Question Number 139059 by sahnaz last updated on 21/Apr/21 Answered by mr W last updated on 21/Apr/21 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}×\left(\frac{\mathrm{2}}{\mathrm{4}}\right)^{{n}} +\mathrm{9}×\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} +\mathrm{64}}{\left(\frac{\mathrm{2}}{\mathrm{4}}\right)^{{n}} +\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} +\mathrm{1}} \\ $$$$=\frac{\mathrm{2}×\mathrm{0}+\mathrm{9}×\mathrm{0}+\mathrm{64}}{\mathrm{0}+\mathrm{0}+\mathrm{1}}…
Question Number 73518 by aliesam last updated on 13/Nov/19 Commented by mathmax by abdo last updated on 13/Nov/19 $${let}\:{A}\left({x}\right)=\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}{x}−\mathrm{1}} \:\Rightarrow{A}\left({x}\right)={e}^{\left(\mathrm{2}{x}−\mathrm{1}\right){ln}\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)} \\ $$$$={e}^{\left(\mathrm{2}{x}−\mathrm{1}\right){ln}\left(\frac{{x}−\mathrm{1}+\mathrm{2}}{{x}−\mathrm{1}}\right)} \:={e}^{\left(\mathrm{2}{x}−\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{1}}\right)} \:\:{we}\:{have}\:{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{1}}\right)\sim\frac{\mathrm{2}}{{x}−\mathrm{1}}\left({x}\rightarrow+\infty\right) \\…
Question Number 139001 by bramlexs22 last updated on 21/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}−\sqrt[{\mathrm{4}}]{\mathrm{1}−\mathrm{2tan}\:\mathrm{x}}}{\mathrm{sin}\:\mathrm{x}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:=? \\ $$ Answered by EDWIN88 last updated on 21/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{x}\left(\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}}…
Question Number 138996 by bramlexs22 last updated on 21/Apr/21 $$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}}{\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:=? \\ $$$$ \\ $$ Answered by EDWIN88 last updated on 21/Apr/21 $$\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)}.\underset{{x}\rightarrow\pi/\mathrm{2}}…
Question Number 138997 by bramlexs22 last updated on 21/Apr/21 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{7}+\mathrm{x}^{\mathrm{3}} }−\sqrt{\mathrm{3}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}−\mathrm{1}}=? \\ $$ Answered by EDWIN88 last updated on 21/Apr/21 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{7}+\mathrm{x}^{\mathrm{3}}…
Question Number 138959 by mathlove last updated on 20/Apr/21 Answered by mathmax by abdo last updated on 20/Apr/21 $$\mathrm{a}^{\mathrm{x}^{\mathrm{2}} −\mathrm{1}} \:=\mathrm{e}^{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{loga}} \:\sim\mathrm{1}+\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{loga}\:\mathrm{and} \\…
Question Number 73406 by Rio Michael last updated on 11/Nov/19 $${Use}\:{the}\:{Sandwich}\left(\:{Pinchin}\:{or}\:{Squeez}\:\right)\:{theorem}\:{to}\:{prove} \\ $$$${that}\: \\ $$$$\:\underset{{x}\rightarrow{a}} {\mathrm{Lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 73251 by aliesam last updated on 09/Nov/19 Answered by mind is power last updated on 09/Nov/19 $$\mathrm{E}\left(\mathrm{u}_{\mathrm{n}} \right)\leqslant\mathrm{u}_{\mathrm{n}} <\mathrm{E}\left(\mathrm{u}_{\mathrm{n}} \right)+\mathrm{1}=\mathrm{U}_{\mathrm{n}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}} \mathrm{is}\:\mathrm{a}\:\mathrm{creasing}\:\mathrm{squances}…
Question Number 138765 by 676597498 last updated on 18/Apr/21 Commented by 676597498 last updated on 18/Apr/21 $${please}\:{help} \\ $$ Answered by physicstutes last updated on…
Question Number 138767 by bramlexs22 last updated on 18/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi\mathrm{cos}\:^{\mathrm{2}} {x}\right)}{\mathrm{sin}\:\left(\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}} {x}\right)}\:=? \\ $$ Answered by EDWIN88 last updated on 18/Apr/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\pi\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{sin}\:\left(\mathrm{2}\pi\mathrm{sin}\:^{\mathrm{2}}…