Question Number 176808 by cortano1 last updated on 27/Sep/22 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}=? \\ $$ Answered by TheHoneyCat last updated on 08/Oct/22 $$\mathrm{let}\:{n}\in\mathbb{Z} \\ $$$$\frac{\mathrm{sin}\left(\mathrm{2}\pi{n}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2}\pi{n}\right)}=\frac{\mathrm{0}}{\mathrm{1}+\mathrm{1}^{\mathrm{2}}…
Question Number 111259 by bobhans last updated on 03/Sep/20 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}}{\:\sqrt{\mathrm{3x}}}\:? \\ $$ Commented by bobhans last updated on 03/Sep/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}}{\:\sqrt{\mathrm{3x}}}\:.\:\frac{\sqrt{\mathrm{3x}}}{\:\sqrt{\mathrm{3x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} }…
Question Number 111264 by bobhans last updated on 03/Sep/20 Commented by bobhans last updated on 03/Sep/20 Answered by abdomsup last updated on 03/Sep/20 $$=\mathrm{0}\:{due}\:{to}\:\mid{sin}\left(\frac{\mathrm{2}}{{x}}\right)\mid\leqslant\mathrm{1}\:{for}\:{x}\neq\mathrm{0} \\…
Question Number 176721 by youssefelaour last updated on 25/Sep/22 Answered by MJS_new last updated on 25/Sep/22 $$=\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:{x}}{\mathrm{2cos}\:{x}}\:=−\mathrm{1} \\ $$ Answered by cortano1 last updated…
Question Number 176710 by eka last updated on 25/Sep/22 Answered by a.lgnaoui last updated on 26/Sep/22 $$\:\mathrm{3}.\:\:\:{lim}_{{x}\rightarrow−\mathrm{1}} \:\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{{x}^{\mathrm{2}} −\mathrm{2}}=\frac{\sqrt{\mathrm{4}}}{−\mathrm{1}}=−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:{anser}\:\left(\boldsymbol{\mathrm{D}}\right) \\ $$$$\mathrm{4}.\:\:\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}}{{x}−\mathrm{2}}=\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{4}\right)}{{x}−\mathrm{2}}={x}+\mathrm{4} \\ $$$$\:\:\:{lim}\frac{{x}^{\mathrm{2}}…
Question Number 176692 by mnjuly1970 last updated on 25/Sep/22 Commented by cortano1 last updated on 25/Sep/22 $$\:\Omega\:=\:\mathrm{8}\pi^{\mathrm{4}} \\ $$ Answered by blackmamba last updated on…
Question Number 176672 by youssefelaour last updated on 24/Sep/22 Answered by Mathspace last updated on 25/Sep/22 $${we}\:{do}\:{the}\:{changement}\:{x}−\frac{\pi}{\mathrm{2}}={t} \\ $$$${so}\:{l}\left({x}\right)=\frac{\left(\mathrm{1}−{cost}\right)\left(\mathrm{1}−{cos}^{\mathrm{2}} {t}\right)….\left(\mathrm{1}−{cos}^{{n}} {t}\right)}{{sin}^{\mathrm{2}{n}} {t}} \\ $$$${but}\:\mathrm{1}−{cost}\sim\frac{{t}^{\mathrm{2}} }{\mathrm{2}}…
Question Number 176673 by cortano1 last updated on 24/Sep/22 Answered by mr W last updated on 24/Sep/22 $$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{14}\left({x}^{\mathrm{6}} −\mathrm{1}\right)^{\frac{\mathrm{6}}{\mathrm{7}}} }{\mathrm{15}{x}^{\frac{\mathrm{26}}{\mathrm{5}}} }=\mathrm{0} \\ $$ Terms…
Question Number 111103 by bemath last updated on 02/Sep/20 $$\:\:\sqrt{\mathrm{bemath}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{3x}\right)}{\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{8x}\right)}\:? \\ $$$$\left[\:\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital}\:\right] \\ $$$$ \\ $$ Answered by Lordose last updated on…
Question Number 111100 by bobhans last updated on 02/Sep/20 $$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}\:? \\ $$ Answered by 1549442205PVT last updated on 02/Sep/20 $$\underset{\mathrm{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}}…