1))×(x^2 +x+1)×((sin (x^3 −1))/( x^3 −1))=0×1×1=0   ≥−lim_(x→1^+ ) (√(x−1))×(x^2 +x+1)×((sin (x^3 −1))/( x^3 −1))=−0×1×1=0   ⇒lim_(x→1^+ ) ((sin (x^3 −1) cos (1/(1−x)))/( (√(x−1))))=0
$${due}\:{to}\:\sqrt{{x}−\mathrm{1}},\:{x}>\mathrm{1}. \\ $$$${that}\:{means}\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\Box\:{doesn}'{t}\:{exist} \\ $$$${and}\:{the}\:{question}\:{is}\:{wrong}. \\ $$$$ \\ $$$${assume}\:{the}\:{question}\:{is} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}}{\:\sqrt{{x}−\mathrm{1}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{\:\sqrt{{x}−\mathrm{1}}}×\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{\:{x}^{\mathrm{3}} −\mathrm{1}}×\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\: \\ $$$$=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\sqrt{{x}−\mathrm{1}}×\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)×\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{\:{x}^{\mathrm{3}} −\mathrm{1}}×\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\: \\ $$$$\leqslant\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\sqrt{{x}−\mathrm{1}}×\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)×\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{\:{x}^{\mathrm{3}} −\mathrm{1}}=\mathrm{0}×\mathrm{1}×\mathrm{1}=\mathrm{0}\: \\ $$$$\geqslant−\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\sqrt{{x}−\mathrm{1}}×\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)×\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)}{\:{x}^{\mathrm{3}} −\mathrm{1}}=−\mathrm{0}×\mathrm{1}×\mathrm{1}=\mathrm{0}\: \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} −\mathrm{1}\right)\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}}{\:\sqrt{{x}−\mathrm{1}}}=\mathrm{0} \\ $$
Commented by mehdee42 last updated on 16/Apr/23
thats right
$${thats}\:{right} \\ $$
Commented by mathlove last updated on 16/Apr/23
thanks
$${thanks} \\ $$

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