Question Number 39635 by math khazana by abdo last updated on 09/Jul/18 $${calculate}\:\:{lim}_{{x}\rightarrow\mathrm{1}} \:\frac{\mathrm{1}+{cos}\left(\pi{x}\right)}{{x}^{\mathrm{2}} −\:{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)} \\ $$ Commented by abdo mathsup 649 cc last updated…
Question Number 105157 by bramlex last updated on 26/Jul/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)−\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}+\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{2}}\:? \\ $$ Answered by bramlex last updated on 26/Jul/20 $$\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\mathrm{3}{x}+\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\pi/\mathrm{6}}…
Question Number 105126 by bemath last updated on 26/Jul/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}−\sqrt{\mathrm{2}}\mathrm{tan}\:{x}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}} \\ $$ Answered by Dwaipayan Shikari last updated on 26/Jul/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\mathrm{cosx}−\mathrm{sinx}−\sqrt{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\mathrm{cosx}+\mathrm{sinx}}=\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\sqrt{\mathrm{2}}.\mathrm{2}}{\:\sqrt{\mathrm{2}}}=−\mathrm{2} \\…
Question Number 105114 by john santu last updated on 26/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}−\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{sin}\:\:^{\mathrm{2}} \left(\mathrm{3}{x}\right)}−\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{sin}\:\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)}}{{x}^{\mathrm{2}} } \\ $$ Answered by bemath last updated on 26/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 105104 by bemath last updated on 26/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{5}} \left({x}\right)\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{2}{x}\right)\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{3}{x}\right)}{\mathrm{22}{x}^{\mathrm{2}} }? \\ $$ Answered by bramlex last updated on 26/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 105106 by bemath last updated on 26/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\pi\mathrm{cos}\:^{\mathrm{2}} {x}\right)}{\mathrm{3}{x}^{\mathrm{2}} }\:? \\ $$ Answered by bramlex last updated on 26/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\pi\:\mathrm{cos}\:^{\mathrm{2}} {x}\right)}{\mathrm{3}{x}^{\mathrm{2}}…
Question Number 105036 by bobhans last updated on 25/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}.\left[\:\frac{\mathrm{1}}{{x}}\:\right]\:? \\ $$$${note}\:\left[\:\right]\:=\:{greatest}\:{integer}\:{function} \\ $$ Answered by john santu last updated on 25/Jul/20 $${the}\:{limit}\:{as}\:{x}\rightarrow\mathrm{0}^{+} \:{can}\:{be}…
Question Number 170565 by cortano1 last updated on 27/May/22 Answered by Mathspace last updated on 27/May/22 $${f}^{'} \left({x}\right)={x}^{\mathrm{3}} \frac{\mathrm{2}\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{3}\left({x}+\mathrm{2}\right)^{\mathrm{4}} }−{x}^{\mathrm{3}} \frac{\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\…
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Question Number 170551 by 2407 last updated on 26/May/22 Answered by JDamian last updated on 26/May/22 $$\mathrm{3} \\ $$ Answered by Rasheed.Sindhi last updated on…