Question Number 130181 by Adel last updated on 23/Jan/21 Answered by Olaf last updated on 23/Jan/21 $$ \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{{xx}−\mathrm{33}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{\mathrm{11}\left({x}−\mathrm{3}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}}…
Question Number 64640 by Tawa1 last updated on 19/Jul/19 $$\mathrm{I}\:\:=\:\:\underset{\mathrm{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\left(−\:\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{6n}\:+\:\mathrm{1}} \\ $$ Commented by mathmax by abdo last updated on 20/Jul/19 $${let}\:\:{S}\left({x}\right)\:=\sum_{{n}=\mathrm{0}}…
Question Number 130176 by liberty last updated on 23/Jan/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6x}+\sqrt{\mathrm{8x}^{\mathrm{2}} −\sqrt{\mathrm{2x}^{\mathrm{3}} +\sqrt{\mathrm{4x}^{\mathrm{5}} }}}}{\:\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} −\mathrm{x}^{\mathrm{3}} +\sqrt{\mathrm{4x}^{\mathrm{5}} }}\:+\:\mathrm{x}}=? \\ $$ Answered by EDWIN88 last updated on…
Question Number 64612 by Tawa1 last updated on 19/Jul/19 Commented by Tawa1 last updated on 19/Jul/19 $$\mathrm{Just}\:\mathrm{the}\:\left(\mathrm{a}\right)\:\mathrm{part}\:\mathrm{please} \\ $$ Commented by kaivan.ahmadi last updated on…
Question Number 130120 by Study last updated on 22/Jan/21 $${li}\underset{{x}\rightarrow\mathrm{2}} {{m}}\sqrt{{x}−\mathrm{2}}=?\:\:\:\:{or}\:{limit}\:{has}? \\ $$ Answered by Olaf last updated on 22/Jan/21 $$ \\ $$$$\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\sqrt{{x}−\mathrm{2}}\:\mathrm{is}\:\mathrm{undefined}…
Question Number 130119 by Study last updated on 22/Jan/21 $${li}\underset{{x}\rightarrow\mathrm{15}} {{m}log}\left({x}−\mathrm{15}\right)=??\:{or}\:{limit}\:{has}? \\ $$ Answered by Olaf last updated on 22/Jan/21 $$\underset{{x}\rightarrow\mathrm{15}^{−} } {\mathrm{lim}}\:\mathrm{ln}\left({x}−\mathrm{15}\right)\:\mathrm{is}\:\mathrm{undefined} \\ $$$$\mathrm{because}\:{x}−\mathrm{15}\:\mathrm{canno}'\mathrm{t}\:\mathrm{be}\:\mathrm{negative}…
Question Number 64539 by mmkkmm000m last updated on 19/Jul/19 $${lim}_{{xat}\:\mathrm{0}} \left[{cos}^{\mathrm{2}} \left(\mathrm{4}{x}\right)\right]/{x}^{\mathrm{2}} \:\:−{lim}_{{x}\:{at}\:\mathrm{0}} \left[{cos}^{\mathrm{3}} \left(\mathrm{6}{x}\right)\right]/{x}^{\mathrm{2}} \\ $$ Commented by kaivan.ahmadi last updated on 19/Jul/19 $${lim}_{{x}\rightarrow\mathrm{0}}…
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Question Number 129908 by Adel last updated on 20/Jan/21 Answered by mathmax by abdo last updated on 20/Jan/21 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{1}−\sqrt{\mathrm{sinx}}\right)\left(\mathrm{1}−^{\mathrm{3}} \sqrt{\mathrm{sinx}}\right)….\left(\mathrm{1}−^{\mathrm{2020}} \sqrt{\mathrm{sinx}}\right)}{\left(\mathrm{1}−\mathrm{sinx}\right)^{\mathrm{2019}} }\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement} \\ $$$$\mathrm{x}=\frac{\pi}{\mathrm{2}}+\mathrm{t}\:\:\left(\mathrm{so}\:\mathrm{t}\rightarrow\mathrm{0}^{+} \right)\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{1}−\left(\mathrm{cost}\right)^{\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 129819 by Adel last updated on 19/Jan/21 $$\underset{{x}\rightarrow\mathrm{100}} {\mathrm{lim}}\left[\left(\mathrm{x}^{\mathrm{100}} −\mathrm{100}^{\mathrm{x}} \right)\frac{\mathrm{1}}{\mathrm{x}−\mathrm{100}}\right]=? \\ $$$$\mathrm{whit}\:\mathrm{out}\:\mathrm{Hopital}\:\mathrm{Rols} \\ $$$$\mathrm{pleas}\:\mathrm{answer} \\ $$ Terms of Service Privacy Policy Contact:…