Question Number 116872 by bemath last updated on 07/Oct/20 Answered by bobhans last updated on 07/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{x}}\right)}{\mathrm{3}\:\mathrm{tan}\:\mathrm{2x}} \\ $$$$\mathrm{letting}\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{z}\:\mathrm{with}\:\mathrm{z}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{z}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{5z}\right)}{\mathrm{3}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{z}}\right)}\:=\:\underset{\mathrm{z}\rightarrow\mathrm{0}}…
Question Number 116859 by Study last updated on 07/Oct/20 $${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}^{{x}} }=? \\ $$ Commented by JDamian last updated on 07/Oct/20 As exponential function increases more rapidly than parabola, you even can get this limit intuitively Answered by…
Question Number 182346 by cortano1 last updated on 08/Dec/22 Commented by SANOGO last updated on 08/Dec/22 $${thank}\:{you} \\ $$ Answered by CrispyXYZ last updated on…
Question Number 116797 by ravisoni505 last updated on 06/Oct/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116756 by bemath last updated on 06/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{x}−\mathrm{2}}}{\mathrm{x}−\mathrm{2}}\:=? \\ $$ Answered by bobhans last updated on 06/Oct/20 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{x}−\mathrm{2}}}{\mathrm{x}−\mathrm{2}}=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} }}\:=\:\infty \\…
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…