Question Number 213744 by efronzo1 last updated on 15/Nov/24 Answered by golsendro last updated on 15/Nov/24 $$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{4x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} +\left(\mathrm{1}−\mathrm{4x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{1}+\mathrm{4x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{3}\left(\mathrm{1}−\mathrm{4x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}}…
Question Number 213642 by universe last updated on 12/Nov/24 Answered by Berbere last updated on 12/Nov/24 $${a}_{{n}} =\underset{{N}\rightarrow\infty} {\mathrm{lim}}\underset{{k}={n}} {\overset{{N}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} };\forall{k}\geqslant{n}>\mathrm{1}\:\:{k}\left({k}−\mathrm{1}\right)\:\leqslant{k}^{\mathrm{2}} \leqslant{k}\left({k}+\mathrm{1}\right)\Rightarrow \\ $$$$\underset{{N}\rightarrow\infty}…
Question Number 213548 by universe last updated on 08/Nov/24 $$\mathrm{0}<{c}<\mathrm{1}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{recursive}\:\mathrm{sequence} \\ $$$$\left\{{a}_{{n}} \right\}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{setting}\: \\ $$$$\:\mathrm{a}_{\mathrm{1}\:} =\:\frac{\mathrm{c}}{\mathrm{2}}\:\:,\:{a}_{\mathrm{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{c}+\mathrm{a}_{\mathrm{n}} ^{\mathrm{2}} \right)\:\:\mathrm{for}\:\mathrm{n}\in\:\mathbb{N} \\ $$$$\mathrm{monotonic}\:\mathrm{and}\:\mathrm{convergent} \\ $$ Answered by…
Question Number 213511 by universe last updated on 07/Nov/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\left[\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{r}} }\right] \\ $$$$\:\:\:\mathrm{where}\:\left[\bullet\right]\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{finction} \\ $$ Answered by issac last updated on 07/Nov/24…
Question Number 213468 by universe last updated on 06/Nov/24 $$\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}\:\mathrm{defined}\: \\ $$$$\mathrm{recurssively}\:\mathrm{by}\:\mathrm{a}_{\mathrm{1}} =\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{n}\:} =\:\sqrt{\mathrm{3a}_{\mathrm{n}−\mathrm{1}\:} \:−\mathrm{2}\:}\:\:\:\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{2}\:\:\mathrm{converges}\: \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{its}\:\mathrm{limit}. \\ $$ Answered by issac…
Question Number 213503 by Spillover last updated on 06/Nov/24 Answered by mr W last updated on 06/Nov/24 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 213241 by RoseAli last updated on 01/Nov/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}−\mathrm{tan}\:{x}}{{x}^{\mathrm{3}} } \\ $$ Answered by ajfour last updated on 01/Nov/24 $$=−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{sin}\:{x}}{{x}}×\left(\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{4}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)×\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right\} \\…
Question Number 213169 by MrGaster last updated on 31/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{prove} \\ $$$$\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{e}^{{x}^{\mathrm{2}} } {dx}=\frac{\pi}{\mathrm{2}}. \\ $$ Answered by Berbere last…
Question Number 213082 by liuxinnan last updated on 30/Oct/24 Answered by MrGaster last updated on 30/Oct/24 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\frac{\mathrm{1}}{{n}}} \approx\int_{\mathrm{1}} ^{{n}} {x}^{\frac{\mathrm{1}}{{n}}} {dx} \\ $$$$\int{x}^{\frac{\mathrm{1}}{{x}}}…
Question Number 213109 by mathlove last updated on 30/Oct/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}{\:\sqrt{{x}+\mathrm{1}}}=? \\ $$ Answered by MrGaster last updated on 30/Oct/24 $$=\frac{\sqrt{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}\right)}}{\:\sqrt{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}} \\ $$$$=\frac{\sqrt{{x}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}}}{\:\sqrt{{x}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}} \\ $$$$=\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{1}}{{x}}}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}}}…