Question Number 83690 by jagoll last updated on 05/Mar/20 $$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{\pi{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\right)}{{x}+\mathrm{1}}\:=\:? \\ $$ Commented by mathmax by abdo last updated on 05/Mar/20 $${let}\:{f}\left({x}\right)=\frac{{tan}\left(\frac{\pi{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\right)}{{x}+\mathrm{1}}\:\Rightarrow{f}\left({x}\right)=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}×\frac{{x}+\frac{\mathrm{1}}{\pi}}{{x}+\mathrm{1}}\right)}{{x}+\mathrm{1}} \\ $$$$=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}×\left(\frac{{x}+\mathrm{1}\:+\frac{\mathrm{1}}{\pi}−\mathrm{1}}{{x}+\mathrm{1}}\right)\right)}{{x}+\mathrm{1}}\:=\frac{{tan}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}−\pi}{\pi\left({x}+\mathrm{1}\right)}\right)\right.}{{x}+\mathrm{1}}…
Question Number 83668 by jagoll last updated on 05/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}−{x}}{{x}^{\mathrm{2}} \:\mathrm{sin}\:\left(\mathrm{2}{x}\right)}\:=\: \\ $$ Answered by john santu last updated on 05/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)−{x}}{{x}^{\mathrm{2}} \:\mathrm{sin}\:\left(\mathrm{2}{x}\right)}\:=\:…
Question Number 149184 by liberty last updated on 03/Aug/21 $$\:\:\:\Omega\:=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{tan}\:\sqrt{\mathrm{3x}+\mathrm{2}}−\mathrm{tan}\:\sqrt{\mathrm{2x}+\mathrm{3}}}{\mathrm{tan}\:\sqrt{\mathrm{5x}+\mathrm{4}}−\mathrm{tan}\:\sqrt{\mathrm{4x}+\mathrm{5}}}\:=? \\ $$ Answered by EDWIN88 last updated on 03/Aug/21 $$\Omega=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{3}{x}+\mathrm{2}}}\:\mathrm{sec}\:^{\mathrm{2}} \:\sqrt{\mathrm{3}{x}+\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}+\mathrm{3}}}\mathrm{sec}\:^{\mathrm{2}} \sqrt{\mathrm{2}{x}+\mathrm{3}}}{\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5}{x}+\mathrm{4}}}\mathrm{sec}\:^{\mathrm{2}} \sqrt{\mathrm{5}{x}+\mathrm{4}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{x}+\mathrm{5}}}\mathrm{sec}\:^{\mathrm{2}}…
Question Number 83587 by jagoll last updated on 04/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3sin}\:\pi\mathrm{x}−\mathrm{sin}\:\mathrm{3}\pi\mathrm{x}}{\mathrm{x}^{\mathrm{3}} } \\ $$ Commented by john santu last updated on 04/Mar/20 $$\mathrm{let}\:\mathrm{u}\:=\:\pi\mathrm{x}\:\Rightarrow\:\mathrm{x}=\:\frac{\mathrm{u}}{\pi} \\ $$$$\underset{\mathrm{u}\rightarrow\mathrm{0}}…
Question Number 83492 by jagoll last updated on 03/Mar/20 $$\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{f}\left(\mathrm{2x}−\mathrm{t}\right)+\mathrm{f}\left(\mathrm{2x}−\mathrm{2t}\right)−\mathrm{2f}\left(\mathrm{2x}+\mathrm{t}\right)}{\mathrm{t}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 83493 by jagoll last updated on 03/Mar/20 $$\mathrm{If}\:\underset{{x}\rightarrow\mathrm{p}} {\mathrm{lim}}\:\frac{\mid\mathrm{x}+\mathrm{p}\mid^{\mathrm{2}} −\mid\mathrm{2p}\mid^{\mathrm{2}} }{\mathrm{p}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{L} \\ $$$$\mathrm{find}\:\underset{{x}\rightarrow\mathrm{p}} {\mathrm{lim}}\:\frac{\mid\mathrm{2p}\mid^{\mathrm{3}} −\mid\mathrm{x}+\mathrm{p}\mid^{\mathrm{3}} }{\mathrm{x}−\mathrm{p}} \\ $$ Terms of Service…
Question Number 17957 by Arnab Maiti last updated on 13/Jul/17 $$\mathrm{prove}\:\mathrm{that}\:\left(\mathrm{1}+\frac{\mathrm{a}}{\mathrm{x}}\right)^{\mathrm{n}} =\left(\mathrm{1}+\frac{\mathrm{an}}{\mathrm{x}}\right)\:,\:\:\mathrm{x}\gg\mathrm{a} \\ $$ Answered by 42 last updated on 13/Jul/17 $$\mathrm{Expand}\:\mathrm{using}\:\mathrm{binomial}\:\mathrm{theorem}: \\ $$$$\mathrm{1}\:+\:{n}\left(\frac{{a}}{{x}}\right)\:+\:\frac{{n}\left({n}\:−\:\mathrm{1}\right)}{\mathrm{2}}\left(\frac{{a}}{{x}}\right)^{\mathrm{2}} \:+\:……
Question Number 83491 by john santu last updated on 03/Mar/20 $$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\sqrt{\mathrm{1}+\mathrm{sin}\:\left(\mathrm{2x}\right)}−\sqrt{\mathrm{1}−\mathrm{2sin}\:\left(\mathrm{x}\right)}}{\mathrm{x}} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\:=\:\mathrm{2x}+\sqrt{\mathrm{2x}} \\ $$$$\mathrm{find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\: \\ $$ Commented by john santu last updated on…
Question Number 148992 by mathlove last updated on 02/Aug/21 Answered by gsk2684 last updated on 02/Aug/21 $$\approx\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{sin}\:{x}+\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{2}!}+..\right)−\left(\mathrm{sin}\:\mathrm{3}{x}+\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{3}{x}}{\mathrm{2}!}+..\right)}{\mathrm{2}{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\mathrm{1}+\mathrm{0}+…\right)−\left(\mathrm{3}+\mathrm{0}+..\right)\right\} \\ $$$$=−\mathrm{1} \\…
Question Number 148990 by gsk2684 last updated on 02/Aug/21 $${let}\:{f}:{R}\rightarrow{R}\:{be}\:{a}\:{continuius}\:{function} \\ $$$${such}\:{that}\:{for}\:{any}\:{two}\:{real}\:{numbers} \\ $$$${x}\:{and}\:{y}\:\mid{f}\left({x}\right)−{f}\left({y}\right)\mid\leqslant\mathrm{10}\mid{x}−{y}\mid^{\mathrm{201}} \\ $$$${then}\:{prove}\:{that} \\ $$$${f}\left(\mathrm{2019}\right)+{f}\left(\mathrm{2022}\right)=\mathrm{2}\:{f}\left(\mathrm{2021}\right) \\ $$ Terms of Service Privacy Policy…