Question Number 149637 by mnjuly1970 last updated on 06/Aug/21 Answered by iloveisrael last updated on 06/Aug/21 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{1}} +\mathrm{x}^{−\mathrm{4}} }\:−\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{−\mathrm{1}} \right)}{\mathrm{x}} \\ $$$$=\:\underset{{x}\rightarrow\infty}…
Question Number 84083 by john santu last updated on 09/Mar/20 $$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{x}−\mathrm{a}\:}\right)}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:= \\ $$ Commented by mathmax by abdo last updated on 09/Mar/20…
Question Number 149596 by iloveisrael last updated on 06/Aug/21 $$\:\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\:\underset{{x}\rightarrow\pi/\mathrm{7}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{sin}\:\mathrm{3x}−\frac{\sqrt{\mathrm{7}}}{\mathrm{8}}}{\mathrm{x}−\frac{\pi}{\mathrm{7}}}\:=? \\ $$ Answered by EDWIN88 last updated on 07/Aug/21 $$\: \\ $$$$\mathrm{let}\:{x}−\frac{\pi}{\mathrm{7}}={t}\:;\:{x}={t}+\frac{\pi}{\mathrm{7}}\:\wedge\:{t}\rightarrow\mathrm{0}…
Question Number 149547 by iloveisrael last updated on 06/Aug/21 $$\:\mathrm{Evaluate}\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{7}}} {\mathrm{lim}}\frac{\mathrm{8cos}\:\mathrm{xcos}\:\mathrm{2xcos}\:\mathrm{4x}+\mathrm{1}}{\mathrm{x}−\frac{\pi}{\mathrm{7}}} \\ $$ Answered by EDWIN88 last updated on 06/Aug/21 Terms of Service Privacy Policy…
Question Number 83943 by jagoll last updated on 08/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:\mathrm{6x}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{2x}\:\mathrm{tan}\:\mathrm{5x}} \\ $$$$ \\ $$ Answered by jagoll last updated on 08/Mar/20 Terms of Service…
Question Number 83919 by john santu last updated on 08/Mar/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{3}}{\pi}\mathrm{arc}\:\mathrm{tan}\:{x}\right)^{\mathrm{2}{x}} \:=\:? \\ $$ Commented by john santu last updated on 08/Mar/20 $$=\:\mathrm{e}\:^{\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\left(\mathrm{ln}\left(\frac{\mathrm{3}}{\pi}\mathrm{tan}^{−\mathrm{1}}…
Question Number 83865 by jagoll last updated on 07/Mar/20 $$\underset{{x}\rightarrow−\infty\:} {\mathrm{lim}}\:\left(\mathrm{x}\sqrt{\mathrm{2x}+\mathrm{2}}−\mathrm{x}\sqrt{\mathrm{2x}+\mathrm{3}}\right) \\ $$ Commented by mr W last updated on 07/Mar/20 $$\mathrm{2}{x}+\mathrm{2}\geqslant\mathrm{0} \\ $$$$\Rightarrow{x}\geqslant−\mathrm{1} \\…
Question Number 83859 by john santu last updated on 06/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\:\mathrm{cot}\:^{\mathrm{2}} {x}\right)=\:? \\ $$ Commented by john santu last updated on 06/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 149377 by fotosy2k last updated on 05/Aug/21 Answered by john_santu last updated on 05/Aug/21 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}.\mathrm{3}^{\mathrm{n}} }{\mathrm{3}^{\mathrm{n}} \left(\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}} \right)}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}}{\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}} }=\mathrm{3} \\ $$…
Question Number 149372 by fotosy2k last updated on 05/Aug/21 Answered by john_santu last updated on 05/Aug/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\mathrm{x}+\mathrm{2sin}\:\mathrm{x}\right)}{\mathrm{sin}\:\mathrm{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{sin}\:\left(\mathrm{x}+\mathrm{2sin}\:\mathrm{x}\right)}{\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}\:.\:\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}.\:\mathrm{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}{\mathrm{x}}\:=\:\mathrm{1}+\mathrm{2}=\mathrm{3} \\…