Question Number 169527 by Giantyusuf last updated on 01/May/22 $$\boldsymbol{{evaluate}}\:\boldsymbol{{the}}\:\boldsymbol{{limit}}\left(\boldsymbol{{if}}\:\boldsymbol{{it}}\:\boldsymbol{{exists}}\right) \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{n}}\rightarrow\infty} \left[\frac{\sqrt{\boldsymbol{{n}}^{\mathrm{4}} −\mathrm{2}\boldsymbol{{n}}^{\mathrm{3}} }−\boldsymbol{{n}}^{\mathrm{2}} }{\boldsymbol{{n}}+\mathrm{2}}\right]^{\mathrm{3}} \\ $$ Commented by infinityaction last updated on 02/May/22…
Question Number 103979 by Study last updated on 18/Jul/20 $$\:\:{show}\:{right}\:{and}\:{left}\:{sid}\:\:\:\:{limitation}\:\:\:\: \\ $$$${of}\:\:\:{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{ln}\left({b}−{x}\right)}{{ax}} \\ $$$$ \\ $$ Commented by Study last updated on 18/Jul/20 $${help}\:{me}…
Question Number 169514 by Giantyusuf last updated on 01/May/22 $$\boldsymbol{{evaluate}}\:\boldsymbol{{the}}\:\boldsymbol{{following}}\:\boldsymbol{{limit}} \\ $$$$\left(\boldsymbol{{if}}\:\boldsymbol{{it}}\:\boldsymbol{{exists}}\right) \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{n}}\rightarrow\infty} \frac{\boldsymbol{{ln}}^{\mathrm{2}} \left(\boldsymbol{{n}}+\mathrm{1}\right)}{\left(\boldsymbol{{n}}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by Giantyusuf last updated on…
Question Number 169508 by Giantyusuf last updated on 01/May/22 $${Evaluate}\:{if}\:{the}\:{limit}\:{exist} \\ $$$${lim}_{\boldsymbol{{n}}\rightarrow\infty} \left(\frac{\mathrm{3}^{\boldsymbol{{n}}} +\left(−\mathrm{2}\right)^{\boldsymbol{{n}}+\mathrm{1}} }{\mathrm{3}^{\boldsymbol{{n}}−\mathrm{2}} −\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}} }\right) \\ $$ Answered by qaz last updated on…
Question Number 169501 by Giantyusuf last updated on 01/May/22 Answered by qaz last updated on 01/May/22 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\:\mathrm{n}\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}narctan}\:\frac{\mathrm{1}}{\mathrm{n}}=\mathrm{1} \\ $$ Terms of Service Privacy…
Question Number 169498 by bagjagugum123 last updated on 01/May/22 Commented by infinityaction last updated on 01/May/22 $$\frac{\mathrm{8}}{\mathrm{3}} \\ $$ Commented by bagjagugum123 last updated on…
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Question Number 169480 by TOTTI last updated on 01/May/22 Answered by MikeH last updated on 01/May/22 $$\left(\mathrm{a}\right)\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:{x}\:=\:\mathrm{1} \\ $$$$\left(\mathrm{b}\right)\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\mathrm{2}−{x}^{\mathrm{2}}…
Question Number 169479 by cortano1 last updated on 01/May/22 $$\:\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:{a}\right)^{{x}} −\left(\mathrm{sin}\:{a}\right)^{{x}} −\mathrm{cos}\:\mathrm{2}{a}}{{x}−\mathrm{4}}\:=? \\ $$ Commented by infinityaction last updated on 01/May/22 $${use}\:{l}\:{hospital}\:{rule} \\ $$$$\:\:\:\:\:{p}\left({let}\right)\:\:=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 169412 by mathlove last updated on 29/Apr/22 Commented by mathlove last updated on 30/Apr/22 $${any}\:{one}\:{is}\:{answer}\:{for}\:{this} \\ $$ Terms of Service Privacy Policy Contact:…