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Author: Tinku Tara

lim-x-1-1-x-x-2-e-x-

Question Number 212798 by MrGaster last updated on 24/Oct/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{{x}^{\mathrm{2}} }{{e}^{{x}} }} =? \\ $$ Answered by mehdee7396 last updated on 24/Oct/24 $${lim}_{{x}\rightarrow\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}\:\:\:\:\&\:\:\:\:{lim}_{{x}\rightarrow\infty}…

Question-212827

Question Number 212827 by Spillover last updated on 25/Oct/24 Answered by A5T last updated on 25/Oct/24 Commented by A5T last updated on 25/Oct/24 $$\frac{{y}}{\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}}=\frac{{r}}{\frac{\mathrm{4}{r}}{\:\sqrt{\mathrm{3}}}}\Rightarrow{y}=\frac{\mathrm{2}{r}\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{{r}\sqrt{\mathrm{3}}}{\mathrm{4}{r}}=\frac{{r}}{\mathrm{2}} \\…

Let-f-x-There-is-a-secondorder-continuoust-derivaive-t-x-2-y-2-g-x-y-f-1-r-ask-2-g-x-2-2-g-y-2-

Question Number 212788 by MrGaster last updated on 24/Oct/24 $$ \\ $$$${Let}\:{f}\left({x}\right)\:\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{secondorder}\:\mathrm{continuoust} \\ $$$$\mathrm{derivaive},{t}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} },{g}\left({x},{y}\right)={f}\left(\frac{\mathrm{1}}{{r}}\right),\mathrm{ask}\:\frac{\partial^{\mathrm{2}} {g}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {g}}{\partial{y}^{\mathrm{2}} }. \\ $$ Terms of Service…

Question-212790

Question Number 212790 by RojaTaniya last updated on 24/Oct/24 Answered by A5T last updated on 24/Oct/24 $${x}−{y}=\left({y}−\mathrm{1}+{x}−\mathrm{1}\right)\left({y}−\mathrm{1}−{x}+\mathrm{1}\right) \\ $$$$\Rightarrow{x}={y}\:{or}\:−\mathrm{1}={x}+{y}−\mathrm{2}\Rightarrow{x}+{y}=\mathrm{1} \\ $$$$\Rightarrow{x}+\mathrm{1}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:{or}\:{x}+\mathrm{1}={x}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={x}+{y}+\mathrm{2}=\mathrm{3}…

Question-212816

Question Number 212816 by Akayx last updated on 24/Oct/24 Answered by mahdipoor last updated on 24/Oct/24 $${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}\left({t}−\mathrm{4}{k}\right)\left({u}_{\mathrm{4}{k}} −{u}_{\mathrm{4}{k}+\mathrm{2}} \right)= \\ $$$$\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left({t}−\mathrm{4}{k}\right){u}_{\mathrm{4}{k}}…