Question Number 221348 by RoseAli last updated on 31/May/25 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}−\mathrm{2}^{{x}} }{{x}−\mathrm{2}} \\ $$ Answered by mr W last updated on 31/May/25 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{2}^{{x}} \right)}{{x}}…
Question Number 221347 by RoseAli last updated on 31/May/25 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}−\mathrm{2}} \\ $$ Answered by AntonCWX8 last updated on 31/May/25 $$\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}−\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}+{x}\right)\left(\mathrm{2}−{x}\right)}{{x}−\mathrm{2}}…
Question Number 221260 by mnjuly1970 last updated on 28/May/25 Answered by maths2 last updated on 30/May/25 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{{n}}.\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${n}!\sim\sqrt{\mathrm{2}\pi{n}}.\left(\frac{{n}}{{e}}\right)^{{n}} ..…
Question Number 221047 by universe last updated on 23/May/25 Answered by breniam last updated on 24/May/25 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{r}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \left({k}−\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)!}−\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{k}=\mathrm{0}}…
Question Number 221034 by Nicholas666 last updated on 23/May/25 Answered by vnm last updated on 23/May/25 $${let}\:{lim}={a} \\ $$$$\mathrm{sin}\frac{{a}}{\:\sqrt{{n}}}=\frac{{a}}{\:\sqrt{{n}}}−\frac{\mathrm{1}}{\mathrm{6}}\frac{{a}^{\mathrm{3}} }{{n}\sqrt{{n}}}+{o}\left(\frac{\mathrm{1}}{{n}\sqrt{{n}}}\right)=\frac{{a}}{\:\sqrt{{n}}}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{6}{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$$\sqrt{{n}+\mathrm{1}}=\sqrt{{n}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}=\sqrt{{n}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$$\mathrm{sin}\frac{{a}}{\:\sqrt{{n}}}\centerdot\sqrt{{n}+\mathrm{1}}=\frac{{a}}{\:\sqrt{{n}}}\sqrt{{n}}\left(\left(\mathrm{1}−\frac{{a}^{\mathrm{2}}…
Question Number 220852 by fantastic last updated on 20/May/25 $$\underset{{x}\rightarrow\mathrm{0}} {{Lim}}\left\{\frac{{xe}^{{x}} −{log}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} }\right\} \\ $$ Answered by SdC355 last updated on 20/May/25 $$\frac{\frac{\mathrm{d}\:\:}{\mathrm{d}{x}}\left({xe}^{{x}} −\mathrm{ln}\left({x}+\mathrm{1}\right)\right)}{\frac{\mathrm{d}\:}{\mathrm{d}{x}}\:{x}^{\mathrm{2}} }=\frac{\left({x}+\mathrm{1}\right){e}^{{x}}…
Question Number 220843 by Nicholas666 last updated on 20/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha\:\in\:\mathbb{R} \\ $$$$\:\:\:\:\:\mathrm{lim}_{{x}\rightarrow\mathrm{1}} \:\frac{\left(\mathrm{1}\:−\:{x}\right)^{\alpha} }{\:^{\mathrm{3}} \sqrt{\mathrm{1}\:−\:{x}^{\mathrm{4}} }}\:\:\:\:\:\:\:\:\in\left(\mathrm{0},\infty\right) \\ $$$$ \\ $$ Commented by SdC355…
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Question Number 220764 by Nicholas666 last updated on 18/May/25 $$ \\ $$$$\:\:\boldsymbol{\mathrm{L}}=\:\boldsymbol{\mathrm{lim}}\underset{\:\boldsymbol{{n}}\rightarrow\infty} {\:}\left(\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\:\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} }\right).\left(\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} } \boldsymbol{{dx}}\overset{−\mathrm{1}} {\right)}.\left(\underset{\boldsymbol{{m}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{m}}} }{\left(\mathrm{2}\boldsymbol{{m}}+\mathrm{1}\right)\mathrm{3}^{\boldsymbol{{m}}}…
Question Number 220380 by MrGaster last updated on 12/May/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}tan}\left[\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right]^{{n}} =? \\ $$ Answered by SdC355 last updated on 12/May/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\mathrm{tan}\left(\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \right)=\mathrm{tan}\left(\mathrm{0}\right)=\mathrm{0} \\…