Question Number 160268 by mathlove last updated on 27/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94730 by i jagooll last updated on 20/May/20 Answered by john santu last updated on 20/May/20 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\mathrm{e}^{\mathrm{ln}\left(\mathrm{x}\right)^{\mathrm{sin}\:\mathrm{x}} } =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{e}^{\mathrm{sin}\:\mathrm{x}.\:\mathrm{ln}\left(\mathrm{x}\right)} \\…
Question Number 160256 by qaz last updated on 26/Nov/21 $$\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}^{\mathrm{x}} −\mathrm{sin}\:\mathrm{2}^{\mathrm{x}} }{\mathrm{2}^{\mathrm{x}^{\mathrm{x}} } −\mathrm{2}^{\mathrm{2}^{\mathrm{x}} } }=? \\ $$ Commented by cortano last updated on…
Question Number 94723 by john santu last updated on 20/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{\mathrm{0}} {\overset{{x}^{\mathrm{2}} } {\int}}\:\sqrt{\mathrm{4}+{t}^{\mathrm{3}} \:}\:{dt}}{{x}^{\mathrm{2}} }\:?\: \\ $$ Answered by i jagooll last updated…
Question Number 29180 by A1B1C1D1 last updated on 05/Feb/18 $$\underset{\mathrm{x}\:\rightarrow\:\mathrm{a}} {\mathrm{lim}}\:\left(\frac{\sqrt[{\mathrm{m}}]{\mathrm{x}}\:−\:\sqrt[{\mathrm{m}}]{\mathrm{a}}}{\:\sqrt[{\mathrm{n}}]{\mathrm{x}}\:−\:\sqrt[{\mathrm{n}}]{\mathrm{a}}}\right) \\ $$$$\left.\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{L}'\mathrm{hospital}\:\mathrm{rules}\::-\right) \\ $$ Answered by Rasheed.Sindhi last updated on 05/Feb/18 $$\mathrm{Formula} \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{a}}…
Question Number 160244 by cortano last updated on 26/Nov/21 $$\:\:\:\:\:{L}\underset{{x}\rightarrow\mathrm{0}} {{i}m}\:\left(\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)−\mathrm{tan}\:\left({x}\right)}{\mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\mathrm{sin}\:\left({x}\right)}\right)=? \\ $$ Answered by FongXD last updated on 26/Nov/21 $$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}−\mathrm{tanx}}{\mathrm{x}^{\mathrm{3}}…
Question Number 160241 by mathlove last updated on 26/Nov/21 Commented by blackmamba last updated on 26/Nov/21 $$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{9}}]{\mathrm{7}{x}−\mathrm{6}}−\mathrm{1}}{{nx}^{\mathrm{3}} −{n}}\: \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{{n}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{9}}]{\mathrm{7}{x}−\mathrm{6}}−\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{3}{n}}\:\underset{{x}\rightarrow\mathrm{1}}…
Question Number 29158 by abdo imad last updated on 04/Feb/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{1}} \frac{\sqrt{\mathrm{3}+\sqrt{\mathrm{2}{x}−\mathrm{1}}}\:−\mathrm{2}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}{x}+\mathrm{1}}}\:\:−\sqrt{{x}+\mathrm{3}}}\:\:. \\ $$ Commented by abdo imad last updated on 09/Feb/18 $${in}\:{this}\:{case}\:{its}\:{better}\:{to}\:{use}\:{hospital}\:{theorem}\:{let}\:{put} \\ $$$${u}\left({x}\right)=\sqrt{\mathrm{3}+\sqrt{\mathrm{2}{x}−\mathrm{1}}}\:−\mathrm{2}\:{and}\:{v}\left({x}\right)=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}{x}+\mathrm{1}}}\:\:−\sqrt{{x}+\mathrm{3}}…
Question Number 29157 by abdo imad last updated on 04/Feb/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{1}} \:\:\:\frac{\left({x}−\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:+\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:. \\ $$ Commented by abdo imad last updated on…
Question Number 160218 by qaz last updated on 26/Nov/21 $$\underset{\mathrm{x}\rightarrow+\infty} {\mathrm{lim}}\frac{\int_{\mathrm{0}} ^{\mathrm{x}} \left(\mathrm{arctan}\:\mathrm{t}\right)^{\mathrm{2}} \mathrm{dt}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}=? \\ $$ Answered by mathmax by abdo last updated on…