Question Number 195066 by alcohol last updated on 23/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195075 by tri26112004 last updated on 23/Jul/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{3}} {x}}{{x}} \\ $$ Answered by MM42 last updated on 23/Jul/23 $$\exists\:{m}>\mathrm{0}\:;\:\forall\:{x}\in\mathbb{R}\:\Rightarrow\:\mid{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{3}} {x}\mid\leqslant{m}…
Question Number 195033 by mathlove last updated on 22/Jul/23 $${any}\:{point}\:{is}\:{the}\:{function}\:{is} \\ $$$${not}\:{continous} \\ $$$${f}\left({x}\right)=\left(\mathrm{4}{x}+\mathrm{8}\right)^{\frac{{ln}\mathrm{45}}{\mathrm{8}}} \\ $$$$\left.{a}\left.\right)\:−\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\right)\:−\mathrm{2} \\ $$$$\left.{c}\left.\right)\:{no}\:{one}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}\right)\:\mathrm{5} \\ $$ Answered by alephzero last updated…
Question Number 195029 by cortano12 last updated on 22/Jul/23 $$\:\:\:\:\underbrace{ } \\ $$ Answered by MM42 last updated on 22/Jul/23 $${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}{ln}\left(\mathrm{1}+{x}\right)}\:=\overset{{hop}} {\rightarrow}\: \\…
Question Number 195021 by tri26112004 last updated on 22/Jul/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−{sin}\:{x}}{{x}^{{n}} }=¿\:\left({n}\in{N}^{\ast} \right) \\ $$ Answered by MM42 last updated on 22/Jul/23 $${if}\:\:{x}\rightarrow\mathrm{0}\:\Rightarrow\:{x}−{sinx}\:\sim\:\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \\ $$$$\Rightarrow{if}\:\:\:{n}<\mathrm{3}\:\Rightarrow\:{lim}=\mathrm{0}…
Question Number 195013 by horsebrand11 last updated on 22/Jul/23 $$\:\:\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\left(\frac{\mathrm{2}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2x}}\:+\sqrt{\mathrm{6}−\mathrm{2x}}\right)}{\:\sqrt{\mathrm{36}−\mathrm{4x}^{\mathrm{2}} }}\:\right) \\ $$ Answered by cortano12 last updated on 22/Jul/23 $$\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\frac{\sqrt{\mathrm{6}−\mathrm{2}{x}}}{\:\sqrt{\mathrm{6}}\:+\sqrt{\mathrm{2}{x}}}\:+\mathrm{1}\:\right)}{\:\sqrt{\mathrm{6}+\mathrm{2}{x}}} \\ $$$$\:\:\:\:\:=\:\:\begin{array}{|c|}{\frac{\mathrm{2}}{\:\sqrt{\mathrm{12}}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{3}}}\\\hline\end{array}…
Question Number 194968 by cortano12 last updated on 21/Jul/23 Answered by tri26112004 last updated on 21/Jul/23 $$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{1}−{x}}}.−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{1}−{x}}−\mathrm{2}}{{x}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}.\left\{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{4}{x}+\mathrm{4}}−\left({x}+\mathrm{2}\right)\right]+\sqrt{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{1}}{{x}^{\mathrm{2}} }\right\} \\…
Question Number 194928 by cortano12 last updated on 19/Jul/23 $$\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{9}}−\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{8}}\:=?\right. \\ $$ Answered by horsebrand11 last updated on 20/Jul/23…
Question Number 194879 by tri26112004 last updated on 18/Jul/23 Answered by cortano12 last updated on 18/Jul/23 $$\:\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{2}{u}\right)^{{u}} −\mathrm{1}}{\left(\mathrm{1}+\mathrm{3}{u}\right)^{{u}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\left[\:{u}\:=\frac{\mathrm{1}}{{x}}\:\right] \\ $$$$\:{then}\:{apply}\:{L}'{Hopital}\: \\…
Question Number 194891 by cortano12 last updated on 18/Jul/23 $$\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{3}^{+} } {\mathrm{lim}}\:\left(\frac{\sqrt{{x}}−\sqrt{{x}−\mathrm{3}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}\:\right)=? \\ $$ Answered by MM42 last updated on 18/Jul/23 $${lim}_{{x}\rightarrow\mathrm{3}} \:\left(\frac{\sqrt{{x}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:−\:\frac{\sqrt{{x}−\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:\right) \\…