Question Number 173225 by cortano1 last updated on 08/Jul/22 $$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}+\mathrm{4tan}\:\mathrm{2}{x}−\mathrm{3tan}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} \:\mathrm{tan}\:{x}}\:=?\: \\ $$ Answered by floor(10²Eta[1]) last updated on 08/Jul/22 $$\mathrm{tg}\left(\mathrm{2x}\right)=\frac{\mathrm{2tgx}}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{tg3x}=\frac{\frac{\mathrm{2tgx}}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}}…
Question Number 173192 by mathlove last updated on 08/Jul/22 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)……\infty}}}=? \\ $$ Answered by Frix last updated on 08/Jul/22 $${u}=\sqrt{{t}\sqrt{{t}\sqrt{{t}\sqrt{…}}}}\:\Leftrightarrow\:{u}=\sqrt{{tu}}\:\Leftrightarrow\:{u}={t} \\ $$$$ \\…
Question Number 173190 by mathlove last updated on 08/Jul/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{2}−{e}^{{x}} \right)}{{x}+{lnx}}=? \\ $$ Answered by thfchristopher last updated on 08/Jul/22 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{−{e}^{{x}} }{\mathrm{2}−{e}^{{x}} }}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}…
Question Number 107598 by mathDivergent last updated on 11/Aug/20 $$\:\:\:\:\mathrm{Calculate}:\:\:\sqrt{\mathrm{1}\:+\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{3}\:+\:\sqrt{\mathrm{4}\:+\:\sqrt{\mathrm{5}\:+\:…}}}}} \\ $$ Commented by Her_Majesty last updated on 11/Aug/20 $$\approx\mathrm{1}.\mathrm{75793275662} \\ $$ Commented by Dwaipayan…
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Question Number 107550 by ajfour last updated on 11/Aug/20 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{2}^{{x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}/{x}} \:\:=\:? \\ $$ Answered by hgrocks last updated on 11/Aug/20 Commented by ajfour…
Question Number 107536 by bemath last updated on 11/Aug/20 $$\:\spadesuit\mathcal{B}{e}\mathcal{M}{ath}\bigstar \\ $$$${Without}\:{L}'{Hopital}\:{and}\:{series}\: \\ $$$${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−{x}}{{x}^{\mathrm{3}} } \\ $$ Commented by bobhans last updated on 12/Aug/20…
Question Number 41998 by Joel578 last updated on 16/Aug/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}\:+\:\mathrm{cos}\:{x}}{{x}\:+\:\mathrm{sin}\:{x}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}+\frac{{cosx}}{{x}}}{\mathrm{1}+\frac{{sinx}}{{x}}} \\ $$$${the}\:{value}\:{of}\:{sinx}\:{lies}\:{in}\:{between}\:\pm\mathrm{1}\:{whatever} \\…
Question Number 107516 by bemath last updated on 11/Aug/20 $$\:\:\:\:\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc \\ $$$$\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{2}{t}}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{t}\right)−\mathrm{cos}\:\mathrm{2}{t}}\:? \\ $$ Answered by bemath last updated on 11/Aug/20 $$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{t}}{\mathrm{cos}\:{t}−\mathrm{cos}\:\mathrm{2}{t}}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{cos}\:\mathrm{2}{t}}}…
Question Number 173053 by mathlove last updated on 06/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com