Question Number 159388 by amin96 last updated on 16/Nov/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{4}}+\frac{\mathrm{3}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{9}}+\ldots\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 93824 by student work last updated on 15/May/20 Answered by john santu last updated on 15/May/20 $$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{lny}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{tan}\:\left(\mathrm{3t}+\frac{\pi}{\mathrm{2}}\right)\mathrm{ln}\left(\mathrm{tan}\:\left(\frac{\mathrm{3t}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{tan}\:\left(\frac{\mathrm{3t}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)}{\mathrm{cot}\:\left(\mathrm{3t}+\frac{\pi}{\mathrm{2}}\right)} \\…
Question Number 93821 by john santu last updated on 15/May/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\mathrm{ln}\left(\mathrm{e}^{{x}} +\mathrm{1}\right)−{x}\:=\: \\ $$ Commented by JDamian last updated on 15/May/20 $$\mathrm{0} \\ $$…
Question Number 159349 by cortano last updated on 15/Nov/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8sec}\:{x}−\mathrm{8}+\mathrm{tan}\:^{\mathrm{4}} {x}−\mathrm{4tan}\:^{\mathrm{2}} {x}}{{x}^{\mathrm{6}} }\:=? \\ $$ Commented by bobhans last updated on 16/Nov/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8}\left(\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}}…
Question Number 159318 by tounghoungko last updated on 15/Nov/21 $$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}}{\:\sqrt{{x}−\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:−\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:=? \\ $$ Commented by bobhans last updated on 16/Nov/21 $$\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{+}…
Question Number 93695 by i jagooll last updated on 14/May/20 Commented by mathmax by abdo last updated on 14/May/20 $${f}\left({x}\right)\:=\frac{\mathrm{6}{x}−^{\mathrm{3}} \sqrt{\mathrm{27}{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}}}{\left(^{\mathrm{3}} \sqrt{\mathrm{8}{x}^{\mathrm{3}} −{x}}\:+{x}\right)}\:\Rightarrow{f}\left({x}\right)\:=\frac{\mathrm{6}{x}−\mathrm{3}{x}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{27}{x}^{\mathrm{3}}…
Question Number 159229 by tounghoungko last updated on 14/Nov/21 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}+{x}−\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}}\:\right)=? \\ $$ Answered by FongXD last updated on 14/Nov/21 $$\mathrm{L}=\underset{\mathrm{x}\rightarrow+\infty} {\mathrm{lim}x}\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}}−\mathrm{x}+\mathrm{2x}−\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 28151 by tawa tawa last updated on 21/Jan/18 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\left(\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{e}} \mathrm{x}}\:−\:\frac{\mathrm{x}}{\mathrm{x}\:−\:\mathrm{1}}\right) \\ $$ Commented by çhëý böý last updated on 21/Jan/18 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\left({x}−\mathrm{1}\right)−{xlnx}}{\left({x}−\mathrm{1}\right){lnx}}\right)=\frac{\mathrm{0}}{\mathrm{0}}…
Question Number 159217 by mathlove last updated on 14/Nov/21 Answered by mnjuly1970 last updated on 15/Nov/21 $$\:\:\:\:\:\:\:{solution}.. \\ $$$$\:\:\:\Omega:\overset{{hopital}\:{rule}} {=}{lim}_{\:{x}\rightarrow\frac{\pi}{\mathrm{6}}} \:\frac{\:\mathrm{2}\:.\:\frac{\left(\Gamma\left({sin}\left({x}\right)\right)'\right.}{\Gamma\left({sin}\left({x}\right)\right)}}{\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}\overset{\:\:'} {\right)}\Gamma'\left(\:\frac{\mathrm{1}}{{sin}\left({x}\right)}\right)}\: \\ $$$$\:\:=\mathrm{2}\:{lim}_{\:{x}\rightarrow\frac{\pi}{\mathrm{6}}} \frac{\:{cos}\left({x}\right)\Gamma'\left({sin}\left({x}\right)\right)}{\Gamma\left(\:{sin}\left({x}\right)\right)}\:.\frac{{sin}^{\:\mathrm{2}}…
Question Number 93676 by john santu last updated on 14/May/20 $$\mathrm{solve}\:\mathrm{without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{x}+\mathrm{4}}−\mathrm{1}}{\mathrm{x}}\:? \\ $$ Commented by mathmax by abdo last updated on 14/May/20…