Question Number 176638 by cortano1 last updated on 23/Sep/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{cos}\:\left(\mathrm{x}^{\mathrm{2}} \right)\right)}=? \\ $$ Answered by a.lgnaoui last updated on 23/Sep/22…
Question Number 111092 by bemath last updated on 02/Sep/20 $$\:\:\sqrt{\mathrm{bemath}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arctan}\:\mathrm{x}}{\mathrm{arc}\:\mathrm{sin}\:\mathrm{x}−\mathrm{x}} \\ $$ Commented by imsahil last updated on 02/Sep/20 How to write cube root? Commented by…
Question Number 176589 by Ar Brandon last updated on 22/Sep/22 Answered by Peace last updated on 23/Sep/22 $$\mid{u}_{{n}} \left({z}\right)\mid=\frac{\mid{z}\mid^{{n}} }{\mid\mathrm{1}+{z}^{\mathrm{2}{n}+\mathrm{1}} \mid}\leqslant\frac{\mid{z}\mid^{{n}} }{\mathrm{1}−\mid{z}\mid^{\mathrm{2}{n}+\mathrm{1}} },\:\mid{z}\mid<\mathrm{1} \\ $$$$\exists{n}\in\mathbb{N}\mid\forall{m}\geqslant{n}\:\:\:\:\mid{z}\mid\:\:\:<\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow.\mid{U}_{{m}}…
Question Number 176571 by mathlove last updated on 21/Sep/22 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{lnx}}{\mathrm{1}+{lnx}−\mathrm{1}}=? \\ $$ Answered by Peace last updated on 21/Sep/22 $$=\mathrm{1} \\ $$ Terms of…
Question Number 45494 by Meritguide1234 last updated on 13/Oct/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 110988 by Rio Michael last updated on 01/Sep/20 $$\:\mathrm{Evaluate}\:\mathrm{without}\:\mathrm{using}\:\mathrm{L}'\mathrm{hopital}'\mathrm{s}\:\mathrm{rule} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}} \\ $$ Answered by Rasheed.Sindhi last updated on 01/Sep/20 $$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}}…
Question Number 45443 by Meritguide1234 last updated on 13/Oct/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 176486 by cortano1 last updated on 20/Sep/22 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}^{\mathrm{5}} }\:−\frac{\mathrm{1}}{\mathrm{3x}^{\mathrm{2}} }\:\right]=? \\ $$ Answered by blackmamba last updated on 20/Sep/22 $$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{tan}\:{x}−{x}\right)−{x}^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{5}}…
Question Number 110869 by Study last updated on 31/Aug/20 Answered by mathdave last updated on 31/Aug/20 $${let}\:{I}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}!^{\frac{\mathrm{1}}{{x}}} \\ $$$${by}\:{loggin}\:{both}\:{side}\:{we}\:{have} \\ $$$$\mathrm{ln}{I}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}{x}!}{{x}}\:\:\:\:\:\:{but}\:\:{x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{ln}{I}=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 110861 by apriani last updated on 31/Aug/20 Answered by bemath last updated on 31/Aug/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} }−\left(\mathrm{x}+\mathrm{2}\right)= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} \right)−\left(\mathrm{x}^{\mathrm{2}}…