Question Number 121576 by oustmuchiya@gmail.com last updated on 09/Nov/20 Answered by TANMAY PANACEA last updated on 09/Nov/20 $${f}\left({x}\right)=\frac{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:−\mathrm{5}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{15}−\mathrm{25}}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)}=\frac{{x}^{\mathrm{2}}…
Question Number 121574 by oustmuchiya@gmail.com last updated on 09/Nov/20 Answered by TANMAY PANACEA last updated on 09/Nov/20 $${t}={li}\underset{{n}\rightarrow\infty} {{m}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \\ $$$${lnt}={li}\underset{{n}\rightarrow\infty} {{m}nln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right) \\ $$$${lnt}={li}\underset{{y}\rightarrow\mathrm{0}} {{m}}\:\frac{{ln}\left(\mathrm{1}+{y}\right)}{{y}}=\mathrm{1}…
Question Number 121577 by oustmuchiya@gmail.com last updated on 09/Nov/20 Answered by TANMAY PANACEA last updated on 09/Nov/20 $${f}\left({x}\right)=\frac{\mathrm{3}+\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }} \\ $$$${li}\underset{{x}\rightarrow\infty} {{m}f}\left({x}\right)=\frac{\mathrm{3}+\mathrm{0}+\mathrm{0}}{\mathrm{1}}=\mathrm{3} \\ $$$$…
Question Number 121575 by oustmuchiya@gmail.com last updated on 09/Nov/20 Commented by Dwaipayan Shikari last updated on 09/Nov/20 $${R}−\left\{\mathrm{7},−\mathrm{7}\right\} \\ $$ Answered by TANMAY PANACEA last…
Question Number 187092 by cortano12 last updated on 13/Feb/23 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}^{\mathrm{3}} −\mathrm{sin}\:^{\mathrm{3}} {x}}{{x}^{\mathrm{3}} \left(\mathrm{cos}\:{x}^{\mathrm{3}} −\mathrm{cos}\:^{\mathrm{3}} {x}\right)}\:=? \\ $$ Answered by Farhadazizi last updated on 13/Feb/23…
Question Number 121527 by benjo_mathlover last updated on 09/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\pi+\mathrm{2sin}\:\mathrm{x}}\:−\sqrt{\pi}}\:? \\ $$ Answered by Dwaipayan Shikari last updated on 09/Nov/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{xcosx}}{\pi+\mathrm{2}{sinx}−\pi}\left(\sqrt{\pi+\mathrm{2}{sinx}}+\sqrt{\pi}\right) \\ $$$$=\frac{{x}}{\mathrm{2}{sinx}}\left(\sqrt{\pi}+\sqrt{\pi}\right)\:\:\:\:\:\:\:\:\:\left({sinx}\rightarrow{x}\:\:\:{and}\:{x}\rightarrow\mathrm{0}\right)…
Question Number 121506 by benjo_mathlover last updated on 08/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{x}+\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} \:?\: \\ $$ Answered by liberty last updated on 09/Nov/20 $$\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{x}+\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}}…
Question Number 121502 by bramlexs22 last updated on 08/Nov/20 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}\sqrt{\mathrm{2}}}\:? \\ $$$$\left(\mathrm{2}\right)\:\int\:\mathrm{cot}\:^{\mathrm{3}} \mathrm{x}\:\mathrm{csc}\:^{\mathrm{3}} \mathrm{x}\:\mathrm{dx}\: \\ $$$$ \\ $$ Commented by liberty last updated on…
Question Number 121460 by bramlexs22 last updated on 08/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(\mathrm{cot}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{x}}} \:?\: \\ $$ Answered by Dwaipayan Shikari last updated on 08/Nov/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({cotx}\right)^{\frac{\mathrm{1}}{{logx}}}…
Question Number 186983 by cortano12 last updated on 12/Feb/23 Answered by CElcedricjunior last updated on 12/Feb/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{x}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{\left(\boldsymbol{{x}}+\mathrm{1}\right)}−\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\boldsymbol{{x}}+\mathrm{1}}+\sqrt[{\mathrm{4}}]{\boldsymbol{{x}}+\mathrm{1}}−\mathrm{1}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{k}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{x}}\left(−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}+\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{x}}^{\mathrm{2}} \right)+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{x}}−\frac{\mathrm{3}}{\mathrm{32}}\boldsymbol{{x}}^{\mathrm{2}} \right)−\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}−\frac{\:\:\mathrm{1}}{\mathrm{9}}\boldsymbol{{x}}^{\mathrm{2}}…