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) \\…
Question Number 194685 by cortano12 last updated on 13/Jul/23 Answered by qaz last updated on 14/Jul/23 $$\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}\right)} ={e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…} \\ $$$$={e}\left(\mathrm{1}+\left(−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…\right)^{\mathrm{2}} +…\right.…
Question Number 194640 by cortano12 last updated on 12/Jul/23 $$\:\:\:\:\underbrace{ } \\ $$ Answered by horsebrand11 last updated on 12/Jul/23 $$\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:=\:\mathrm{y} \\ $$$$\:\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}}{\mathrm{y}\:\mathrm{sin}\:\mathrm{3y}}\:=\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}}…
Question Number 194462 by Erico last updated on 08/Jul/23 $$\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:+\infty^{} } \frac{\mathrm{1}−\boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} } }{\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}}=\sqrt{\boldsymbol{\pi}} \\ $$ Answered by mnjuly1970 last updated on 08/Jul/23…
Question Number 194482 by horsebrand11 last updated on 08/Jul/23 $$\:\:\:\:\:\begin{array}{|c|}{\:\cancel{\underline{\underbrace{\Subset}}}}\\\hline\end{array} \\ $$ Answered by cortano12 last updated on 08/Jul/23 $$\:\:\:\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:\left(\frac{\mathrm{2}}{\mathrm{cos}\:\mathrm{x}}\:+\:\mathrm{cos}\:\mathrm{x}−\mathrm{3}\right)\right] \\ $$$$\:\:\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{cos}\:^{\mathrm{2}}…
Question Number 194466 by horsebrand11 last updated on 08/Jul/23 $$\:\:\:\:\Subset \\ $$ Answered by cortano12 last updated on 08/Jul/23 $$\:\:\:\underbrace{\Subset} \\ $$ Answered by witcher3…