Question Number 51218 by gunawan last updated on 25/Dec/18 $$\mathrm{show}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{functions}\:\mathrm{a}\:\mathrm{entire}\:\mathrm{functions} \\ $$$$\mathrm{a}.\:{f}\left({z}\right)={e}^{−{y}} \mathrm{sin}\:{x}−{i}\:{e}^{−{y}} \mathrm{cos}\:{x} \\ $$$${b}.\:{f}\left({z}\right)=\left({z}^{\mathrm{2}} −\mathrm{2}\right){e}^{−{x}} {e}^{−{iy}} \\ $$ Terms of Service…
Question Number 51217 by gunawan last updated on 25/Dec/18 $$\mathrm{show}\:\mathrm{that}\:{f}\left({z}\right)={z}^{\mathrm{2}} \:\mathrm{continuous}\:\mathrm{at}\:{z}={z}_{\mathrm{0}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 51216 by gunawan last updated on 25/Dec/18 $$\mathrm{show}\:\mathrm{lim}\:{f}\left({z}\right)\:\mathrm{for}\:{z}\rightarrow\mathrm{0}\:\mathrm{along}\:\mathrm{the}\:\mathrm{line}\:{y}={x} \\ $$$$\mathrm{where}:\:{f}\left({z}\right)=\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }−{i}\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$ Answered by kaivan.ahmadi last updated on 11/Jan/19…
Question Number 51214 by gunawan last updated on 25/Dec/18 $$\mathrm{1}.\underset{{x}\rightarrow−\frac{\mathrm{i}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\left({z}−{i}\right)^{\mathrm{2}} }{\left(\mathrm{2}{z}−{i}\right)\left(\mathrm{3}−{z}\right)} \\ $$$$\mathrm{2}.\underset{{x}\rightarrow{e}^{\frac{\pi{i}}{\mathrm{4}}} } {\mathrm{lim}}\:\frac{\mathrm{2}{z}^{\mathrm{2}} }{{z}^{\mathrm{3}} −{z}−\mathrm{1}} \\ $$$$\mathrm{3}.\underset{{x}\rightarrow\mathrm{2}{i}} {\mathrm{lim}}\:\frac{\mathrm{2}{z}^{\mathrm{2}} +\mathrm{8}}{\:\sqrt{{z}^{\mathrm{4}} }−^{\mathrm{3}} \sqrt{\mathrm{64}}} \\…
Question Number 51190 by Abdo msup. last updated on 24/Dec/18 $${find}\:{lim}_{\xi\rightarrow\mathrm{0}^{+} } ^{} \:\:\:\:\:\int_{\mathrm{0}} ^{\xi} \:\:\:\:\:\:\frac{{x}}{{sinx}\:−\sqrt{{sin}^{\mathrm{2}} {x}\:+\xi^{\mathrm{2}} }}{dx} \\ $$ Terms of Service Privacy Policy…
Question Number 182256 by mathlove last updated on 06/Dec/22 Answered by Ar Brandon last updated on 06/Dec/22 $$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{k}=\mathrm{1}} {\overset{{mr}} {\sum}}\frac{{mn}^{\mathrm{2}} }{\left({m}^{\mathrm{2}}…
Question Number 51095 by Tinkutara last updated on 23/Dec/18 Commented by prakash jain last updated on 25/Dec/18 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[{x}\right]^{\mathrm{2}} −\left[{x}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} }…
Question Number 116588 by bemath last updated on 05/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4x}}{\mathrm{2}\:\mathrm{cosec}\:\mathrm{x}\:\left(\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{x}}\right)}\:=? \\ $$ Answered by bobhans last updated on 05/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4x}.\mathrm{sin}\:\mathrm{x}}{\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 116576 by bemath last updated on 05/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:=? \\ $$ Commented by mathmax by abdo last updated on 05/Oct/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…
Question Number 51005 by Tawa1 last updated on 23/Dec/18 Commented by maxmathsup by imad last updated on 23/Dec/18 $${let}\:{A}\left(\theta\right)\:=\frac{{sin}\left(\theta−\frac{\pi}{\mathrm{6}}\right)}{\:\sqrt{\mathrm{3}}\:−\mathrm{2}{cos}\theta}\:\:{changement}\:\:\theta\:−\frac{\pi}{\mathrm{6}}\:={x}\:{give} \\ $$$${A}\left(\theta\right)\:=\frac{{sinx}}{\:\sqrt{\mathrm{3}}−\mathrm{2}{cos}\left({x}+\frac{\pi}{\mathrm{6}}\right)}\:=\frac{{sinx}}{\:\sqrt{\mathrm{3}}−\mathrm{2}\left({cosx}\:{cos}\left(\frac{\pi}{\mathrm{6}}\right)−{sinx}\:{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)} \\ $$$$=\:\frac{{sinx}}{\:\sqrt{\mathrm{3}}−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}−\frac{\mathrm{1}}{\mathrm{2}}{sinx}\right)}\:=\frac{{sinx}}{\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}\:+{sinx}}\:{but}\:{lim}_{\theta\rightarrow\frac{\pi}{\mathrm{6}}} \:\:={lim}_{{x}\rightarrow\mathrm{0}} \frac{{sinx}}{\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}+{sinx}}…