Question Number 55334 by Mikael_Marshall last updated on 21/Feb/19 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{8}^{{n}} }{\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{3}^{{n}+\mathrm{2}} } \\ $$ Answered by mr W last updated on 21/Feb/19 $$=\underset{{n}\rightarrow\infty}…
Question Number 120862 by bemath last updated on 03/Nov/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{x}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{1}}{\mathrm{sec}\:\left(\frac{\mathrm{3}}{\mathrm{x}}\right)\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{x}}\right)}\:? \\ $$ Commented by liberty last updated on 03/Nov/20 $$\mathrm{take}\:\mathrm{X}=\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{then}\:\:\underset{\mathrm{X}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2X}−\mathrm{cos}\:\mathrm{X}+\mathrm{1}}{\mathrm{sec}\:\mathrm{3X}.\:\mathrm{tan}\:^{\mathrm{2}}…
Question Number 120843 by bramlexs22 last updated on 03/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}+\mathrm{9}−\mathrm{6}\sqrt{\mathrm{x}}}}{\:\sqrt{\mathrm{x}}−\mathrm{3}}\:? \\ $$ Commented by Dwaipayan Shikari last updated on 03/Nov/20 $$\frac{\sqrt{\mathrm{12}−\mathrm{6}\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{3}}−\mathrm{3}}=−\mathrm{1} \\ $$ Answered…
Question Number 120823 by bramlexs22 last updated on 03/Nov/20 Answered by liberty last updated on 03/Nov/20 $$\:\rho\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{cos}\:\mathrm{x}+\sqrt{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{x}^{\mathrm{3}} }}−\mathrm{1}}{\mathrm{xsin}\:\mathrm{x}} \\ $$$$\:\rho=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)+\sqrt{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{x}^{\mathrm{3}} }}{\mathrm{x}\:\mathrm{sin}\:\mathrm{x}}\:.\left(\frac{\mathrm{1}}{\mathrm{2}}\right)…
Question Number 186327 by cortano1 last updated on 03/Feb/23 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\mathrm{tan}\:^{\mathrm{2}} {x}}{{x}^{\mathrm{6}} }\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186267 by TUN last updated on 02/Feb/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{2016}{x}\right)^{\mathrm{2017}} −\left(\mathrm{1}+\mathrm{2017}{x}\right)^{\mathrm{2016}} }{{x}^{\mathrm{2}} } \\ $$$${without}\:{L}'{H}\:{or}\:{series} \\ $$ Commented by JDamian last updated on 02/Feb/23…
Question Number 120720 by bramlexs22 last updated on 02/Nov/20 $$\:\mathrm{Given}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{x}+\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{3}}\:−\sqrt{\mathrm{x}}+\mathrm{b}\:=\:\mathrm{4} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{b}^{\mathrm{2}} +\mathrm{1} \\ $$ Answered by Dwaipayan Shikari last updated on 02/Nov/20 $$\sqrt{{x}}\left(\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{{x}}}+\frac{\mathrm{3}}{{x}}}−\mathrm{1}+\frac{{b}}{\:\sqrt{{x}}}\right)=\mathrm{4}…
Question Number 120686 by bobhans last updated on 01/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{tan}\:\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\:? \\ $$ Answered by john santu last updated on 02/Nov/20 $${standart}\:{way}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 55149 by pooja24 last updated on 18/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 18/Feb/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{li}} \\ $$$${x}={tan}\theta\:\:\:\:{as}\:{x}\rightarrow\mathrm{0}\:\:\theta\rightarrow\mathrm{0} \\ $$$$\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}}{{ln}\left({tan}\theta+{sec}\theta\right)}−\frac{\mathrm{1}}{{ln}\left(\mathrm{1}−{tan}\theta\right)} \\ $$$$\underset{\theta\rightarrow\mathrm{0}}…
Question Number 186065 by TUN last updated on 31/Jan/23 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{7}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{9}}−\sqrt{\mathrm{6}{x}+\mathrm{3}}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$ Commented by Frix last updated on 01/Feb/23…