Question Number 112900 by bemath last updated on 10/Sep/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\right)^{\mathrm{x}} ? \\ $$ Commented by kaivan.ahmadi last updated on 10/Sep/20…
Question Number 112881 by bemath last updated on 10/Sep/20 Answered by Dwaipayan Shikari last updated on 10/Sep/20 $$\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}}… \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\mathrm{40}} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{41}}\right)=\frac{\mathrm{80}}{\mathrm{41}} \\…
Question Number 112866 by bemath last updated on 10/Sep/20 Answered by bobhans last updated on 10/Sep/20 $$\left(\mathrm{7}\right)\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{tan}\:\mathrm{x}} \\ $$$$\:\:\:\mathrm{L}\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}tan}\:\mathrm{x}.\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} =\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\mathrm{tan}\:\mathrm{x}.\mathrm{ln}\:\mathrm{x}} \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}}…
Question Number 112863 by bemath last updated on 10/Sep/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}+\mathrm{3}}{\mathrm{2x}−\mathrm{1}}\right)^{\mathrm{4x}+\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{3x}+\mathrm{1}}{\mathrm{3x}−\mathrm{1}}\right)^{\mathrm{4x}−\mathrm{2}} \\ $$ Commented by bobhans last updated on 22/Sep/20 $$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}{x}+\mathrm{3}}{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{4}{x}+\mathrm{2}}…
Question Number 112855 by ruwedkabeh last updated on 10/Sep/20 $$ \\ $$$$\mathrm{1}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}{x}+\mathrm{4}}−\sqrt{\mathrm{3}{x}+\mathrm{9}}}{\mathrm{4}{x}}=… \\ $$$$ \\ $$$$\mathrm{2}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}−\mathrm{4}{x}+\mathrm{3}{x}^{\mathrm{2}} }−\sqrt{\mathrm{4}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}{x}}=… \\ $$$$ \\ $$ Answered…
Question Number 112851 by bemath last updated on 10/Sep/20 $$\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{a}}{\mathrm{x}}\:−\:\mathrm{cot}\:\frac{\mathrm{x}}{\mathrm{a}}\right)\:? \\ $$ Answered by bobhans last updated on 10/Sep/20 $$\:\mathrm{set}\:\frac{\mathrm{x}}{\mathrm{a}}\:=\:\mathrm{t}\:\rightarrow\frac{\mathrm{a}}{\mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{t}} \\ $$$$\:\mathrm{L}=\:\underset{\mathrm{t}\rightarrow\mathrm{0}}…
Question Number 112782 by bemath last updated on 09/Sep/20 $$\mathrm{without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{x}−\pi} \\ $$ Answered by john santu last updated on 09/Sep/20 $${setting}\:{x}\:=\:\pi+{s}\:,\:{s}\rightarrow\mathrm{0} \\…
Question Number 47174 by vajpaithegrate@gmail.com last updated on 05/Nov/18 $$\mathrm{y}=\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{2}} ^{\mathrm{x}} \right)\mathrm{then}\:\:\frac{\mathrm{dy}}{\mathrm{dx}}= \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18 $${y}={log}_{\mathrm{2}} {t}=\frac{{lnt}}{{ln}\mathrm{2}} \\…
Question Number 112651 by bemath last updated on 09/Sep/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\left(\mathrm{x}−\mathrm{a}\right)\:\mathrm{cosec}\:\left(\frac{\pi\mathrm{x}}{\mathrm{a}}\right)\:? \\ $$$$\left(\mathrm{2}\right)\:\int\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx}\:,\:\mathrm{by}\:\mathrm{using}\:\mathrm{Euler}'\mathrm{s} \\ $$$$\mathrm{substitution} \\ $$ Answered by john santu last updated on…
Question Number 112650 by bemath last updated on 09/Sep/20 $$\mathrm{Determine}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{and}\:\mathrm{e}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{ax}+\mathrm{bx}^{\mathrm{3}} +\mathrm{cx}^{\mathrm{2}} +\mathrm{dx}+\mathrm{e}}{\mathrm{x}^{\mathrm{4}} }\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{find}\:\mathrm{general}\:\mathrm{solution}\::\:\mathrm{3xy}^{\mathrm{2}} \:\mathrm{y}'\:=\:\mathrm{4y}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \: \\ $$ Answered by…