Question Number 187270 by MikeH last updated on 15/Feb/23 $$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}\:+\:\mathrm{sin}\left(\mathrm{2}{x}\right)\:+\mathrm{1}}{{x}^{\mathrm{2}} −\pi^{\mathrm{2}} }\:= \\ $$$$\mathrm{A}.\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\:\:\:\:\:\:\:\:{C}.\:\mathrm{1} \\ $$$${B}.\:\frac{\mathrm{1}}{\pi}\:\:\:\:\:\:\:\:\:{D}.\:\mathrm{Does}\:\mathrm{Not}\:\mathrm{exist} \\ $$ Answered by horsebrand11 last updated on…
Question Number 121727 by oustmuchiya@gmail.com last updated on 11/Nov/20 Answered by TANMAY PANACEA last updated on 11/Nov/20 $$\left.\mathrm{1}\right){b}^{\mathrm{2}} =\frac{{a}+\mathrm{1}}{\mathrm{2}}\rightarrow\mathrm{2}{b}×\frac{{db}}{{da}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{when}\:{a}=\mathrm{4}\:\:\:{b}=\sqrt{\frac{\mathrm{5}}{\mathrm{2}}}\: \\ $$$${gradint}\:\left(\frac{{db}}{{da}}\right)_{{a}=\mathrm{4}\:} =\left(\frac{\mathrm{1}}{\mathrm{4}{b}}\right)_{{b}=\sqrt{\frac{\mathrm{5}}{\mathrm{2}}}\:} =\frac{\sqrt{\mathrm{2}}}{\mathrm{4}×\sqrt{\mathrm{5}}} \\ $$$$\left.{ii}\right)\frac{{dt}}{{dx}}=\frac{{x}×\mathrm{2}{x}−\left({x}^{\mathrm{2}}…
Question Number 121726 by oustmuchiya@gmail.com last updated on 11/Nov/20 Answered by TANMAY PANACEA last updated on 11/Nov/20 $$\mathrm{6}{n}^{\mathrm{3}} >\mathrm{4}{n}^{\mathrm{3}} \:\:{also}\:\mathrm{2}^{{n}} >\mathrm{2} \\ $$$${so}\:\:\mathrm{6}{n}^{\mathrm{3}} +\mathrm{2}^{{n}} >\mathrm{4}{n}^{\mathrm{3}}…
Question Number 121725 by oustmuchiya@gmail.com last updated on 11/Nov/20 Answered by TANMAY PANACEA last updated on 11/Nov/20 $${f}\left({n}\right)=\frac{{n}^{\mathrm{2}} −\mathrm{2}{n}−\mathrm{15}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{5}{n}−\mathrm{3}}=\frac{{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{3}{n}−\mathrm{15}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{6}{n}−{n}−\mathrm{3}}=\frac{{n}\left({n}−\mathrm{5}\right)+\mathrm{3}\left({n}−\mathrm{5}\right)}{\mathrm{2}{n}\left({n}+\mathrm{3}\right)−\mathrm{1}\left({n}+\mathrm{3}\right)} \\ $$$${f}\left({n}\right)=\frac{\left({n}−\mathrm{5}\right)\left({n}+\mathrm{3}\right)}{\left({n}+\mathrm{3}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}=\frac{{n}−\mathrm{5}}{\mathrm{2}{n}−\mathrm{1}} \\…
Question Number 121723 by oustmuchiya@gmail.com last updated on 11/Nov/20 Answered by TANMAY PANACEA last updated on 11/Nov/20 $${f}\left({x}\right)=\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}} \\ $$$${li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\mathrm{3}+\frac{\mathrm{2}}{{x}\:}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}=\frac{\mathrm{3}+\mathrm{0}+\mathrm{0}}{\mathrm{1}−\mathrm{0}}=\mathrm{3}…
Question Number 121716 by bemath last updated on 11/Nov/20 $$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)…\left(\mathrm{3}{n}\right)}{{n}^{\mathrm{2}{n}} }\:\right)^{\frac{\mathrm{1}}{{n}}} =? \\ $$ Answered by Dwaipayan Shikari last updated on 11/Nov/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)…\left(\mathrm{1}+\frac{\mathrm{2}{n}}{{n}}\right)\right)^{\frac{\mathrm{1}}{{n}}}…
Question Number 187184 by Safiullah_21 last updated on 14/Feb/23 $$ ×/ \left\{ \left\{ \left\{\left\{×=\right.\right.\right.\right. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121638 by bemath last updated on 10/Nov/20 $$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{xf}\left({x}\right)−{af}\left({a}\right)}{{x}−{a}}\:? \\ $$ Answered by Dwaipayan Shikari last updated on 10/Nov/20 $$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{xf}^{'} \left({x}\right)+{f}\left({x}\right)}{\mathrm{1}}=\:{af}'\left({a}\right)+{f}\left({a}\right) \\…
Question Number 121618 by SOMEDAVONG last updated on 10/Nov/20 Answered by bemath last updated on 10/Nov/20 $$\:\mathrm{let}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{u}^{\mathrm{mn}\:} \\ $$$$\:\underset{\mathrm{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{1}−\mathrm{u}^{\mathrm{m}} \right)}\:−\:\frac{\mathrm{1}}{\mathrm{m}\left(\mathrm{1}−\mathrm{u}^{\mathrm{n}} \right)}\right)\:= \\ $$$$\:\underset{\mathrm{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{m}\left(\mathrm{1}−\mathrm{u}^{\mathrm{n}}…
Question Number 121576 by oustmuchiya@gmail.com last updated on 09/Nov/20 Answered by TANMAY PANACEA last updated on 09/Nov/20 $${f}\left({x}\right)=\frac{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:−\mathrm{5}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{15}−\mathrm{25}}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)}=\frac{{x}^{\mathrm{2}}…