Question Number 117898 by bobhans last updated on 14/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\:\mathrm{1}}{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}}\:\left(\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)−\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}\:\right)\:=? \\ $$ Answered by Lordose last updated on 14/Oct/20 $$\frac{\mathrm{1}}{\mathrm{6}}\:\left(\mathrm{rough}\:\mathrm{work}\right) \\ $$ Terms…
Question Number 183376 by TUN last updated on 25/Dec/22 Commented by CElcedricjunior last updated on 25/Dec/22 $$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\boldsymbol{{nx}}^{\boldsymbol{{n}}} =\boldsymbol{{x}}+\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}^{\mathrm{3}} +…\infty\boldsymbol{{x}}^{\infty} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{{x}}\left(\mathrm{1}+\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +….+\infty\boldsymbol{{x}}^{\infty−\mathrm{1}}…
Question Number 117820 by islam last updated on 13/Oct/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left({x}.\mathrm{sin}\:\frac{\mathrm{1}}{{x}}\right)^{{x}^{\mathrm{2}} } \\ $$ Answered by Dwaipayan Shikari last updated on 13/Oct/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left({x}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{6}{x}^{\mathrm{3}} }\right)\right)^{{x}^{\mathrm{2}}…
Question Number 183269 by TUN last updated on 24/Dec/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 183260 by TUN last updated on 24/Dec/22 Answered by greougoury555 last updated on 24/Dec/22 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$ Answered by TheSupreme last…
Question Number 117687 by bemath last updated on 13/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{ln}\:\left(\mathrm{cosh}\:\mathrm{x}\right)−\mathrm{ln}\:\left(\mathrm{cos}\:\mathrm{x}\right)\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{cosh}\:\mathrm{x}}+\sqrt{\mathrm{cos}\:\mathrm{x}}−\mathrm{2}}\:=?\: \\ $$ Answered by 1549442205PVT last updated on 13/Oct/20 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}\:}\frac{\left(\mathrm{ln}\:\left(\mathrm{cosh}\:\mathrm{x}\right)−\mathrm{ln}\:\left(\mathrm{cos}\:\mathrm{x}\right)\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{cosh}\:\mathrm{x}}+\sqrt{\mathrm{cos}\:\mathrm{x}}−\mathrm{2}}\:=^{\frac{\mathrm{0}}{\mathrm{0}}} \\…
Question Number 117673 by huotpat last updated on 13/Oct/20 Answered by bemath last updated on 13/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3x}−\left(\mathrm{3x}−\frac{\mathrm{27x}^{\mathrm{3}} }{\mathrm{6}}\right)}{\mathrm{2x}−\left(\mathrm{2x}−\frac{\mathrm{8x}^{\mathrm{3}} }{\mathrm{6}}\right)}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{27x}^{\mathrm{3}} }{\mathrm{8x}^{\mathrm{3}} }\:=\:\:\frac{\mathrm{27}}{\mathrm{8}}…
Question Number 52123 by 786786AM last updated on 03/Jan/19 $$\mathrm{Find}\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}. \\ $$ Commented by turbo msup by abdo last updated on 03/Jan/19 $${we}\:{have}\:\mid\frac{{sinx}}{{x}}\mid\leqslant\frac{\mathrm{1}}{\mid{x}\mid}\:\forall{x}\neq\mathrm{0}\:\Rightarrow \\…
Question Number 117646 by bemath last updated on 13/Oct/20 Commented by bemath last updated on 13/Oct/20 $$\mathrm{without}\:\mathrm{L}'\mathrm{Hopital} \\ $$ Answered by bobhans last updated on…
Question Number 117637 by bemath last updated on 13/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:\mathrm{x}\:\right)^{\mathrm{cot}\:\mathrm{x}} \:=? \\ $$ Answered by AbduraufKodiriy last updated on 13/Oct/20 $$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\boldsymbol{{cosx}}^{\boldsymbol{{cotx}}} \right)=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\boldsymbol{{cosx}}^{\frac{\boldsymbol{{cosx}}}{\boldsymbol{{sinx}}}}…