Question Number 34540 by rahul 19 last updated on 07/May/18 $$\left.{a}\right)\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)^{\mathrm{3}} −\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}}\:=? \\ $$$$\left.{b}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{sin}\:{x}\right)^{\frac{\mathrm{1}}{{x}}} \:=\:? \\ $$ Commented by abdo mathsup 649 cc…
Question Number 34522 by rahul 19 last updated on 07/May/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{log}\:_{{e}} \left\{\frac{\mathrm{sin}\:\left({a}+\frac{\mathrm{1}}{{x}}\right)}{\mathrm{sin}\:{a}}\right\}^{{x}} ,\:\mathrm{0}<{a}<\frac{\pi}{\mathrm{2}}\:. \\ $$ Commented by rahul 19 last updated on 08/May/18 $$?…
Question Number 34516 by rahul 19 last updated on 07/May/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{log}\:_{\mathrm{tan}\:^{\mathrm{2}} {x}} \left(\mathrm{tan}\:^{\mathrm{2}} \mathrm{2}{x}\right)\:=\:? \\ $$ Commented by math khazana by abdo last updated…
Question Number 100036 by bobhans last updated on 24/Jun/20 $$\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2x}}−\sqrt{\mathrm{1}−\mathrm{2sin}\:\mathrm{x}}}{\mathrm{x}} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\:=\:\mathrm{2x}+\:\sqrt{\mathrm{2x}}\:.\:\mathrm{find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right) \\ $$ Answered by john santu last updated on 24/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\:=\:\mathrm{g}\left(\underset{{x}\rightarrow\mathrm{0}}…
Question Number 100037 by bobhans last updated on 24/Jun/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\sqrt{\mathrm{x}}\:\mathrm{sin}\:\left(\mathrm{2x}\right)\:? \\ $$ Commented by bramlex last updated on 24/Jun/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\sqrt{{x}}\:=\:\infty\:\wedge\:−\mathrm{1}\leqslant\:\mathrm{sin}\:\mathrm{2}{x}\:\leqslant\:\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\sqrt{{x}}\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\infty\:…
Question Number 100027 by bemath last updated on 24/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{xsin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:?\: \\ $$ Commented by john santu last updated on 24/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{x}\:=\:\mathrm{0}\:\&\:−\mathrm{1}\leqslant\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{then}\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 34464 by rahul 19 last updated on 06/May/18 $$\boldsymbol{{E}}{valuate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{x}^{{m}\:} \left({log}\:{x}\:\right)^{{n}} \:,\:{m},{n}\:\in\:\mathbb{N} \\ $$ Answered by MJS last updated on…
Question Number 34458 by rahul 19 last updated on 06/May/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}\:−\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)}{{x}^{\mathrm{3}} }\:=\:? \\ $$ Commented by math khazana by abdo last updated on 06/May/18…
Question Number 34439 by rahul 19 last updated on 06/May/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(−\mathrm{1}\right)^{{x}−\mathrm{1}} \mathrm{sin}\:\left(\pi\sqrt{{x}^{\mathrm{2}} +\mathrm{0}.\mathrm{5}{x}+\mathrm{1}}\right), \\ $$$${where}\:{x}\in\mathbb{N}. \\ $$ Answered by MJS last updated on 07/May/18…
Question Number 99972 by bobhans last updated on 24/Jun/20 Commented by bemath last updated on 24/Jun/20 $$\mathrm{b}\:=\:\mathrm{9}\:\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{a}}{\mathrm{2}\sqrt{\mathrm{ax}+\mathrm{9}}}\:=\:\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{a}}{\mathrm{6}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{a}\:=\:\mathrm{2}\:.\:\Rightarrow\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt{\mathrm{9x}−\mathrm{4}}−\sqrt{\mathrm{5}}}{\mathrm{x}−\mathrm{1}}\:=\:\frac{\mathrm{9}}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$ \\ $$…