Question Number 206069 by MathematicalUser2357 last updated on 06/Apr/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}^{\mathrm{3}} +{x}}{\mathrm{sin}\:{x}} \\ $$ Answered by MetaLahor1999 last updated on 06/Apr/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−{x}^{\mathrm{3}} }{{sin}\left({x}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{2}}…
Question Number 206095 by RoseAli last updated on 06/Apr/24 Answered by Frix last updated on 07/Apr/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}−\mathrm{tan}\:{x}}\:\:\overset{\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right]} {=}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left[{x}−\mathrm{sin}\:{x}\right]}{\frac{{d}}{{dx}}\left[{x}−\mathrm{tan}\:{x}\right]}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{−\mathrm{tan}^{\mathrm{2}} \:{x}}…
Question Number 205916 by mathzup last updated on 02/Apr/24 $${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left({e}^{{x}} −\mathrm{1}\right) \\ $$ Answered by mathzup last updated on 03/Apr/24 $${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left({e}^{{x}}…
Question Number 205774 by mnjuly1970 last updated on 30/Mar/24 $$ \\ $$$$\:\:\:\:\:\:−−−−−−− \\ $$$$\:\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}} }{\left(−\mathrm{1}\right)^{\:{n}} \:−{n}}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:−−−−−−− \\ $$ Answered by MathedUp…
Question Number 205716 by universe last updated on 28/Mar/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)…\left(\mathrm{4}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)…\left(\mathrm{4}{n}\right)}\:\:=\:\:? \\ $$ Answered by MM42 last updated on 28/Mar/24 $$\frac{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)…\left(\mathrm{4}{n}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)…\left(\mathrm{4}{n}−\mathrm{1}\right)}<{A}<\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{4}\right)…\left(\mathrm{4}{n}+\mathrm{2}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)…\left(\mathrm{4}{n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\mathrm{4}{n}+\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right){A}}<{A}<\frac{\mathrm{4}{n}+\mathrm{2}}{\left(\mathrm{2}{n}\right){A}} \\ $$$$\Rightarrow\frac{\mathrm{4}{n}+\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}<{A}^{\mathrm{2}}…
Question Number 205727 by mathzup last updated on 28/Mar/24 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{e}^{{x}} −{cosx}}{{x}^{\mathrm{2}} } \\ $$ Commented by lepuissantcedricjunior last updated on 29/Mar/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\boldsymbol{{cosx}}}{\boldsymbol{{x}}^{\mathrm{2}}…
Question Number 205580 by universe last updated on 25/Mar/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)…\left(\mathrm{4}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)…\left(\mathrm{4}{n}\right)}\:\:=\:\:? \\ $$ Commented by lepuissantcedricjunior last updated on 26/Mar/24 $$\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)×\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{3}\right)×…×\left(\mathrm{4}\boldsymbol{{n}}+\mathrm{1}\right)}{\left(\mathrm{2}\boldsymbol{{n}}\right)×\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{2}\right)×..×\left(\mathrm{4}\boldsymbol{{n}}\right)} \\ $$$$=\underset{\boldsymbol{{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{2}\boldsymbol{{n}}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{2}\right)…\left(\mathrm{4}\boldsymbol{{n}}\right)\left(\mathrm{4}\boldsymbol{{n}}+\mathrm{1}\right)}{\left[\left(\mathrm{2}\boldsymbol{{n}}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{2}\right)….\left(\mathrm{4}\boldsymbol{{n}}\right)\right]^{\mathrm{2}}…
Question Number 205448 by mathlove last updated on 21/Mar/24 $${A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sinx}}{{x}^{\mathrm{3}} }=? \\ $$ Answered by namphamduc last updated on 21/Mar/24 $${A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\left({x}\right)}{{x}^{\mathrm{3}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left({x}\right)}{{x}}.{x}^{\mathrm{4}}…
Question Number 205379 by cortano12 last updated on 19/Mar/24 Answered by MM42 last updated on 19/Mar/24 $${hop}\rightarrow={lim}_{{x}\rightarrow\mathrm{0}} \frac{−{cosx}×{sin}\left({sinx}\right)+{sinx}}{\mathrm{4}{x}^{\mathrm{3}} } \\ $$$${hop}\rightarrow=\frac{{sinx}×{sin}\left({sinx}\right)−{cos}^{\mathrm{2}} {x}×{cos}\left({sinx}\right)+{cosx}}{\mathrm{12}{x}^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 205339 by depressiveshrek last updated on 17/Mar/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{tan}\left(\mathrm{tan}{x}\right)}{\mathrm{sin}\left(\mathrm{1}−\mathrm{cos}{x}\right)} \\ $$ Answered by MM42 last updated on 18/Mar/24 $$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{tan}\left({x}\right)}{{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \right)} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}}…