Question Number 104838 by bramlex last updated on 24/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\:^{\mathrm{3}} \left(\mathrm{8}{x}\right)−\mathrm{1}}{\mathrm{6}{x}^{\mathrm{2}} }\:? \\ $$ Answered by john santu last updated on 24/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:\mathrm{8}{x}−\mathrm{1}\right)\left(\mathrm{cos}\:^{\mathrm{2}}…
Question Number 104812 by bobhans last updated on 24/Jul/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\frac{\mathrm{3}}{{x}}\right)}{\left(\frac{\mathrm{2}}{{x}}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{25}}\:? \\ $$ Answered by bemath last updated on 24/Jul/20 $${set}\:\frac{\mathrm{1}}{{x}}\:=\:{a}\:,\:{a}\rightarrow\mathrm{0} \\ $$$$\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}{a}}{\left(\mathrm{2}{a}+\mathrm{5}\right)^{\mathrm{2}}…
Question Number 170321 by amin96 last updated on 20/May/22 Answered by aleks041103 last updated on 21/May/22 $$\left.{a}\right) \\ $$$$\frac{\mathrm{2}^{{x}} }{\mathrm{3}^{{x}^{\mathrm{2}} } }=\mathrm{2}^{{x}} \mathrm{3}^{−{x}^{\mathrm{2}} } ={e}^{{ln}\left(\mathrm{2}\right){x}−{ln}\left(\mathrm{3}\right){x}^{\mathrm{2}}…
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Question Number 104746 by bobhans last updated on 23/Jul/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}{x}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \: \\ $$ Answered by Dwaipayan Shikari last updated on 23/Jul/20…
Question Number 170221 by mathlove last updated on 18/May/22 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(\frac{\left(\sqrt{{x}}\right)^{\sqrt{{x}}} −{x}^{{x}} }{\left(\sqrt{{x}}\right)^{{x}} −{x}^{\sqrt{{x}}} }\right)=? \\ $$$${help}\:{me} \\ $$ Commented by mathlove last updated…
Question Number 170211 by mathlove last updated on 18/May/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{tan}^{−\mathrm{1}} \left(\frac{{x}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} }\right)=? \\ $$$${pleas}\:{solve}\:{this} \\ $$ Answered by aleks041103 last updated…
Question Number 104667 by bobhans last updated on 23/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{3}^{\pi{x}} }{\mathrm{tan}\:{x}}\:−\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}\right]\:? \\ $$ Answered by bramlex last updated on 23/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}^{\pi{x}} .\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}−\:\frac{\mathrm{cos}\:^{\mathrm{2}}…
Question Number 104664 by ~blr237~ last updated on 23/Jul/20 $$\:\:\left(\mathrm{1}−{lna}\right)\left(\mathrm{1}−{lnb}\right)…….\left(\mathrm{1}−{lnx}\right)\left(\mathrm{1}−{lny}\right)\left(\mathrm{1}−{lnz}\right)=\mathrm{0}\:\:???? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170197 by mathlove last updated on 18/May/22 Answered by qaz last updated on 18/May/22 $$\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2018}} \right)=\mathrm{x}^{\mathrm{2018}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{4036}} +… \\ $$$$\left(\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\right)^{\mathrm{2018}} =\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +…\right)^{\mathrm{2018}} =\mathrm{x}^{\mathrm{2018}}…