Question Number 69212 by aliesam last updated on 21/Sep/19 $${use}\:{limit}\:{comparison}\:{test}\:{to}\:{determine} \\ $$$${the}\:{series}\:{converge}\:{or}\:{diverge} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$ Answered by mind is power last…
Question Number 134741 by zahaku last updated on 06/Mar/21 $$\underset{\rightarrow−\mathrm{3}} {{lim}}\frac{{x}+\mathrm{3}}{\:\sqrt{\mid{x}^{\mathrm{2}} +{x}−\mathrm{6}\mid}}\:\:? \\ $$ Answered by mathmax by abdo last updated on 07/Mar/21 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+\mathrm{3}}{\:\sqrt{\mid\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{6}\mid}}\:\mathrm{changement}\:\mathrm{x}+\mathrm{3}=\mathrm{t}\:\mathrm{give}…
Question Number 134746 by zahaku last updated on 06/Mar/21 $${How}\:{to}\:{calculate} \\ $$$${li}\underset{{x}\rightarrow−\mathrm{3}} {{m}}\frac{{x}+\mathrm{3}}{\:\sqrt{\mid{x}^{\mathrm{2}} +{x}−\mathrm{6}\mid}}\:\:\:\:? \\ $$ Answered by mathmax by abdo last updated on 06/Mar/21…
Question Number 69174 by Aditya789 last updated on 21/Sep/19 $${find}\:{the}\:{value}\:{of}\: \\ $$$${lim}\frac{\mathrm{1}+{cosx}}{{tan}^{\mathrm{2}{x}} } \\ $$$${x}\hat {−}\upuparrows \\ $$ Commented by Henri Boucatchou last updated on…
Question Number 134665 by EDWIN88 last updated on 06/Mar/21 $$\mathscr{N}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:+\:\sqrt{\mathrm{x}−\mathrm{1}}}{\:\sqrt{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}}\:=? \\ $$ Answered by benjo_mathlover last updated on 06/Mar/21 $$\mathscr{N}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}−\mathrm{1}}\:\left\{\sqrt{\mathrm{x}+\mathrm{1}}+\mathrm{1}\:\right\}}{\:\sqrt{\mathrm{x}−\mathrm{1}}\:\left\{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:\right\}}…
Question Number 3546 by prakash jain last updated on 15/Dec/15 $$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{p}_{{i}} },\:{p}_{{i}} \:{is}\:{i}^{{th}} \:{prime}\:{diverges}. \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{s}_{{i}} },\:{s}_{{i}} \:{is}\:{i}^{{th}} \:{whole}\:{square}\:{converges}. \\ $$$$\mathrm{Why}?…
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Question Number 69075 by Aditya789 last updated on 19/Sep/19 Commented by Prithwish sen last updated on 19/Sep/19 $$\mathrm{Divide}\:\mathrm{both}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{denominator} \\ $$$$\mathrm{by}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{will}\:\mathrm{be}\:\mathrm{1} \\ $$ Commented by…
Question Number 134549 by liberty last updated on 05/Mar/21 $$\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\left(\frac{{xe}^{{ax}} }{{a}−{b}}\:+\:\frac{{e}^{{bx}} }{\left({a}−{b}\right)^{\mathrm{2}} }\:−\:\frac{{e}^{{ax}} }{\left({a}−{b}\right)^{\mathrm{2}} }\right)\:=? \\ $$ Answered by EDWIN88 last updated on 05/Mar/21…
Question Number 3460 by prakash jain last updated on 14/Dec/15 $$\mathrm{If}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \:\mathrm{is}\:\mathrm{convergent}\:\mathrm{and}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{b}_{{n}} \:\mathrm{is}\:\mathrm{convergent}. \\ $$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sufficient}\:\mathrm{condition}\:\mathrm{so}\:\mathrm{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}}…