Question Number 142721 by mathlove last updated on 04/Jun/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:{x}}}=? \\ $$ Answered by Ar Brandon last updated on 04/Jun/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}−\mathrm{cosx}}}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)}}=\underset{\mathrm{x}\rightarrow\mathrm{0}}…
Question Number 77182 by jagoll last updated on 04/Jan/20 $$ \\ $$$$ \\ $$$$\mathrm{evaluate}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\underset{\mathrm{a}} {\overset{\mathrm{x}} {\int}}\left(\frac{\mathrm{cos}\:\mathrm{t}}{\mathrm{t}}\right)\mathrm{dt}}{\mathrm{x}}\:. \\ $$ Commented by mathmax by abdo last…
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Question Number 11549 by Nayon last updated on 28/Mar/17 $$\:\:\:\: \\ $$$${proof} \\ $$$${lim} \\ $$$${h}\rightarrow\mathrm{0}^{\frac{{x}^{{h}} −\mathrm{1}}{{h}}={ln}\left({x}\right)} \\ $$ Answered by mrW1 last updated on…
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Question Number 11444 by Joel576 last updated on 26/Mar/17 $$\mathrm{If}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{px}\:+\:{q}}\:−\:\mathrm{2}}{{x}}\:=\:\mathrm{1} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:{p}\:+\:{q}\:? \\ $$ Answered by ajfour last updated on 26/Mar/17 $$\mathrm{then}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{q}}\left(\mathrm{1}+\mathrm{px}/\mathrm{q}\right)^{\mathrm{1}/\mathrm{2}} \:−\mathrm{2}}{\mathrm{x}}\:=\mathrm{1}…
Question Number 142492 by leesjyons last updated on 01/Jun/21 $$\mathrm{Prove}\:\mathrm{that}\:\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{3}{n}^{\mathrm{2}} \:+\:\mathrm{2}}\:\mathrm{diverges}. \\ $$ Answered by mathmax by abdo last updated on 02/Jun/21 $$\mathrm{dvergence}\:\mathrm{clear}\:\mathrm{due}\left[\mathrm{to}\:\:\frac{\mathrm{n}}{\mathrm{3n}^{\mathrm{2}}…
Question Number 11416 by anisa last updated on 25/Mar/17 Answered by Joel576 last updated on 25/Mar/17 $$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{2}{x}\:\mathrm{sin}\:\mathrm{2}{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}}{\mathrm{2}}\:.\:\frac{\mathrm{sin}\:{x}}{{x}}\:.\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{2}{x}} \\ $$$$=\:\mathrm{1}\:.\:\mathrm{1}\:.\:\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 11408 by anisa last updated on 24/Mar/17 Answered by sm3l2996 last updated on 24/Mar/17 $$=\frac{\mathrm{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)+\mathrm{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)}{\mathrm{tg}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)}=\frac{\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{3}}} \\ $$$$=−\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$ Terms of Service Privacy…