Question Number 12946 by 433 last updated on 07/May/17 $${a}_{{n}} =\sqrt{\mathrm{3}{a}_{{n}−\mathrm{1}} +\mathrm{2}}\:\:\:{a}_{\mathrm{1}} =\mathrm{1} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =? \\ $$ Answered by ajfour last updated on…
Question Number 143995 by liberty last updated on 20/Jun/21 $$\:\:\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\pi−\mathrm{4}{x}}{\:\sqrt{\mathrm{1}−\sqrt{\mathrm{sin}\:\mathrm{2}{x}}}}\:=? \\ $$ Answered by mathmax by abdo last updated on 20/Jun/21 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\pi−\mathrm{4x}}{\:\sqrt{\mathrm{1}−\sqrt{\mathrm{sin2x}}}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=_{\frac{\pi}{\mathrm{4}}−\mathrm{x}=\mathrm{t}} \:\:\:\frac{\pi−\mathrm{4}\left(\frac{\pi}{\mathrm{4}}−\mathrm{t}\right)}{\:\sqrt{\mathrm{1}−\sqrt{\mathrm{sin}\left(\mathrm{2}\left(\frac{\pi}{\mathrm{4}}−\mathrm{t}\right)\right.}}} \\…
Question Number 143960 by bobhans last updated on 20/Jun/21 $$\:\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\:=? \\ $$ Answered by lapache last updated on 20/Jun/21 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\sqrt{\mathrm{1}−\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:}}{\mathrm{1}−\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}={li}\underset{{x}\rightarrow\mathrm{0}}…
Question Number 12881 by kunalshukla95040 last updated on 05/May/17 $$\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:{x}}}{{x}} \\ $$$${is}\:{equals}\:{to}. \\ $$ Answered by nume1114 last updated on 05/May/17 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:{x}}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 78392 by jagoll last updated on 17/Jan/20 Answered by john santu last updated on 17/Jan/20 $${let}\:{h}=\:\mathrm{cos}\:{x}\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\sqrt[{\:\:\mathrm{3}}]{\mathrm{cos}\:\mathrm{3}{x}}\:\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{4}{x}}…\:\sqrt[{{n}}]{\mathrm{cos}\:{nx}} \\ $$$${ln}\left({h}\right)=\:{ln}\left(\mathrm{cos}\:{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{cos}\:\mathrm{2}{x}\right)+\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{cos}\:\mathrm{3}{x}\right)+…+\frac{\mathrm{1}}{{n}}{ln}\left(\mathrm{cos}\:{nx}\right) \\ $$$$\frac{{dh}}{{dx}}=\frac{−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}−\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:\mathrm{2}{x}}−\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{3}{x}}−…−\frac{\mathrm{sin}\:{nx}}{\mathrm{cos}\:{nx}} \\ $$$${now}\:{we}\:{use}\:{L}'{Hopital}\:{rule} \\…
Question Number 78343 by balbayrak last updated on 16/Jan/20 Commented by MJS last updated on 16/Jan/20 $$\mathrm{syntax}\:\mathrm{error}.\:\mathrm{type}\:\mathrm{it}\:\mathrm{correctly}\:\mathrm{please} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 78324 by jagoll last updated on 16/Jan/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{4}{x}−\mathrm{cos}\:\mathrm{2}{x}}{\:\sqrt{{x}+\mathrm{1}}−\sqrt{\mathrm{1}−{x}}}= \\ $$ Commented by john santu last updated on 16/Jan/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{1}−{x}}\right)×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} {x}+\mathrm{2sin}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}{x}}…
Question Number 78325 by jagoll last updated on 16/Jan/20 $$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)^{\mathrm{3}{x}−\mathrm{1}} \\ $$ Commented by john santu last updated on 16/Jan/20…
Question Number 12732 by Joel577 last updated on 30/Apr/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:−\:{x}}{\:\sqrt{{x}}\:+\:{x}} \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/Apr/17 $$=\mathrm{1} \\ $$ Commented by…
Question Number 12651 by okhemafrancis last updated on 28/Apr/17 $${find}\:{the}\:{values}\:{of}\:{x}\:{for}\:{which}\:\frac{{x}^{\mathrm{3}} +\mathrm{8}}{{x}^{\mathrm{2}} −\mathrm{4}\:}{is}\:{discontinuous}\:{and}\:{state}\:{each}\:{kinds}\:{of}\:{dicontinuity} \\ $$ Answered by FilupS last updated on 28/Apr/17 $${f}\left({x}\right)=\frac{{x}^{\mathrm{3}} +\mathrm{8}}{{x}^{\mathrm{2}} −\mathrm{4}} \\…